https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jdong2006&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:51:54ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_21&diff=1629042003 AMC 12A Problems/Problem 212021-09-30T01:12:39Z<p>Jdong2006: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
The graph of the [[polynomial]]<br />
<br />
<cmath>P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e</cmath><br />
<br />
has five distinct <math>x</math>-intercepts, one of which is at <math>(0,0)</math>. Which of the following [[coefficient]]s cannot be zero?<br />
<br />
<math>\text{(A)}\ a \qquad \text{(B)}\ b \qquad \text{(C)}\ c \qquad \text{(D)}\ d \qquad \text{(E)}\ e</math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Let the roots be <math>r_1=0, r_2, r_3, r_4, r_5</math>. According to [[Vieta's formulas | Viète's formulae]], we have <math>d=r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5</math>. The first four terms contain <math>r_1=0</math> and are therefore zero, thus <math>d=r_2r_3r_4r_5</math>. This is a product of four non-zero numbers, therefore <math>d</math> must be non-zero <math>\Longrightarrow \mathrm{(D)}</math>.<br />
<br />
=== Solution 2 ===<br />
Clearly, since <math>(0,0)</math> is an intercept, <math>e</math> must be <math>0</math>. But if <math>d</math> was <math>0</math>, <math>x^2</math> would divide the polynomial, which means it would have a double root at <math>0</math>, which is impossible, since all five roots are distinct.<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_19&diff=1617052021 AMC 12B Problems/Problem 192021-09-06T20:06:19Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Two fair dice, each with at least <math>6</math> faces are rolled. On each face of each dice is printed a distinct integer from <math>1</math> to the number of faces on that die, inclusive. The probability of rolling a sum of <math>7</math> is <math>\frac34</math> of the probability of rolling a sum of <math>10,</math> and the probability of rolling a sum of <math>12</math> is <math>\frac{1}{12}</math>. What is the least possible number of faces on the two dice combined?<br />
<br />
<math>\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18\qquad \textbf{(D) }19 \qquad \textbf{(E) }20</math><br />
<br />
==Solution 1==<br />
Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. Since each die has at least <math>6</math> faces, there will always be <math>6</math> ways to sum to <math>7</math>. As a result, there must be <math>\tfrac{4}{3}\cdot6=8</math> ways to sum to <math>10</math>. There are at most nine distinct ways to get a sum of <math>10</math>, which are possible whenever <math>a,b\geq{9}</math>. To achieve exactly eight ways, <math>b</math> must have <math>8</math> faces, and <math>a\geq9</math>. Let <math>n</math> be the number of ways to obtain a sum of <math>12</math>, then <math>\tfrac{n}{8a}=\tfrac{1}{12}\implies n=\tfrac{2}{3}a</math>. Since <math>b=8</math>, <math>n\leq8\implies a\leq{12}</math>. In addition to <math>3\mid{a}</math>, we only have to test <math>a=9,12</math>, of which both work. Taking the smaller one, our answer becomes <math>a+b=9+8=\boxed{\textbf{(B)}\ 17}</math>.<br />
<br />
==Solution 2==<br />
Suppose the dice have <math>a</math> and <math>b</math> faces, and WLOG <math>a\geq{b}</math>. Note that if <math>a+b=12</math> since they are both <math>6</math>, there is one way to make <math>12</math>, and incrementing <math>a</math> or <math>b</math> by one will add another way. This gives us the probability of making a 12 as <br />
<cmath>\frac{a+b-11}{ab}=\frac{1}{12}</cmath><br />
Cross-multiplying, we get<br />
<cmath>12a+12b-132=ab</cmath><br />
Simon's Favorite Factoring Trick now gives<br />
<cmath>(a-12)(b-12)=12</cmath><br />
This narrows the possibilities down to 3 ordered pairs of <math>(a,b)</math>, which are <math>(13,24)</math>, <math>(6,10)</math>, and <math>(8,9)</math>. We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer:<br />
<cmath>\frac{6}{72}=\frac{3}{4}\left(\frac{9+8-9}{72}\right)</cmath><br />
The answer is then <math>a+b=8+9=\boxed{\textbf{(B)}\ 17}</math>.<br />
<br />
~Hyprox1413<br />
<br />
== Video Solution by OmegaLearn (Using Probability) ==<br />
https://youtu.be/geEDrsV5Glw<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
{{AMC12 box|year=2021|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_3&diff=1616412018 AIME I Problems/Problem 32021-09-05T21:37:24Z<p>Jdong2006: /* Solution 6 (Official MAA) */</p>
<hr />
<div>==Problem==<br />
Kathy has <math>5</math> red cards and <math>5</math> green cards. She shuffles the <math>10</math> cards and lays out <math>5</math> of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy, but RRRGR will not. The probability that Kathy will be happy is <math> \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
==Solution 1==<br />
<br />
We have <math>2+4\cdot 2</math> cases total.<br />
<br />
The two are all red and all green. Then, you have 4 of one, 1 of other. 3 of one, 2 of other. 2 of one, 3 of other. 1 of one, 4 of other. Then flip the order, so times two.<br />
<br />
Obviously the denominator is <math>10\cdot 9\cdot 8\cdot 7\cdot 6</math>, since we are choosing a card without replacement.<br />
<br />
Then, we have for the numerator for the two of all red and green: <cmath>5\cdot 4\cdot 3\cdot 2\cdot 1.</cmath><br />
<br />
For the 4 and 1, we have: <cmath>5\cdot 4\cdot 3\cdot 2\cdot 5.</cmath><br />
<br />
For the 3 and 2, we have: <cmath>5\cdot 4\cdot 3\cdot 5\cdot 4.</cmath><br />
<br />
For the 2 and 3, we have: <cmath>5\cdot 4\cdot 5\cdot 4\cdot 3.</cmath><br />
<br />
For the 1 and 4, we have: <cmath>5\cdot 5\cdot 4\cdot 3\cdot 2.</cmath><br />
<br />
Adding up and remembering to double all of them, since they can be reversed and the 5's can be red or green, we get, after simplifying: <math>\dfrac{31}{126}</math><br />
<br />
Thus the answer is <math>31 + 126 = \boxed{157}</math>.<br />
-gorefeebuddie<br />
<br />
==Solution 2==<br />
<br />
Our probability will be <math>\dfrac{\text{number of "happy" configurations of cards}}{\text{total number of configurations of cards}}.</math><br />
<br />
First of all, we have <math>10</math> choices for the first card, <math>9</math> choices for the second card, <math>8</math> choices for the third card, <math>7</math> choices for the fourth card, and <math>6</math> choices for the last card. This gives a total of <math>10\cdot 9\cdot 8\cdot 7\cdot 6</math> possible ways for five cards to be chosen.<br />
<br />
Finding the number of configurations that make Kathy happy is a more difficult task, however, and we will resort to casework to do it.<br />
<br />
First, let's look at the appearances of the "happy configurations" that Kathy likes. Based on the premise of the problem, we can realize that there are ten cases for the appearance of the configurations: <cmath>\text{RRRRR},</cmath> <cmath>\text{GGGGG},</cmath> <cmath>\text{RRRRG},</cmath> <cmath>\text{GGGGR},</cmath> <cmath>\text{RRRGG},</cmath> <cmath>\text{GGGRR},</cmath> <cmath>\text{RRGGG},</cmath> <cmath>\text{GGRRR},</cmath> <cmath>\text{RGGGG},</cmath> <cmath>\text{GRRRR}.</cmath> But this doesn't mean there are 10 "happy configurations" in total-- remember that we've been treating these cards as distinguishable, so we must continue to do so.<br />
<br />
To lighten the load of 10 cases on the human brain, we can note that in the eyes of what we will soon do, <math>RRRRR</math> and <math>GGGGG</math> are effectively equivalent, and therefore may be treated in the same case. We will have to multiply by 2 at the end, though.<br />
<br />
Similarly, we can equate <math>RRRRG,</math> <math>GGGGR,</math> <math>RGGGG,</math> and <math>GRRRR,</math> as well as <math>RRRGG,</math> <math>GGGRR,</math> <math>RRGGG,</math> and <math>GGRRR,</math> so that we just have three cases. We can approach each of these cases with constructive counting.<br />
<br />
Case 1: <math>RRRRR</math>-type.<br />
<br />
For this case, there are <math>5</math> available choices for the first card, <math>4</math> available choices for the second card, <math>3</math> for the third card, <math>2</math> for the fourth card, and <math>1</math> for the last card. This leads to a total of <math>5\cdot 4\cdot 3\cdot 2\cdot 1=120</math> configurations for this case. There are <math>2</math> cases of this type.<br />
<br />
Case 2: <math>RRRRG</math>-type.<br />
<br />
For this case, there are <math>5</math> available choices for the first card, <math>4</math> available choices for the second card, <math>3</math> for the third card, <math>2</math> for the fourth card, and <math>5</math> choices for the last card (not <math>1</math>, because we're doing a new color). This leads to a total of <math>5\cdot 4\cdot 3\cdot 2\cdot 5=600</math> configurations for this case. There are <math>4</math> cases of this type.<br />
<br />
Case 3: <math>RRRGG</math>-type.<br />
<br />
For this case, there are <math>5</math> available choices for the first card, <math>4</math> available choices for the second card, <math>3</math> for the third card, <math>5</math> for the fourth card, and <math>4</math> choices for the last card. This leads to a total of <math>5\cdot 4\cdot 3\cdot 5\cdot 4=1200</math> configurations for this case. There are <math>4</math> cases of this type.<br />
<br />
Adding the cases up gives <math>2\cdot 120+4\cdot 600+4\cdot 1200=7440</math> "happy" configurations in total.<br />
<br />
This means that the probability that Kathy is happy will be <math>\dfrac{7440}{10\cdot 9\cdot 8\cdot 7\cdot 6},</math> which simplifies to <math>\dfrac{31}{126}.</math><br />
<br />
So the answer is <math>31+126=\boxed{157}</math><br />
<br />
==Solution 3==<br />
As the problem states, some examples of valid are <math>RRGGG</math>, <math>GGGGR</math>, and <math>RRRRR</math>. Let's use each of these as more general cases. <br />
<br />
Let <math>RRGGG</math> be the case when there are 2 adjacents of one color, and 3 adjacents of the other color. This yields <math>4</math> combinations (<math>RRGGG</math>, <math>GGRRR</math>, <math>RRRGG</math>, and <math>GGGRR</math>). The probability of each of these is equal, equating to <math>\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}=\frac{5}{126}</math>, and since there are <math>4</math> combinations, the probability of this case is <math>4\cdot \frac{5}{126}=\frac{10}{63}</math><br />
<br />
Next case is <math>GGGGR</math>. Let this be when there are 4 adjacents of one color, and 1 individual color. Once again, this yields <math>4</math> combinations (<math>GGGGR</math>, <math>RRRRG</math>, <math>RGGGG</math>, and <math>GRRRR</math>). The probability of each is the same, equating to <math>\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\cdot \frac{2}{7}\cdot \frac{5}{6}=\frac{5}{252}</math>, and since there are <math>4</math> combinations, the probability of this case is <math>4\cdot \frac{5}{252}=\frac{5}{63}</math><br />
<br />
The final case is <math>RRRRR</math>, in which there is just an adjacent block of <math>5</math> colors. There are only <math>2</math> combinations this time, each equating to the probability <math>\frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\cdot \frac{2}{7}\cdot \frac{1}{6}=\frac{1}{252}</math>, and since there are <math>2</math> combinations, the probability of this case is <math>2\cdot \frac{1}{252}=\frac{1}{126}</math>. <br />
<br />
Thus, the total probability is <math>\frac{10}{63}+\frac{5}{63}+\frac{1}{126}=\frac{31}{126} \implies m+n=\boxed{157}</math><br />
<br />
==Solution 4==<br />
Kathy will draw the 10 cards in some order, but the restriction of having all greens in a row and all reds in a row only applies to the first 5 cards drawn. The total number of ways the 10 cards can be drawn is simply 10 choose 5 which is 252. Now we just count the number of possible successful configurations of the ten cards. The first 5 cards can start either be <math>GRRRR</math>, <math>GGRRR</math>, <math>GGGRR</math>, <math>GGGGR</math>, <math>GGGGG</math> or the same thing except starting with a red. The number of ways to order <math>GRRRR</math> is the number of ways to order the last 5 cards, which is 5C1. Doing all of the other cases, the total is <math>(5C1+5C2+5C3+5C4+5C5)*2 = 62</math>. <math>\frac{62}{252} = \frac{31}{126}</math> <math>31 + 126 = \boxed{157}</math><br />
<br />
-bradleyguo<br />
<br><br />
Minor LaTeX edits by fasterthanlight<br />
<br />
==Solution 5==<br />
Suppose <math>k</math> of the first <math>5</math> cards are red. There are <math>\binom{5}{k}</math> ways to order the last five cards, and either <math>1</math> or <math>2</math> ways to order the first five cards: <math>1</math> if <math>k=0</math> or <math>k=5</math>, otherwise <math>2</math>. So, the number of orderings that work is<br />
<cmath>\binom{5}{0} + 2\binom{5}{1} + 2\binom{5}{2} + 2\binom{5}{3} + 2\binom{5}{4} + \binom{5}{5} = 2\sum_{k=0}^{5}\binom{5}{k} - 2 = 2^6 - 2.</cmath><br />
<br />
The total number of orderings is <math>\binom{10}{5} = 252,</math> so the answer is <math>\frac{62}{252} = \boxed{\frac{31}{126}}</math><br />
<br />
==Solution 6 (Official MAA)==<br />
Assume without loss of generality that the first card laid out is red. Then the arrangements that satisfy Kathy’s requirements are RRRRR, RRRRG, RRRGG, RRGGG, and RGGGG. The probability that Kathy will lay out one of these<br />
arrangements is <cmath>\frac49\cdot\frac38\cdot\frac27\cdot\frac16</cmath> <cmath>\frac49\cdot\frac38\cdot\frac27\cdot\frac56</cmath> <cmath>\frac49\cdot\frac38\cdot\frac57\cdot\frac46</cmath> <cmath>\frac49\cdot\frac58\cdot\frac47\cdot\frac36</cmath> <cmath>+\frac59\cdot\frac48\cdot\frac37\cdot\frac26</cmath> <cmath>\overline{..........................}</cmath> <cmath>\frac{31}{126}</cmath> The requested sum is <math>31+126=\boxed{157}.</math><br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=WVtbD8x9fCM<br />
~Shreyas S<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_15&diff=1616402020 AIME II Problems/Problem 152021-09-05T20:51:02Z<p>Jdong2006: /* Solution 2 (Official MAA) */</p>
<hr />
<div>==Problem==<br />
Let <math>\triangle ABC</math> be an acute scalene triangle with circumcircle <math>\omega</math>. The tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at <math>T</math>. Let <math>X</math> and <math>Y</math> be the projections of <math>T</math> onto lines <math>AB</math> and <math>AC</math>, respectively. Suppose <math>BT = CT = 16</math>, <math>BC = 22</math>, and <math>TX^2 + TY^2 + XY^2 = 1143</math>. Find <math>XY^2</math>.<br />
<br />
==Solution==<br />
Assume <math>O</math> to be the center of triangle <math>ABC</math>, <math>OT</math> cross <math>BC</math> at <math>M</math>, link <math>XM</math>, <math>YM</math>. Let <math>P</math> be the middle point of <math>BT</math> and <math>Q</math> be the middle point of <math>CT</math>, so we have <math>MT=3\sqrt{15}</math>. Since <math>\angle A=\angle CBT=\angle BCT</math>, we have <math>\cos A=\frac{11}{16}</math>. Notice that <math>\angle XTY=180^{\circ}-A</math>, so <math>\cos XYT=-\cos A</math>, and this gives us <math>1143-2XY^2=\frac{-11}{8}XT\cdot YT</math>. Since <math>TM</math> is perpendicular to <math>BC</math>, <math>BXTM</math> and <math>CYTM</math> cocycle (respectively), so <math>\theta_1=\angle ABC=\angle MTX</math> and <math>\theta_2=\angle ACB=\angle YTM</math>. So <math>\angle XPM=2\theta_1</math>, so <cmath>\frac{\frac{XM}{2}}{XP}=\sin \theta_1</cmath>, which yields <math>XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.</math> So same we have <math>YM=XT</math>. Apply Ptolemy theorem in <math>BXTM</math> we have <math>16TY=11TX+3\sqrt{15}BX</math>, and use Pythagoras theorem we have <math>BX^2+XT^2=16^2</math>. Same in <math>YTMC</math> and triangle <math>CYT</math> we have <math>16TX=11TY+3\sqrt{15}CY</math> and <math>CY^2+YT^2=16^2</math>. Solve this for <math>XT</math> and <math>TY</math> and submit into the equation about <math>\cos XYT</math>, we can obtain the result <math>XY^2=\boxed{717}</math>.<br />
<br />
(Notice that <math>MXTY</math> is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)<br />
<br />
-Fanyuchen20020715<br />
<br />
==Solution 2 (Official MAA)==<br />
Let <math>M</math> denote the midpoint of <math>\overline{BC}</math>. The critical claim is that <math>M</math> is the orthocenter of <math>\triangle AXY</math>, which has the circle with diameter <math>\overline{AT}</math> as its circumcircle. To see this, note that because <math>\angle BXT = \angle BMT = 90^\circ</math>, the quadrilateral <math>MBXT</math> is cyclic, it follows that<br />
<cmath>\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,</cmath> implying that <math>\overline{MX} \perp \overline{AC}</math>. Similarly, <math>\overline{MY} \perp \overline{AB}</math>. In particular, <math>MXTY</math> is a parallelogram.<br />
<asy><br />
defaultpen(fontsize(8pt));<br />
unitsize(0.8cm);<br />
<br />
pair A = (0,0); <br />
pair B = (-1.26,-4.43);<br />
pair C = (-1.26+3.89, -4.43);<br />
pair M = (B+C)/2; <br />
pair O = circumcenter(A,B,C); <br />
pair T = (0.68, -6.49);<br />
pair X = foot(T,A,B); <br />
pair Y = foot(T,A,C);<br />
path omega = circumcircle(A,B,C);<br />
real rad = circumradius(A,B,C);<br />
<br />
<br />
<br />
filldraw(A--B--C--cycle, rgb(0/255,0/255,255/255));<br />
label("$\omega$", O + rad*dir(45), SW);<br />
filldraw(T--Y--M--X--cycle, rgb(0/255,255/255,0/255));<br />
draw(M--T); <br />
draw(X--Y);<br />
draw(B--T--C);<br />
draw(A--X--Y--cycle);<br />
draw(omega);<br />
dot("$X$", X, W); <br />
dot("$Y$", Y, E);<br />
dot("$O$", O, W);<br />
dot("$T$", T, S); <br />
dot("$A$", A, N); <br />
dot("$B$", B, W); <br />
dot("$C$", C, E); <br />
dot("$M$", M, N);<br />
<br />
<br />
</asy><br />
Hence, by the Parallelogram Law,<br />
<cmath> TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).</cmath> But <math>TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135</math>. Therefore <cmath>XY^2 = \frac13(2 \cdot 1143-135) = 717.</cmath><br />
<br />
==Solution 3 (Law of Cosines)==<br />
Let <math>H</math> be the orthocenter of <math>\triangle AXY</math>. <br />
<br />
<b>Lemma 1:</b> <math>H</math> is the midpoint of <math>BC</math>.<br />
<br />
<b>Proof:</b> Let <math>H'</math> be the midpoint of <math>BC</math>, and observe that <math>XBH'T</math> and <math>TH'CY</math> are cyclical. Define <math>H'Y \cap BA=E</math> and <math>H'X \cap AC=F</math>, then note that:<br />
<cmath>\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.</cmath><br />
That implies that <math>\angle H'XB=\angle H'YC=90^\circ-\angle A</math>, <math>\angle CH'Y=\angle EH'B=90^\circ-\angle B</math>, and <math>\angle BH'Y=\angle FH'C=90^\circ-\angle C</math>. Thus <math>YH'\perp AX</math> and <math>XH' \perp AY</math>; <math>H'</math> is indeed the same as <math>H</math>, and we have proved lemma 1.<br />
<br />
Since <math>AXTY</math> is cyclical, <math>\angle XTY=\angle XHY</math> and this implies that <math>XHYT</math> is a paralelogram.<br />
By the Law of Cosines:<br />
<cmath>XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)</cmath><br />
<cmath>XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)</cmath><br />
<cmath>HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)</cmath><br />
<cmath>HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).</cmath><br />
We add all these equations to get:<br />
<cmath>HT^2+XY^2=2(XT^2+TY^2) \qquad (1).</cmath><br />
We have that <math>BH=HC=11</math> and <math>BT=TC=16</math> using our midpoints. Note that <math>HT \perp BC</math>, so by the Pythagorean Theorem, it follows that <math>HT^2=135</math>. We were also given that <math>XT^2+TY^2=1143-XY^2</math>, which we multiply by <math>2</math> to use equation <math>(1)</math>. <cmath>2(XT^2+TY^2)=2286-2 \cdot XY^2</cmath> Since <math>2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2</math>, we have<br />
<cmath>135+XY^2=2286-2 \cdot XY^2</cmath> <cmath>3 \cdot XY^2=2151.</cmath><br />
Therefore, <math>XY^2=\boxed{717}</math>. ~ MathLuis<br />
<br />
==See Also==<br />
{{AIME box|year=2020|n=II|num-b=14|after=Last Problem}}<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=Infinite_Defenestration&diff=161625Infinite Defenestration2021-09-05T14:58:57Z<p>Jdong2006: </p>
<hr />
<div>'''Infinite Defenestration''' is a method of proof which utilizes repeatedly thowing members of a party out a window. It relies on that given <math>n</math> people at a party, it is possible to throw <math>1</math> member out the window, leaving <math>n - 1</math> members remaining.<br />
<br />
==Problem 1==<br />
Prove that given <math>n</math> attendees of a party, one can defenestrate a member <math>n</math> times.<br />
===Solution===<br />
'''Proof'''<br />
<br />
<math>\textbf{Lemma:} \textit{ Given integer }k; 0\le k\le n, k \textit{ members of a party of size }n.</math><br />
<br />
We use [[mathematical induction]]. We start with <math>k = 0</math>, which is trivial.<br />
We induct; give <math>i < n</math> members defenestrated, <math>n - i</math> remain. We throw out <math>1</math> member, leaving us with <math>i + 1</math> members defenestrated and <math>n - (i + 1)</math> members remaining.<br />
<br />
Thus lemma is true. Plugging in <math>k = 0,1,2,3\cdots</math>, we have <math>\mathbb{Q.E.D.}</math><br />
<br />
Try your hand at the following problems:<br />
<br />
==Problem 2==<br />
Prove that given <math>n</math> couples, you can defenestrate <math>1</math> couple <math>n</math> times.<br />
<br />
==Problem 3==<br />
Jim and his wife Jeri attend a party with 4 other married couples. As they enter, Jim and Jeri shake hands with some of the guests, but not with each other. During the evening, each person except 1 shakes hands with some of the guests, but not with their spouse. After the party, Jim asks each guest how many people they shook hands with and got answers 0,1,2,3,4,5,6,7,8. How many people did Jeri shake hands with?</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_12&diff=1616222018 AIME II Problems/Problem 122021-09-05T14:01:23Z<p>Jdong2006: /* Solution 1 */ Beautify the 1st solution</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>.<br />
<br />
==Solution 1==<br />
<br />
For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes<br />
<br />
<cmath>\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP</cmath><br />
<br />
What an important finding! Note that the opposite sides <math>\overline{AB}</math> and <math>\overline{CD}</math> have equal length, and note that diagonal <math>\overline{DB}</math> bisects diagonal <math>\overline{AC}</math>. This is very similar to what happens if <math>ABCD</math> were a parallelogram with <math>AB = CD = 10</math>, so let's extend <math>\overline{DB}</math> to point <math>E</math>, such that <math>AECD</math> is a parallelogram. In other words, <cmath>AE = CD = 10</cmath> and <cmath>EC = DA = 2\sqrt{65}</cmath> Now, let's examine <math>\triangle ABE</math>. Since <math>AB = AE = 10</math>, the triangle is isosceles, and <math>\angle ABE \cong \angle AEB</math>. Note that in parallelogram <math>AECD</math>, <math>\angle AED</math> and <math>\angle CDE</math> are congruent, so <math>\angle ABE \cong \angle CDE</math> and thus <cmath>\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB</cmath> Define <math>\alpha := \text{m}\angle CDB</math>, so <math>180^\circ - \alpha = \text{m}\angle ABD</math>. <br />
<br />
We use the Law of Cosines on <math>\triangle DAB</math> and <math>\triangle CDB</math>:<br />
<br />
<cmath>\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha</cmath><br />
<br />
<cmath>14^2 = 10^2 + BD^2 - 20BD\cos\alpha</cmath><br />
<br />
Subtracting the second equation from the first yields<br />
<br />
<cmath>260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}</cmath><br />
<br />
This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <cmath>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</cmath> and hence <cmath>\left[ABCD\right] = 56 + 56 = \boxed{112}</cmath> -kgator<br />
<br />
<br />
Just to be complete -- <math>h1</math> and <math>h2</math> can actually be equal. In this case, <math>AP \neq CP</math>, but <math>BP</math> must be equal to <math>DP</math>. We get the same result. -Mathdummy.<br />
<br />
==Solution 2 (Another way to get the middle point)==<br />
<br />
So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>, <math>\triangle {BCP}=S_{2}</math>, <math>\triangle {CDP}=S_{3}</math>, <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}.</cmath> Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2.</cmath> So <cmath>(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2.</cmath> So <cmath>S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2.</cmath> As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}.</cmath> Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}.</cmath> Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we can find out that <cmath>S_{3}=S_{4},</cmath> so <math>P</math> is the midpoint of <math>\overline {AC}</math><br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
==Solution 3 (With yet another way to get the middle point)==<br />
<br />
Using the formula for the area of a triangle, <cmath>(\frac{1}{2}AP\cdot BP+\frac{1}{2}CP\cdot DP)\sin{APB}=(\frac{1}{2}AP\cdot DP+\frac{1}{2}CP\cdot BP)\sin{APD}</cmath> <br />
But <math>\sin{APB}=\sin{APD}</math>, so <cmath>(AP-CP)(BP-DP)=0</cmath><br />
Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here).<br />
Now, assume that <math>AP=CP=x</math>,<math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for triangles <math>APB</math> and <math>BPC</math>, it is clear that <br />
<br />
<math>x^2+y^2-100=2 \cdot x \cdot y \cdot \cos{APB}=-(2 \cdot x \cdot y \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196)</math>, or <cmath>x^2+y^2=148...(1)</cmath> Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>x^2+z^2=180...(2)</cmath>. It follows that <cmath>z^2-y^2=32...(3)</cmath>. Now, denote angle <math>APB</math> by <math>\alpha</math>. Since <math>\sin\alpha=\sqrt{1-\cos^2\alpha}</math>, <cmath>\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}</cmath> which simplifies to <cmath>\frac{48^2}{y^2}=\frac{80^2}{z^2}</cmath>, giving <cmath>5y=3z</cmath>. Plugging this back to equations (1), (2), and (3), it can be solved that <math>x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}</math>. Then, the area of the quadrilateral is <cmath>x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}</cmath><br />
--Solution by MicGu<br />
<br />
==Solution 4 ==<br />
As in all other solutions, we can first find that either <math>AP=CP</math> or <math>BP=DP</math>, but it's an AIME problem, we can take <math>AP=CP</math>, and assume the other choice will lead to the same result (which is true). <br />
<br />
From <math>AP=CP</math>, we have <math>[DAP]=[DCP]</math>, <math>[BAP]=[BCP]</math> => <math>[ABD] = [CBD]</math>, therefore,<br />
<cmath>1/2AB*AD\sin A = 1/2BC*CD\sin C => 7\sin C = \sqrt{65}\sin A ... (1)</cmath><br />
By Law of Cosine,<br />
<cmath>10^2+14^2-2*10*14\cos C=10^2+4*65-2*10*2\sqrt{65}\cos A</cmath><br />
<cmath>(-8/5)-7\cos C = \sqrt{65}\cos A ...(2) </cmath><br />
Square (1) and (2), add them, we get<br />
<cmath>(8/5)^2 +2(8/5)7\cos C + 7^2 = 65 </cmath><br />
Solve, <math>\cos C = 3/5</math> => <math>\sin C = 4/5</math>,<br />
<cmath>[ABCD] = 2[BCD] = BC*CD*\sin C = 14*10*(4/5) = \boxed{112}</cmath><br />
-Mathdummy<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=II|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_11&diff=1616002019 AIME II Problems/Problem 112021-09-05T04:48:34Z<p>Jdong2006: /* Solution 3 (Death By Trig Bash) */ making the death bash readable</p>
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<div>==Problem==<br />
Triangle <math>ABC</math> has side lengths <math>AB=7, BC=8, </math> and <math>CA=9.</math> Circle <math>\omega_1</math> passes through <math>B</math> and is tangent to line <math>AC</math> at <math>A.</math> Circle <math>\omega_2</math> passes through <math>C</math> and is tangent to line <math>AB</math> at <math>A.</math> Let <math>K</math> be the intersection of circles <math>\omega_1</math> and <math>\omega_2</math> not equal to <math>A.</math> Then <math>AK=\tfrac mn,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math><br />
<br />
==Solution 1==<br />
<asy><br />
unitsize(20);<br />
pair B = (0,0);<br />
pair A = (2,sqrt(45));<br />
pair C = (8,0);<br />
draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));<br />
draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));<br />
draw(B--A--C--cycle);<br />
label("$A$",A,dir(105));<br />
label("$B$",B,dir(-135));<br />
label("$C$",C,dir(-75));<br />
dot((2.68,2.25));<br />
label("$K$",(2.68,2.25),dir(-150));<br />
label("$\omega_1$",(-6,1));<br />
label("$\omega_2$",(14,6));<br />
label("$7$",(A+B)/2,dir(140));<br />
label("$8$",(B+C)/2,dir(-90));<br />
label("$9$",(A+C)/2,dir(60));<br />
</asy><br />
-Diagram by Brendanb4321<br />
<br />
<br />
Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> <br />
<br />
*Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}</math>. This gives us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>.<br />
-franchester<br />
*The motivation for using the Law of Cosines ("LoC") is after finding the similar triangles it's hard to figure out what to do with <math>BK</math> and <math>CK</math> yet we know <math>BC</math> which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealing with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got stuck). To anyone in the future reading this, math is beautiful.<br />
-First<br />
<br />
11111111:L)xiexie<br />
<br />
==Solution 2 (Inversion)==<br />
Consider an inversion with center <math>A</math> and radius <math>r=AK</math>. Then, we have <math>AB\cdot AB^*=AK^2</math>, or <math>AB^*=\frac{AK^2}{7}</math>. Similarly, <math>AC^*=\frac{AK^2}{9}</math>. Notice that <math>AB^*KC^*</math> is a parallelogram, since <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AC</math> and <math>AB</math>, respectively. Thus, <math>AC^*=B^*K</math>. Now, we get that<br />
<cmath>\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}</cmath><br />
so by Law of Cosines on <math>\triangle AB^*K</math> we have <br />
<cmath>(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)</cmath><br />
<cmath>\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}</cmath><br />
<cmath>\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}</cmath><br />
<cmath>\Rightarrow AK=\frac{9}{2}</cmath><br />
Then, our answer is <math>9+2=\boxed{11}</math>. <br />
-brianzjk<br />
<br />
<br />
<br />
== Solution 3 (Death By Trig Bash) ==<br />
<br />
14. Let the centers of the circles be <math>O_{1}</math> and <math>O_{2}</math> where the <math>O_{1}</math> has the side length <math>7</math> contained in the circle. Now let <math>\angle BAC =x.</math> This implies <cmath>\angle AO_{1}B = \angle AO_{2}C = 2x</cmath> by the angle by by tangent. Then we also know that <cmath>\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x</cmath> Now we first find <math>\cos x.</math> We use law of cosines on <math>\bigtriangleup ABC</math> to obtain <cmath>64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}</cmath> <cmath>\implies \cos{x} =\frac{11}{21}</cmath> <cmath>\implies \sin{x} =\frac{8\sqrt{5}}{21}</cmath> Then applying law of sines on <math>\bigtriangleup AO_{1}B</math> we obtain <cmath>\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}}</cmath> <cmath>\implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}}</cmath> <cmath>\implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}</cmath> Using similar logic we obtain <math>OA_{1} =\frac{189}{16\sqrt{5}}.</math> <br />
<br />
Now we know that <math>\angle O_{1}AO_{2}=180^{\circ}-x.</math> Thus using law of cosines on <math>\bigtriangleup O_{1}AO_{2}</math> yields <cmath>O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}</cmath> While this does look daunting we can write the above expression as <cmath>\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}</cmath> Then factoring yields <cmath>\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}</cmath> The area <cmath>[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}</cmath> Now <math>AK</math> is twice the length of the altitude of <math>\bigtriangleup O_{1}AO_{2}</math> so we let the altitude be <math>h</math> and we have <cmath>\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}</cmath> <cmath>\implies h =\frac{9}{4}</cmath> Thus our desired length is <math>\frac{9}{2} \implies n+n = \boxed{11}.</math><br />
<br />
==Solution 4 (Video)==<br />
<br />
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=II|num-b=10|num-a=12}}<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_11&diff=1615992015 AIME II Problems/Problem 112021-09-05T04:40:21Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
The circumcircle of acute <math>\triangle ABC</math> has center <math>O</math>. The line passing through point <math>O</math> perpendicular to <math>\overline{OB}</math> intersects lines <math>AB</math> and <math>BC</math> and <math>P</math> and <math>Q</math>, respectively. Also <math>AB=5</math>, <math>BC=4</math>, <math>BQ=4.5</math>, and <math>BP=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Diagram==<br />
<asy><br />
unitsize(30);<br />
draw(Circle((0,0),3));<br />
pair A,B,C,O, Q, P, M, N;<br />
A=(2.5, -sqrt(11/4));<br />
B=(-2.5, -sqrt(11/4));<br />
C=(-1.96, 2.28);<br />
Q=(-1.89, 2.81);<br />
P=(1.13, -1.68);<br />
O=origin;<br />
M=foot(O,C,B);<br />
N=foot(O,A,B);<br />
draw(A--B--C--cycle);<br />
label("$A$",A,SE);<br />
label("$B$",B,SW);<br />
label("$C$",C,NW);<br />
label("$Q$",Q,NW);<br />
dot(O);<br />
label("$O$",O,NE);<br />
label("$M$",M,W);<br />
label("$N$",N,S);<br />
label("$P$",P,S);<br />
draw(B--O);<br />
draw(C--Q);<br />
draw(Q--O);<br />
draw(O--C);<br />
draw(O--A);<br />
draw(O--P);<br />
draw(O--M, dashed);<br />
draw(O--N, dashed);<br />
draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5));<br />
draw(rightanglemark(O,N,B,5));<br />
draw(rightanglemark(B,O,P,5));<br />
draw(rightanglemark(O,M,C,5));<br />
</asy><br />
<br />
==Solution==<br />
<br />
<br />
===Solution 1===<br />
Call <math>M</math> and <math>N</math> the feet of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> . Notice that <math>\triangle{OMB} \sim \triangle{QOB}</math> because both are right triangles, and <math>\angle{OBQ} \cong \angle{OBM}</math>. By <math>\frac{MB}{BO}=\frac{BO}{BQ}</math>, <math>MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}</math>. However, since <math>O</math> is the circumcenter of triangle <math>ABC</math>, <math>OM</math> is a perpendicular bisector by the definition of a circumcenter. Hence, <math>\frac{r^2}{4.5} = 2 \implies r = 3</math>. Since we know <math>BN=\frac{5}{2}</math> and <math>\triangle BOP \sim \triangle BNO</math>, we have <math>\frac{BP}{3} = \frac{3}{\frac{5}{2}}</math>. Thus, <math>BP = \frac{18}{5}</math>. <math>m + n=\boxed{023}</math>.<br />
<br />
===Solution 2 (fastest)===<br />
Notice that <math>\angle{CBO}=90-A</math>, so <math>\angle{BQO}=A</math>. From this we get that <math>\triangle{BPQ}\sim \triangle{BCA}</math>. So <math>\dfrac{BP}{BC}=\dfrac{BQ}{BA}</math>, plugging in the given values we get <math>\dfrac{BP}{4}=\dfrac{4.5}{5}</math>, so <math>BP=\dfrac{18}{5}</math>, and <math>m+n=\boxed{023}</math>.<br />
<br />
===Solution 3===<br />
Let <math>r=BO</math>. Drawing perpendiculars, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, <cmath>OM=\sqrt{r^2-4}</cmath> Thus, <cmath>OQ=\frac{\sqrt{4r^2+9}}{2}</cmath> Using <math>\triangle{BOQ}</math>, we get <math>r=3</math>. Now let's find <math>NP</math>. After some calculations with <math>\triangle{BON}</math> ~ <math>\triangle{OPN}</math>, <math>{NP=11/10}</math>. Therefore, <cmath>BP=\frac{5}{2}+\frac{11}{10}=18/5</cmath> <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 4===<br />
Let <math>\angle{BQO}=\alpha</math>. Extend <math>OB</math> to touch the circumcircle at a point <math>K</math>. Then, note that <math>\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha</math>. But since <math>BK</math> is a diameter, <math>\angle{KAB}=90^\circ</math>, implying <math>\angle{CAB}=\alpha</math>. It follows that <math>APCQ</math> is a cyclic quadrilateral.<br />
<br />
Let <math>BP=x</math>. By Power of a Point, <cmath>5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.</cmath>The answer is <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 5===<br />
<math>\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}</math><br />
<br />
Denote the circumradius of <math>ABC</math> to be <math>R</math>, the circumcircle of <math>ABC</math> to be <math>O</math>, and the shortest distance from <math>Q</math> to circle <math>O</math> to be <math>x</math>. <br />
<br />
Using Power of a Point on <math>Q</math> relative to circle <math>O</math>, we get that <math>x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}</math>. Using Pythagorean Theorem on triangle <math>QOB</math> to get <math>(x + r)^2 + r^2 = \frac{81}{4}</math>. Subtracting the first equation from the second, we get that <math>2r^2 = 18</math> and therefore <math>r = 3</math>. Now, set <math>\cos{ABC} = y</math>. Using law of cosines on <math>ABC</math> to find <math>AC</math> in terms of <math>y</math> and plugging that into the extended law of sines, we get <math>\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6</math>. Squaring both sides and cross multiplying, we get <math>36x^2 - 40x + 5 = 0</math>. Now, we get <math>x = \frac{10 \pm \sqrt{55}}{18}</math> using quadratic formula. If you drew a decent diagram, <math>B</math> is acute and therefore <math>x = \frac{10 + \sqrt{55}}{18}</math>(You can also try plugging in both in the end and seeing which gives a rational solution). Note that <math>BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.</math> Using the cosine addition formula and then plugging in what we know about <math>QBO</math>, we get that <math>BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}</math>. Now, the hard part is to find what <math>\sin{B}</math> is. We therefore want <math>\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}</math>. For the numerator, by inspection <math>(a + b\sqrt{55})^2</math> will not work for integers <math>a</math> and <math>b</math>. The other case is if there is <math>(a\sqrt{5} + b\sqrt{11})^2</math>. By inspection, <math>5\sqrt{5} - 2\sqrt{11}</math> works. Therefore, plugging all this in yields the answer, <math>\frac{18}{5} \rightarrow \boxed{23}</math>. Solution by hyxue<br />
<br />
===Solution 6===<br />
<asy> <br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(15cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); <br />
draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); <br />
draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */<br />
draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((-1.82,0.87),dotstyle); <br />
label("$A$", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); <br />
dot((3.178984115621537,0.7692140299269852),dotstyle); <br />
label("$B$", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); <br />
dot((2.6857306099246263,4.738685150758791),dotstyle); <br />
label("$C$", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); <br />
dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); <br />
label("$O$", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); <br />
dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); <br />
label("$P$", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); <br />
dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); <br />
label("$Q$", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); <br />
dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); <br />
label("$A'$", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); <br />
dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); <br />
label("$P'$", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); <br />
dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); <br />
label("$D$", (2.603644633462422,5.802794720137042), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
</asy><br />
Reflect <math>A</math>, <math>P</math> across <math>OB</math> to points <math>A'</math> and <math>P'</math>, respectively with <math>A'</math> on the circle and <math>P, O, P'</math> collinear. Now, <math>\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB</math> by parallel lines. From here, <math>\angle P'PB = \angle PP'B = \angle A'P'Q</math> as <math>P, P', Q</math> collinear. From here, <math>A'P'QC</math> is cyclic, and by power of a point we obtain <math>\frac{18}{5} \implies \boxed{023}</math>.<br />
~awang11's sol<br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_11&diff=1615982015 AIME II Problems/Problem 112021-09-05T04:39:54Z<p>Jdong2006: /* Solution 5 */</p>
<hr />
<div>==Problem==<br />
<br />
The circumcircle of acute <math>\triangle ABC</math> has center <math>O</math>. The line passing through point <math>O</math> perpendicular to <math>\overline{OB}</math> intersects lines <math>AB</math> and <math>BC</math> and <math>P</math> and <math>Q</math>, respectively. Also <math>AB=5</math>, <math>BC=4</math>, <math>BQ=4.5</math>, and <math>BP=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Diagram==<br />
<asy><br />
unitsize(30);<br />
draw(Circle((0,0),3));<br />
pair A,B,C,O, Q, P, M, N;<br />
A=(2.5, -sqrt(11/4));<br />
B=(-2.5, -sqrt(11/4));<br />
C=(-1.96, 2.28);<br />
Q=(-1.89, 2.81);<br />
P=(1.13, -1.68);<br />
O=origin;<br />
M=foot(O,C,B);<br />
N=foot(O,A,B);<br />
draw(A--B--C--cycle);<br />
label("$A$",A,SE);<br />
label("$B$",B,SW);<br />
label("$C$",C,NW);<br />
label("$Q$",Q,NW);<br />
dot(O);<br />
label("$O$",O,NE);<br />
label("$M$",M,W);<br />
label("$N$",N,S);<br />
label("$P$",P,S);<br />
draw(B--O);<br />
draw(C--Q);<br />
draw(Q--O);<br />
draw(O--C);<br />
draw(O--A);<br />
draw(O--P);<br />
draw(O--M, dashed);<br />
draw(O--N, dashed);<br />
draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5));<br />
draw(rightanglemark(O,N,B,5));<br />
draw(rightanglemark(B,O,P,5));<br />
draw(rightanglemark(O,M,C,5));<br />
</asy><br />
<br />
==Solution==<br />
<br />
<br />
===Solution 1===<br />
Call <math>M</math> and <math>N</math> the feet of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> . Notice that <math>\triangle{OMB} \sim \triangle{QOB}</math> because both are right triangles, and <math>\angle{OBQ} \cong \angle{OBM}</math>. By <math>\frac{MB}{BO}=\frac{BO}{BQ}</math>, <math>MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}</math>. However, since <math>O</math> is the circumcenter of triangle <math>ABC</math>, <math>OM</math> is a perpendicular bisector by the definition of a circumcenter. Hence, <math>\frac{r^2}{4.5} = 2 \implies r = 3</math>. Since we know <math>BN=\frac{5}{2}</math> and <math>\triangle BOP \sim \triangle BNO</math>, we have <math>\frac{BP}{3} = \frac{3}{\frac{5}{2}}</math>. Thus, <math>BP = \frac{18}{5}</math>. <math>m + n=\boxed{023}</math>.<br />
<br />
===Solution 2===<br />
Notice that <math>\angle{CBO}=90-A</math>, so <math>\angle{BQO}=A</math>. From this we get that <math>\triangle{BPQ}\sim \triangle{BCA}</math>. So <math>\dfrac{BP}{BC}=\dfrac{BQ}{BA}</math>, plugging in the given values we get <math>\dfrac{BP}{4}=\dfrac{4.5}{5}</math>, so <math>BP=\dfrac{18}{5}</math>, and <math>m+n=\boxed{023}</math>.<br />
<br />
===Solution 3===<br />
Let <math>r=BO</math>. Drawing perpendiculars, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, <cmath>OM=\sqrt{r^2-4}</cmath> Thus, <cmath>OQ=\frac{\sqrt{4r^2+9}}{2}</cmath> Using <math>\triangle{BOQ}</math>, we get <math>r=3</math>. Now let's find <math>NP</math>. After some calculations with <math>\triangle{BON}</math> ~ <math>\triangle{OPN}</math>, <math>{NP=11/10}</math>. Therefore, <cmath>BP=\frac{5}{2}+\frac{11}{10}=18/5</cmath> <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 4===<br />
Let <math>\angle{BQO}=\alpha</math>. Extend <math>OB</math> to touch the circumcircle at a point <math>K</math>. Then, note that <math>\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha</math>. But since <math>BK</math> is a diameter, <math>\angle{KAB}=90^\circ</math>, implying <math>\angle{CAB}=\alpha</math>. It follows that <math>APCQ</math> is a cyclic quadrilateral.<br />
<br />
Let <math>BP=x</math>. By Power of a Point, <cmath>5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.</cmath>The answer is <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 5===<br />
<math>\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}</math><br />
<br />
Denote the circumradius of <math>ABC</math> to be <math>R</math>, the circumcircle of <math>ABC</math> to be <math>O</math>, and the shortest distance from <math>Q</math> to circle <math>O</math> to be <math>x</math>. <br />
<br />
Using Power of a Point on <math>Q</math> relative to circle <math>O</math>, we get that <math>x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}</math>. Using Pythagorean Theorem on triangle <math>QOB</math> to get <math>(x + r)^2 + r^2 = \frac{81}{4}</math>. Subtracting the first equation from the second, we get that <math>2r^2 = 18</math> and therefore <math>r = 3</math>. Now, set <math>\cos{ABC} = y</math>. Using law of cosines on <math>ABC</math> to find <math>AC</math> in terms of <math>y</math> and plugging that into the extended law of sines, we get <math>\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6</math>. Squaring both sides and cross multiplying, we get <math>36x^2 - 40x + 5 = 0</math>. Now, we get <math>x = \frac{10 \pm \sqrt{55}}{18}</math> using quadratic formula. If you drew a decent diagram, <math>B</math> is acute and therefore <math>x = \frac{10 + \sqrt{55}}{18}</math>(You can also try plugging in both in the end and seeing which gives a rational solution). Note that <math>BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.</math> Using the cosine addition formula and then plugging in what we know about <math>QBO</math>, we get that <math>BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}</math>. Now, the hard part is to find what <math>\sin{B}</math> is. We therefore want <math>\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}</math>. For the numerator, by inspection <math>(a + b\sqrt{55})^2</math> will not work for integers <math>a</math> and <math>b</math>. The other case is if there is <math>(a\sqrt{5} + b\sqrt{11})^2</math>. By inspection, <math>5\sqrt{5} - 2\sqrt{11}</math> works. Therefore, plugging all this in yields the answer, <math>\frac{18}{5} \rightarrow \boxed{23}</math>. Solution by hyxue<br />
<br />
===Solution 6===<br />
<asy> <br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(15cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); <br />
draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); <br />
draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */<br />
draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((-1.82,0.87),dotstyle); <br />
label("$A$", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); <br />
dot((3.178984115621537,0.7692140299269852),dotstyle); <br />
label("$B$", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); <br />
dot((2.6857306099246263,4.738685150758791),dotstyle); <br />
label("$C$", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); <br />
dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); <br />
label("$O$", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); <br />
dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); <br />
label("$P$", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); <br />
dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); <br />
label("$Q$", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); <br />
dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); <br />
label("$A'$", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); <br />
dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); <br />
label("$P'$", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); <br />
dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); <br />
label("$D$", (2.603644633462422,5.802794720137042), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
</asy><br />
Reflect <math>A</math>, <math>P</math> across <math>OB</math> to points <math>A'</math> and <math>P'</math>, respectively with <math>A'</math> on the circle and <math>P, O, P'</math> collinear. Now, <math>\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB</math> by parallel lines. From here, <math>\angle P'PB = \angle PP'B = \angle A'P'Q</math> as <math>P, P', Q</math> collinear. From here, <math>A'P'QC</math> is cyclic, and by power of a point we obtain <math>\frac{18}{5} \implies \boxed{023}</math>.<br />
~awang11's sol<br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_11&diff=1615972015 AIME II Problems/Problem 112021-09-05T04:39:13Z<p>Jdong2006: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
The circumcircle of acute <math>\triangle ABC</math> has center <math>O</math>. The line passing through point <math>O</math> perpendicular to <math>\overline{OB}</math> intersects lines <math>AB</math> and <math>BC</math> and <math>P</math> and <math>Q</math>, respectively. Also <math>AB=5</math>, <math>BC=4</math>, <math>BQ=4.5</math>, and <math>BP=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Diagram==<br />
<asy><br />
unitsize(30);<br />
draw(Circle((0,0),3));<br />
pair A,B,C,O, Q, P, M, N;<br />
A=(2.5, -sqrt(11/4));<br />
B=(-2.5, -sqrt(11/4));<br />
C=(-1.96, 2.28);<br />
Q=(-1.89, 2.81);<br />
P=(1.13, -1.68);<br />
O=origin;<br />
M=foot(O,C,B);<br />
N=foot(O,A,B);<br />
draw(A--B--C--cycle);<br />
label("$A$",A,SE);<br />
label("$B$",B,SW);<br />
label("$C$",C,NW);<br />
label("$Q$",Q,NW);<br />
dot(O);<br />
label("$O$",O,NE);<br />
label("$M$",M,W);<br />
label("$N$",N,S);<br />
label("$P$",P,S);<br />
draw(B--O);<br />
draw(C--Q);<br />
draw(Q--O);<br />
draw(O--C);<br />
draw(O--A);<br />
draw(O--P);<br />
draw(O--M, dashed);<br />
draw(O--N, dashed);<br />
draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5));<br />
draw(rightanglemark(O,N,B,5));<br />
draw(rightanglemark(B,O,P,5));<br />
draw(rightanglemark(O,M,C,5));<br />
</asy><br />
<br />
==Solution==<br />
<br />
<br />
===Solution 1===<br />
Call <math>M</math> and <math>N</math> the feet of the altitudes from <math>O</math> to <math>BC</math> and <math>AB</math>, respectively. Let <math>OB = r</math> . Notice that <math>\triangle{OMB} \sim \triangle{QOB}</math> because both are right triangles, and <math>\angle{OBQ} \cong \angle{OBM}</math>. By <math>\frac{MB}{BO}=\frac{BO}{BQ}</math>, <math>MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}</math>. However, since <math>O</math> is the circumcenter of triangle <math>ABC</math>, <math>OM</math> is a perpendicular bisector by the definition of a circumcenter. Hence, <math>\frac{r^2}{4.5} = 2 \implies r = 3</math>. Since we know <math>BN=\frac{5}{2}</math> and <math>\triangle BOP \sim \triangle BNO</math>, we have <math>\frac{BP}{3} = \frac{3}{\frac{5}{2}}</math>. Thus, <math>BP = \frac{18}{5}</math>. <math>m + n=\boxed{023}</math>.<br />
<br />
===Solution 2===<br />
Notice that <math>\angle{CBO}=90-A</math>, so <math>\angle{BQO}=A</math>. From this we get that <math>\triangle{BPQ}\sim \triangle{BCA}</math>. So <math>\dfrac{BP}{BC}=\dfrac{BQ}{BA}</math>, plugging in the given values we get <math>\dfrac{BP}{4}=\dfrac{4.5}{5}</math>, so <math>BP=\dfrac{18}{5}</math>, and <math>m+n=\boxed{023}</math>.<br />
<br />
===Solution 3===<br />
Let <math>r=BO</math>. Drawing perpendiculars, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, <cmath>OM=\sqrt{r^2-4}</cmath> Thus, <cmath>OQ=\frac{\sqrt{4r^2+9}}{2}</cmath> Using <math>\triangle{BOQ}</math>, we get <math>r=3</math>. Now let's find <math>NP</math>. After some calculations with <math>\triangle{BON}</math> ~ <math>\triangle{OPN}</math>, <math>{NP=11/10}</math>. Therefore, <cmath>BP=\frac{5}{2}+\frac{11}{10}=18/5</cmath> <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 4===<br />
Let <math>\angle{BQO}=\alpha</math>. Extend <math>OB</math> to touch the circumcircle at a point <math>K</math>. Then, note that <math>\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha</math>. But since <math>BK</math> is a diameter, <math>\angle{KAB}=90^\circ</math>, implying <math>\angle{CAB}=\alpha</math>. It follows that <math>APCQ</math> is a cyclic quadrilateral.<br />
<br />
Let <math>BP=x</math>. By Power of a Point, <cmath>5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.</cmath>The answer is <math>18+5=\boxed{023}</math>.<br />
<br />
===Solution 5===<br />
Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.<br />
<br />
Denote the circumradius of <math>ABC</math> to be <math>R</math>, the circumcircle of <math>ABC</math> to be <math>O</math>, and the shortest distance from <math>Q</math> to circle <math>O</math> to be <math>x</math>. <br />
<br />
Using Power of a Point on <math>Q</math> relative to circle <math>O</math>, we get that <math>x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}</math>. Using Pythagorean Theorem on triangle <math>QOB</math> to get <math>(x + r)^2 + r^2 = \frac{81}{4}</math>. Subtracting the first equation from the second, we get that <math>2r^2 = 18</math> and therefore <math>r = 3</math>. Now, set <math>\cos{ABC} = y</math>. Using law of cosines on <math>ABC</math> to find <math>AC</math> in terms of <math>y</math> and plugging that into the extended law of sines, we get <math>\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6</math>. Squaring both sides and cross multiplying, we get <math>36x^2 - 40x + 5 = 0</math>. Now, we get <math>x = \frac{10 \pm \sqrt{55}}{18}</math> using quadratic formula. If you drew a decent diagram, <math>B</math> is acute and therefore <math>x = \frac{10 + \sqrt{55}}{18}</math>(You can also try plugging in both in the end and seeing which gives a rational solution). Note that <math>BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.</math> Using the cosine addition formula and then plugging in what we know about <math>QBO</math>, we get that <math>BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}</math>. Now, the hard part is to find what <math>\sin{B}</math> is. We therefore want <math>\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}</math>. For the numerator, by inspection <math>(a + b\sqrt{55})^2</math> will not work for integers <math>a</math> and <math>b</math>. The other case is if there is <math>(a\sqrt{5} + b\sqrt{11})^2</math>. By inspection, <math>5\sqrt{5} - 2\sqrt{11}</math> works. Therefore, plugging all this in yields the answer, <math>\frac{18}{5} \rightarrow \boxed{23}</math>. Solution by hyxue<br />
<br />
===Solution 6===<br />
<asy> <br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(15cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); <br />
draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); <br />
draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */<br />
draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */<br />
draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((-1.82,0.87),dotstyle); <br />
label("$A$", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); <br />
dot((3.178984115621537,0.7692140299269852),dotstyle); <br />
label("$B$", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); <br />
dot((2.6857306099246263,4.738685150758791),dotstyle); <br />
label("$C$", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); <br />
dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); <br />
label("$O$", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); <br />
dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); <br />
label("$P$", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); <br />
dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); <br />
label("$Q$", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); <br />
dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); <br />
label("$A'$", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); <br />
dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); <br />
label("$P'$", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); <br />
dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); <br />
label("$D$", (2.603644633462422,5.802794720137042), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
</asy><br />
Reflect <math>A</math>, <math>P</math> across <math>OB</math> to points <math>A'</math> and <math>P'</math>, respectively with <math>A'</math> on the circle and <math>P, O, P'</math> collinear. Now, <math>\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB</math> by parallel lines. From here, <math>\angle P'PB = \angle PP'B = \angle A'P'Q</math> as <math>P, P', Q</math> collinear. From here, <math>A'P'QC</math> is cyclic, and by power of a point we obtain <math>\frac{18}{5} \implies \boxed{023}</math>.<br />
~awang11's sol<br />
<br />
==See also==<br />
{{AIME box|year=2015|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_10&diff=1615892017 AIME II Problems/Problem 102021-09-05T02:23:59Z<p>Jdong2006: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Rectangle <math>ABCD</math> has side lengths <math>AB=84</math> and <math>AD=42</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>, point <math>N</math> is the trisection point of <math>\overline{AB}</math> closer to <math>A</math>, and point <math>O</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{DN}</math>. Point <math>P</math> lies on the quadrilateral <math>BCON</math>, and <math>\overline{BP}</math> bisects the area of <math>BCON</math>. Find the area of <math>\triangle CDP</math>.<br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,D,M,n,O,P;<br />
A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13);<br />
fill(C--D--P--cycle,blue);<br />
draw(A--B--C--D--cycle);<br />
draw(C--M);<br />
draw(D--n);<br />
draw(B--P);<br />
draw(D--P);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$M$",M,W);<br />
label("$N$",n,N);<br />
label("$O$",O,(-0.5,1));<br />
label("$P$",P,N);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(M);<br />
dot(n);<br />
dot(O);<br />
dot(P);<br />
label("28",(0,42)--(28,42),N);<br />
label("56",(28,42)--(84,42),N);<br />
</asy><br />
Impose a coordinate system on the diagram where point <math>D</math> is the origin. Therefore <math>A=(0,42)</math>, <math>B=(84,42)</math>, <math>C=(84,0)</math>, and <math>D=(0,0)</math>. Because <math>M</math> is a midpoint and <math>N</math> is a trisection point, <math>M=(0,21)</math> and <math>N=(28,42)</math>. The equation for line <math>DN</math> is <math>y=\frac{3}{2}x</math> and the equation for line <math>CM</math> is <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, so their intersection, point <math>O</math>, is <math>(12,18)</math>. Using the shoelace formula on quadrilateral <math>BCON</math>, or drawing diagonal <math>\overline{BO}</math> and using <math>\frac12bh</math>, we find that its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2} = 1092</math>. Using <math>A = \frac 12 bh</math>, we get <math>2184 = 42h</math>. Simplifying, we get <math>h = 52</math>. This means that the x-coordinate of <math>P = 84 - 52 = 32</math>. Since P lies on <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, you can solve and get that the y-coordinate of <math>P</math> is <math>13</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}</math>.<br />
<br />
==Solution 2 (No Coordinates)==<br />
Since the problem tells us that segment <math>\overline{BP}</math> bisects the area of quadrilateral <math>BCON</math>, let us compute the area of <math>BCON</math> by subtracting the areas of <math>\triangle{AND}</math> and <math>\triangle{DOC}</math> from rectangle <math>ABCD</math>.<br />
<br />
To do this, drop altitude <math>\overline{OE}</math> onto side <math>\overline{DC}</math> and draw a horizontal segment <math>\overline{MQ}</math> from side <math>\overline{AD}</math> to <math>\overline{ND}</math>. Since <math>M</math> is the midpoint of side <math>\overline{AD}</math>, <cmath>\overline{MQ}=14</cmath> Denote <math>\overline{OE}</math> as <math>a</math>. Noting that <math>\triangle{MOQ}~\triangle{COD}</math>, we can write the statement <cmath>\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}</cmath> <cmath>\implies \frac{84}{a}=\frac{14}{21-a}</cmath> <cmath>\implies a=18</cmath> Using this information, the area of <math>\triangle{DOC}</math> and <math>\triangle{AND}</math> are <cmath>\frac{18\cdot 84}{2}=756</cmath> and <cmath>\frac{28\cdot 42}{2}=588</cmath> respectively. Thus, the area of quadrilateral <math>BCON</math> is <cmath>84\cdot 42-588-756=2184</cmath> Now, it is clear that point <math>P</math> lies on side <math>\overline{MC}</math>, so the area of <math>\triangle{BPC}</math> is <cmath>\frac{2184}{2}=1092</cmath> Given this, drop altitude <math>\overline{PF}</math> (let's call it <math>b</math>) onto <math>\overline{BC}</math>. Therefore, <cmath>\frac{42b}{2}=1092\implies b=52</cmath> From here, drop an altitude <math>\overline{PG}</math> onto <math>\overline{DC}</math>. Recognizing that <math>\overline{PF}=\overline{GC}</math> and that <math>\triangle{MDC}</math> and <math>\triangle{PGC}</math> are similar, we write <cmath>\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}</cmath> <cmath>\implies \frac{\overline{PG}}{52}=\frac{21}{84}</cmath> <cmath>\implies \overline{PG}=13</cmath> The area of <math>\triangle{CDP}</math> is given by <cmath>\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}</cmath> ~blitzkrieg21 and jdong2006<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_10&diff=1615882017 AIME II Problems/Problem 102021-09-05T02:23:31Z<p>Jdong2006: /* Solution 2 (No Coordinates) */</p>
<hr />
<div>==Problem==<br />
Rectangle <math>ABCD</math> has side lengths <math>AB=84</math> and <math>AD=42</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>, point <math>N</math> is the trisection point of <math>\overline{AB}</math> closer to <math>A</math>, and point <math>O</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{DN}</math>. Point <math>P</math> lies on the quadrilateral <math>BCON</math>, and <math>\overline{BP}</math> bisects the area of <math>BCON</math>. Find the area of <math>\triangle CDP</math>.<br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,D,M,n,O,P;<br />
A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13);<br />
fill(C--D--P--cycle,lightgray);<br />
draw(A--B--C--D--cycle);<br />
draw(C--M);<br />
draw(D--n);<br />
draw(B--P);<br />
draw(D--P);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$M$",M,W);<br />
label("$N$",n,N);<br />
label("$O$",O,(-0.5,1));<br />
label("$P$",P,N);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(M);<br />
dot(n);<br />
dot(O);<br />
dot(P);<br />
label("28",(0,42)--(28,42),N);<br />
label("56",(28,42)--(84,42),N);<br />
</asy><br />
Impose a coordinate system on the diagram where point <math>D</math> is the origin. Therefore <math>A=(0,42)</math>, <math>B=(84,42)</math>, <math>C=(84,0)</math>, and <math>D=(0,0)</math>. Because <math>M</math> is a midpoint and <math>N</math> is a trisection point, <math>M=(0,21)</math> and <math>N=(28,42)</math>. The equation for line <math>DN</math> is <math>y=\frac{3}{2}x</math> and the equation for line <math>CM</math> is <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, so their intersection, point <math>O</math>, is <math>(12,18)</math>. Using the shoelace formula on quadrilateral <math>BCON</math>, or drawing diagonal <math>\overline{BO}</math> and using <math>\frac12bh</math>, we find that its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2} = 1092</math>. Using <math>A = \frac 12 bh</math>, we get <math>2184 = 42h</math>. Simplifying, we get <math>h = 52</math>. This means that the x-coordinate of <math>P = 84 - 52 = 32</math>. Since P lies on <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, you can solve and get that the y-coordinate of <math>P</math> is <math>13</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}</math>.<br />
<br />
==Solution 2 (No Coordinates)==<br />
Since the problem tells us that segment <math>\overline{BP}</math> bisects the area of quadrilateral <math>BCON</math>, let us compute the area of <math>BCON</math> by subtracting the areas of <math>\triangle{AND}</math> and <math>\triangle{DOC}</math> from rectangle <math>ABCD</math>.<br />
<br />
To do this, drop altitude <math>\overline{OE}</math> onto side <math>\overline{DC}</math> and draw a horizontal segment <math>\overline{MQ}</math> from side <math>\overline{AD}</math> to <math>\overline{ND}</math>. Since <math>M</math> is the midpoint of side <math>\overline{AD}</math>, <cmath>\overline{MQ}=14</cmath> Denote <math>\overline{OE}</math> as <math>a</math>. Noting that <math>\triangle{MOQ}~\triangle{COD}</math>, we can write the statement <cmath>\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}</cmath> <cmath>\implies \frac{84}{a}=\frac{14}{21-a}</cmath> <cmath>\implies a=18</cmath> Using this information, the area of <math>\triangle{DOC}</math> and <math>\triangle{AND}</math> are <cmath>\frac{18\cdot 84}{2}=756</cmath> and <cmath>\frac{28\cdot 42}{2}=588</cmath> respectively. Thus, the area of quadrilateral <math>BCON</math> is <cmath>84\cdot 42-588-756=2184</cmath> Now, it is clear that point <math>P</math> lies on side <math>\overline{MC}</math>, so the area of <math>\triangle{BPC}</math> is <cmath>\frac{2184}{2}=1092</cmath> Given this, drop altitude <math>\overline{PF}</math> (let's call it <math>b</math>) onto <math>\overline{BC}</math>. Therefore, <cmath>\frac{42b}{2}=1092\implies b=52</cmath> From here, drop an altitude <math>\overline{PG}</math> onto <math>\overline{DC}</math>. Recognizing that <math>\overline{PF}=\overline{GC}</math> and that <math>\triangle{MDC}</math> and <math>\triangle{PGC}</math> are similar, we write <cmath>\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}</cmath> <cmath>\implies \frac{\overline{PG}}{52}=\frac{21}{84}</cmath> <cmath>\implies \overline{PG}=13</cmath> The area of <math>\triangle{CDP}</math> is given by <cmath>\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}</cmath> ~blitzkrieg21 and jdong2006<br />
<br />
=See Also=<br />
{{AIME box|year=2017|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_10&diff=1615762014 AIME I Problems/Problem 102021-09-05T00:02:58Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>== Problem 10 ==<br />
A disk with radius <math>1</math> is externally tangent to a disk with radius <math>5</math>. Let <math>A</math> be the point where the disks are tangent, <math>C</math> be the center of the smaller disk, and <math>E</math> be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of <math>360^\circ</math>. That is, if the center of the smaller disk has moved to the point <math>D</math>, and the point on the smaller disk that began at <math>A</math> has now moved to point <math>B</math>, then <math>\overline{AC}</math> is parallel to <math>\overline{BD}</math>. Then <math>\sin^2(\angle BEA)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
== Solution 1 ==<br />
<br />
<br />
<asy><br />
size(150);<br />
pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0);<br />
draw(circle(e,5));<br />
draw(circle(c,1));<br />
draw(circle(d,1));<br />
dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2));<br />
label("$A$",a,W,fontsize(9));<br />
label("$B$",b,NW,fontsize(9));<br />
label("$C$",c,E,fontsize(9));<br />
label("$D$",d,E,fontsize(9));<br />
label("$E$",e,SW,fontsize(9));<br />
label("$F$",(5/2,5*sqrt(3)/2),SSW,fontsize(9));<br />
</asy><br />
<br />
Let <math>F</math> be the new tangency point of the two disks. The smaller disk rolled along minor arc <math>\overarc{AF}</math> on the larger disk.<br />
<br />
Let <math>\alpha = \angle AEF</math>, in radians. The smaller disk must then have rolled along an arc of length <math>5\alpha</math>, since the larger disk has a radius of <math>5</math>. Since all of the points on major arc <math>\overarc{BF}</math> on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did,<cmath>\overarc{BF}=\overarc{AF}=5\alpha</cmath>. <br />
<br />
Since <math>\overline{AC} || \overline{BD}</math>, <cmath>\angle BDF \cong \angle FEA</cmath> so the angles of minor arc <math>\overarc{BF}</math> and minor arc <math>\overarc{AF}</math> are equal, so minor arc <math>\overarc{BF}</math> has an angle of <math>\alpha</math>. Since the smaller disk has a radius of <math>1</math>, the length of minor arc <math>\overarc{BF}</math> is <math>\alpha</math>. This means that <math>5\alpha + \alpha</math> equals the circumference of the smaller disk, so <math>6\alpha = 2\pi</math>, or <math>\alpha = \frac{\pi}{3}</math>.<br />
<br />
Now, to find <math>\sin^2{\angle BEA}</math>, we construct <math>\triangle BDE</math>. Also, drop a perpendicular from <math>D</math> to <math>\overline{EA}</math>, and call this point <math>X</math>. Since <math>\alpha = \frac{\pi}{3}</math> and <math>\angle DXE</math> is right, <cmath>DE = 6</cmath> <cmath>EX = 3</cmath> and <cmath>DX = 3\sqrt{3}</cmath> <br />
<br />
Now drop a perpendicular from <math>B</math> to <math>\overline{EA}</math>, and call this point <math>Y</math>. Since <math>\overline{BD} || \overline{EA}</math>, <cmath>XY = BD = 1</cmath> and <cmath>BY = DX = 3\sqrt{3}</cmath> Thus, we know that <cmath>EY = EX - XY = 3 - 1 = 2</cmath> and by using the Pythagorean Theorem on <math>\triangle BEY</math>, we get that <cmath>BE = \sqrt{31}</cmath> Thus, <cmath>\sin{\angle BEA} = \frac{\sqrt{27}}{\sqrt{31}}</cmath> so <cmath>\sin^2{\angle BEA} = \frac{27}{31}</cmath> and our answer is <math>27 + 31 = \boxed{058}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate <math>5</math> times (the circumference of the small circle is <math>2\pi</math> while the larger one has a circumference of <math>10\pi</math>) plus the extra rotation the circle gets for rotating around the circle, for a total of <math>6</math> times. Therefore, one rotation will bring point <math>D</math> <math>60^\circ</math> from <math>C</math>.<br />
<br />
Now, draw <math>\triangle DBE</math>, and call <cmath>\angle BED = x^{\circ}</cmath> We know that <math>\overline{ED}</math> is 6, and <math>\overline{BD}</math> is 1. Since <math>EC || BD</math>, <cmath>\angle BDE = 60^\circ</cmath> <br />
<br />
By the Law of Cosines, <cmath>\overline{BE}^2=36+1-2\times 6\times 1\times \cos{60^\circ} = 36+1-6=31</cmath> and since lengths are positive, <cmath>\overline{BE}=\sqrt{31}</cmath> <br />
<br />
By the Law of Sines, we know that <cmath>\frac{1}{\sin{x}}=\frac{\sqrt{31}}{\sin{60^\circ}}</cmath> so <cmath>\sin{x} = \frac{\sin{60^\circ}}{\sqrt{31}} = \frac{\sqrt{93}}{62}</cmath> As <math>x</math> is clearly between <math>0</math> and <math>90^\circ</math>, <math>\cos{x}</math> is positive. As <math>\cos{x}=\sqrt{1-\sin^2{x}}</math>, <cmath>\cos{x} = \frac{11\sqrt{31}}{62}</cmath> <br />
<br />
Now we use the angle sum formula to find the sine of <math>\angle BEA</math>: <cmath>\sin 60^\circ\cos x + \cos 60^\circ\sin x = \frac{\sqrt{3}}{2}\frac{11\sqrt{31}}{62}+\frac{1}{2}\frac{\sqrt{93}}{62}</cmath> <cmath>= \frac{11\sqrt{93}+\sqrt{93}}{124} = \frac{12\sqrt{93}}{124} = \frac{3\sqrt{93}}{31} = \frac{3\sqrt{31}\sqrt{3}}{31} = \frac{3\sqrt{3}}{\sqrt{31}}</cmath><br />
<br />
Finally, we square this to get <cmath>\frac{9\times 3}{31}=\frac{27}{31}</cmath> so our answer is <math>27+31=\boxed{058}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_10&diff=1615752014 AIME I Problems/Problem 102021-09-05T00:00:28Z<p>Jdong2006: /* Solution 1 */</p>
<hr />
<div>== Problem 10 ==<br />
A disk with radius <math>1</math> is externally tangent to a disk with radius <math>5</math>. Let <math>A</math> be the point where the disks are tangent, <math>C</math> be the center of the smaller disk, and <math>E</math> be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of <math>360^\circ</math>. That is, if the center of the smaller disk has moved to the point <math>D</math>, and the point on the smaller disk that began at <math>A</math> has now moved to point <math>B</math>, then <math>\overline{AC}</math> is parallel to <math>\overline{BD}</math>. Then <math>\sin^2(\angle BEA)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
== Solution 1 ==<br />
<br />
<br />
<asy><br />
size(150);<br />
pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0);<br />
draw(circle(e,5));<br />
draw(circle(c,1));<br />
draw(circle(d,1));<br />
dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2));<br />
label("$A$",a,W,fontsize(9));<br />
label("$B$",b,NW,fontsize(9));<br />
label("$C$",c,E,fontsize(9));<br />
label("$D$",d,E,fontsize(9));<br />
label("$E$",e,SW,fontsize(9));<br />
label("$F$",(5/2,5*sqrt(3)/2),SSW,fontsize(9));<br />
</asy><br />
<br />
Let <math>F</math> be the new tangency point of the two disks. The smaller disk rolled along minor arc <math>\overarc{AF}</math> on the larger disk.<br />
<br />
Let <math>\alpha = \angle AEF</math>, in radians. The smaller disk must then have rolled along an arc of length <math>5\alpha</math>, since the larger disk has a radius of <math>5</math>. Since all of the points on major arc <math>\overarc{BF}</math> on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did,<cmath>\overarc{BF}=\overarc{AF}=5\alpha</cmath>. <br />
<br />
Since <math>\overline{AC} || \overline{BD}</math>, <cmath>\angle BDF \cong \angle FEA</cmath> so the angles of minor arc <math>\overarc{BF}</math> and minor arc <math>\overarc{AF}</math> are equal, so minor arc <math>\overarc{BF}</math> has an angle of <math>\alpha</math>. Since the smaller disk has a radius of <math>1</math>, the length of minor arc <math>\overarc{BF}</math> is <math>\alpha</math>. This means that <math>5\alpha + \alpha</math> equals the circumference of the smaller disk, so <math>6\alpha = 2\pi</math>, or <math>\alpha = \frac{\pi}{3}</math>.<br />
<br />
Now, to find <math>\sin^2{\angle BEA}</math>, we construct <math>\triangle BDE</math>. Also, drop a perpendicular from <math>D</math> to <math>\overline{EA}</math>, and call this point <math>X</math>. Since <math>\alpha = \frac{\pi}{3}</math> and <math>\angle DXE</math> is right, <cmath>DE = 6</cmath> <cmath>EX = 3</cmath> and <cmath>DX = 3\sqrt{3}</cmath> <br />
<br />
Now drop a perpendicular from <math>B</math> to <math>\overline{EA}</math>, and call this point <math>Y</math>. Since <math>\overline{BD} || \overline{EA}</math>, <cmath>XY = BD = 1</cmath> and <cmath>BY = DX = 3\sqrt{3}</cmath> Thus, we know that <cmath>EY = EX - XY = 3 - 1 = 2</cmath> and by using the Pythagorean Theorem on <math>\triangle BEY</math>, we get that <cmath>BE = \sqrt{31}</cmath> Thus, <cmath>\sin{\angle BEA} = \frac{\sqrt{27}}{\sqrt{31}}</cmath> so <cmath>\sin^2{\angle BEA} = \frac{27}{31}</cmath> and our answer is <math>27 + 31 = \boxed{058}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate <math>5</math> times (the circumference of the small circle is <math>2\pi</math> while the larger one has a circumference of <math>10\pi</math>) plus the extra rotation the circle gets for rotating around the circle, for a total of <math>6</math> times. Therefore, one rotation will bring point <math>D</math> <math>60^\circ</math> from <math>C</math>.<br />
<br />
Now, draw <math>\triangle DBE</math>, and call <math>\angle BED</math> <math>x</math>, in degrees. We know that <math>\overline{ED}</math> is 6, and <math>\overline{BD}</math> is 1. Since <math>EC ||<br />
BD</math>, <math>\angle BDE = 60^\circ</math>. By the Law of Cosines, <math>\overline{BE}^2=36+1-2\times 6\times 1\times \cos{60^\circ} = 36+1-6=31</math>, and since lengths are positive, <math>\overline{BE}=\sqrt{31}</math>. <br />
<br />
By the Law of Sines, we know that <math>\frac{1}{\sin{x}}=\frac{\sqrt{31}}{\sin{60^\circ}}</math>, so <math>\sin{x} = \frac{\sin{60^\circ}}{\sqrt{31}} = \frac{\sqrt{93}}{62}</math>. As <math>x</math> is clearly between <math>0</math> and <math>90^\circ</math>, <math>\cos{x}</math> is positive. As <math>\cos{x}=\sqrt{1-\sin^2{x}}</math>, <math>\cos{x} = \frac{11\sqrt{31}}{62}</math>. <br />
<br />
Now we use the angle sum formula to find the sine of <math>\angle BEA</math>: <math>\sin 60^\circ\cos x + \cos 60^\circ\sin x = \frac{\sqrt{3}}{2}\frac{11\sqrt{31}}{62}+\frac{1}{2}\frac{\sqrt{93}}{62} = \frac{11\sqrt{93}+\sqrt{93}}{124} = \frac{12\sqrt{93}}{124} = \frac{3\sqrt{93}}{31} = \frac{3\sqrt{31}\sqrt{3}}{31} = \frac{3\sqrt{3}}{\sqrt{31}}</math>. <br />
<br />
Finally, we square this to get <math>\frac{9\times 3}{31}=\frac{27}{31}</math>, so our answer is <math>27+31=\boxed{058}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_10&diff=1615732013 AIME II Problems/Problem 102021-09-04T23:42:32Z<p>Jdong2006: /* Solution 3 (simpler solution) */</p>
<hr />
<div>==Problem==<br />
<br />
Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>.<br />
<br />
==Solution 1==<br />
<asy><br />
import math;<br />
import olympiad;<br />
import graph;<br />
pair A, B, K, L;<br />
B = (sqrt(13), 0); A=(4+sqrt(13), 0);<br />
dot(B);<br />
dot(A);<br />
<br />
draw(Circle((0,0), sqrt(13)));<br />
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);<br />
dot((0,0));<br />
<br />
<br />
<br />
</asy><br />
<br />
<br />
Now we put the figure in the Cartesian plane, let the center of the circle <math>O (0,0)</math>, then <math>B (\sqrt{13},0)</math>, and <math>A(4+\sqrt{13},0)</math><br />
<br />
The equation for Circle O is <math>x^2+y^2=13</math>, and let the slope of the line<math>AKL</math> be <math>k</math>, then the equation for line<math>AKL</math> is <math>y=k(x-4-\sqrt{13})</math>.<br />
<br />
Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math>. According to [[Vieta's Formulas]], we get<br />
<br />
<math>x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}</math>, and <math>x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}</math><br />
<br />
So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math><br />
<br />
Also, the distance between <math>B</math> and <math>LK</math> is <math>\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}</math><br />
<br />
So the area <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</math><br />
<br />
Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math><br />
<br />
So the answer is <math>104+26+13+3=\boxed{146}</math>.<br />
<br />
==Solution 2==<br />
<asy><br />
import math;<br />
import olympiad;<br />
import graph;<br />
pair A, B, K, L;<br />
B = (sqrt(13), 0); A=(4+sqrt(13), 0);<br />
dot(B);<br />
dot(A);<br />
<br />
draw(Circle((0,0), sqrt(13)));<br />
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);<br />
dot((0,0));<br />
<br />
<br />
<br />
</asy><br />
<br />
Draw <math>OC</math> perpendicular to <math>KL</math> at <math>C</math>. Draw <math>BD</math> perpendicular to <math>KL</math> at <math>D</math>. <br />
<br />
<cmath>\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}</cmath><br />
<br />
Therefore, to maximize area of <math>\triangle BKL</math>, we need to maximize area of <math>\triangle OKL</math>.<br />
<br />
<cmath>\triangle OKL = \frac12 r^2 \sin{\angle KOL}</cmath><br />
<br />
So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>.<br />
<br />
Eventually, we get <cmath>\triangle BKL= \frac12 \cdot (\sqrt{13})^2\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}</cmath><br />
<br />
So the answer is <math>104+26+13+3=\boxed{146}</math>.<br />
<br />
==Solution 3 (simpler solution)==<br />
A rather easier solution is presented in the Girls' Angle WordPress:<br />
<br />
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2013|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_10&diff=1615722013 AIME II Problems/Problem 102021-09-04T23:41:49Z<p>Jdong2006: /* Solution 3 (simplest solution) */</p>
<hr />
<div>==Problem==<br />
<br />
Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>.<br />
<br />
==Solution 1==<br />
<asy><br />
import math;<br />
import olympiad;<br />
import graph;<br />
pair A, B, K, L;<br />
B = (sqrt(13), 0); A=(4+sqrt(13), 0);<br />
dot(B);<br />
dot(A);<br />
<br />
draw(Circle((0,0), sqrt(13)));<br />
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);<br />
dot((0,0));<br />
<br />
<br />
<br />
</asy><br />
<br />
<br />
Now we put the figure in the Cartesian plane, let the center of the circle <math>O (0,0)</math>, then <math>B (\sqrt{13},0)</math>, and <math>A(4+\sqrt{13},0)</math><br />
<br />
The equation for Circle O is <math>x^2+y^2=13</math>, and let the slope of the line<math>AKL</math> be <math>k</math>, then the equation for line<math>AKL</math> is <math>y=k(x-4-\sqrt{13})</math>.<br />
<br />
Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math>. According to [[Vieta's Formulas]], we get<br />
<br />
<math>x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}</math>, and <math>x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}</math><br />
<br />
So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math><br />
<br />
Also, the distance between <math>B</math> and <math>LK</math> is <math>\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}</math><br />
<br />
So the area <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</math><br />
<br />
Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math><br />
<br />
So the answer is <math>104+26+13+3=\boxed{146}</math>.<br />
<br />
==Solution 2==<br />
<asy><br />
import math;<br />
import olympiad;<br />
import graph;<br />
pair A, B, K, L;<br />
B = (sqrt(13), 0); A=(4+sqrt(13), 0);<br />
dot(B);<br />
dot(A);<br />
<br />
draw(Circle((0,0), sqrt(13)));<br />
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);<br />
dot((0,0));<br />
<br />
<br />
<br />
</asy><br />
<br />
Draw <math>OC</math> perpendicular to <math>KL</math> at <math>C</math>. Draw <math>BD</math> perpendicular to <math>KL</math> at <math>D</math>. <br />
<br />
<cmath>\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}</cmath><br />
<br />
Therefore, to maximize area of <math>\triangle BKL</math>, we need to maximize area of <math>\triangle OKL</math>.<br />
<br />
<cmath>\triangle OKL = \frac12 r^2 \sin{\angle KOL}</cmath><br />
<br />
So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>.<br />
<br />
Eventually, we get <cmath>\triangle BKL= \frac12 \cdot (\sqrt{13})^2\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}</cmath><br />
<br />
So the answer is <math>104+26+13+3=\boxed{146}</math>.<br />
<br />
==Solution 3 (simpler solution)==<br />
A rather easier solution is presented in the Girls Angle WordPress:<br />
<br />
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2013|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_10&diff=1615712013 AIME II Problems/Problem 102021-09-04T23:41:29Z<p>Jdong2006: </p>
<hr />
<div>==Problem==<br />
<br />
Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>.<br />
<br />
==Solution 1==<br />
<asy><br />
import math;<br />
import olympiad;<br />
import graph;<br />
pair A, B, K, L;<br />
B = (sqrt(13), 0); A=(4+sqrt(13), 0);<br />
dot(B);<br />
dot(A);<br />
<br />
draw(Circle((0,0), sqrt(13)));<br />
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);<br />
dot((0,0));<br />
<br />
<br />
<br />
</asy><br />
<br />
<br />
Now we put the figure in the Cartesian plane, let the center of the circle <math>O (0,0)</math>, then <math>B (\sqrt{13},0)</math>, and <math>A(4+\sqrt{13},0)</math><br />
<br />
The equation for Circle O is <math>x^2+y^2=13</math>, and let the slope of the line<math>AKL</math> be <math>k</math>, then the equation for line<math>AKL</math> is <math>y=k(x-4-\sqrt{13})</math>.<br />
<br />
Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math>. According to [[Vieta's Formulas]], we get<br />
<br />
<math>x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}</math>, and <math>x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}</math><br />
<br />
So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math><br />
<br />
Also, the distance between <math>B</math> and <math>LK</math> is <math>\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}</math><br />
<br />
So the area <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</math><br />
<br />
Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math><br />
<br />
So the answer is <math>104+26+13+3=\boxed{146}</math>.<br />
<br />
==Solution 2==<br />
<asy><br />
import math;<br />
import olympiad;<br />
import graph;<br />
pair A, B, K, L;<br />
B = (sqrt(13), 0); A=(4+sqrt(13), 0);<br />
dot(B);<br />
dot(A);<br />
<br />
draw(Circle((0,0), sqrt(13)));<br />
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);<br />
dot((0,0));<br />
<br />
<br />
<br />
</asy><br />
<br />
Draw <math>OC</math> perpendicular to <math>KL</math> at <math>C</math>. Draw <math>BD</math> perpendicular to <math>KL</math> at <math>D</math>. <br />
<br />
<cmath>\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}</cmath><br />
<br />
Therefore, to maximize area of <math>\triangle BKL</math>, we need to maximize area of <math>\triangle OKL</math>.<br />
<br />
<cmath>\triangle OKL = \frac12 r^2 \sin{\angle KOL}</cmath><br />
<br />
So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>.<br />
<br />
Eventually, we get <cmath>\triangle BKL= \frac12 \cdot (\sqrt{13})^2\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}</cmath><br />
<br />
So the answer is <math>104+26+13+3=\boxed{146}</math>.<br />
<br />
==Solution 3 (simplest solution)==<br />
A rather easier solution is presented in the Girls Angle WordPress:<br />
<br />
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2013|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=Trigonometric_identities&diff=161563Trigonometric identities2021-09-04T21:26:07Z<p>Jdong2006: Added Triple-angle identities</p>
<hr />
<div>In [[trigonometry]], '''trigonometric identities''' are equations involving trigonometric functions that are true for all input values. Trigonometric functions have an abundance of identities, of which only the most widely used are included in this article.<br />
<br />
== Pythagorean identities ==<br />
The Pythagorean identities state that<br />
* <math>\sin^2x + \cos^2x = 1</math><br />
* <math>1 + \cot^2x = \csc^2x</math><br />
* <math>\tan^2x + 1 = \sec^2x</math><br />
<br />
Using the unit circle definition of trigonometry, because the point <math>(\cos (x), \sin (x))</math> is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, <math>\sin^2x + \cos^2x = 1</math>. To derive the other two Pythagorean identities, divide by either <math>\sin^2 (x)</math> or <math>\cos^2 (x)</math> and substitute the respective trigonometry in place of the ratios to obtain the desired result.<br />
<br />
== Angle addition identities ==<br />
The trigonometric angle addition identities state the following identities:<br />
* <math>\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)</math><br />
* <math>\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y) </math><br />
* <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)} </math><br />
There are many proofs of these identities. For the sake of brevity, we list only one here.<br />
<br />
[[Euler's identity]] states that <math>e^{ix} = \cos (x) + i \sin(x)</math>. We have that<br />
<cmath>\begin{align*}<br />
\cos (x+y) + i \sin (x+y) &= e^{i(x+y)} \\<br />
&= e^{ix} \cdot e^{iy} \\<br />
&= (\cos (x) + i \sin (x))(\cos (y) + i \sin (y)) \\<br />
&= (\cos (x) \cos (y) - \sin (x) \sin(y)) + i(\sin (x) \cos(y) + \cos(x) \sin(y))<br />
\end{align*}</cmath><br />
By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.<br />
<br />
To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by <math>\cos (x) \cos (y)</math>, and simplify.<br />
<cmath>\begin{align*}<br />
\tan (x+y) &= \frac{\sin (x+y)}{\cos (x+y)} \\<br />
&= \frac{\sin (x) \cos(y) + \cos(x) \sin(y)}{\cos (x) \cos (y) - \sin (x) \sin(y)} \\<br />
&= \frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(x)}}{1 - \frac{\sin (x) \sin(y)}{\cos (x) \cos(y)}} \\<br />
&= \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan(y)}<br />
\end{align*}</cmath><br />
as desired.<br />
<br />
== Double-angle identities ==<br />
The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting <math>x = y </math>. Doing so yields:<br />
* <math>\sin (2x) = 2\sin (x) \cos (x)</math><br />
* <math>\cos (2x) = \cos^2 (x) - \sin^2 (x)</math><br />
* <math>\tan (2x) = \frac{2\tan (x)}{1-\tan^2 (x)}</math><br />
<br />
=== Cosine double-angle identity ===<br />
Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above.<br />
* <math>\cos (2x) = 1 - 2 \sin^2 (x)</math><br />
* <math>\cos (2x) = 2 \cos^2 (x) - 1</math><br />
<br />
In addition, the following identities are useful in [[integration]] and in deriving the half-angle identities. They are a simple rearrangement of the two above.<br />
* <math>\sin^2 (x) = \frac{1 - \cos (2x)}{2}</math><br />
* <math>\cos^2 (x) = \frac{1 + \cos (2x)}{2}</math><br />
<br />
== Half-angle identities ==<br />
The trigonometric half-angle identities state the following equalities:<br />
* <math>\sin (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{2}}</math><br />
* <math>\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}</math><br />
* <math>\tan (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{1+\cos (x)}} = \frac{\sin (x)}{1 + \cos (x)} = \frac{1-\cos (x)}{\sin (x)}</math><br />
The plus or minus does not mean that there are two answers, but that the sign of the expression depends on the quadrant in which the angle resides.<br />
<br />
Consider the two expressions listed in the cosine double-angle section for <math>\sin^2 (x)</math> and <math>\cos^2 (x)</math>, and substitute <math>\frac{1}{2} x</math> instead of <math>x</math>. Taking the square root then yields the desired half-angle identities for sine and cosine. As for the tangent identity, divide the sine and cosine half-angle identities.<br />
<br />
== Product-to-sum identities ==<br />
The product-to-sum identities are as follows:<br />
* <math>\sin (x) \sin (y) = \frac{1}{2} (\cos (x-y) - \cos (x+y))</math><br />
* <math>\sin (x) \cos (y) = \frac{1}{2} (\sin (x-y) + \sin (x+y))</math><br />
* <math>\cos (x) \cos (y) = \frac{1}{2} (\cos (x-y) + \cos (x+y))</math><br />
They can be derived by expanding out <math>\cos (x+y)</math> and <math>\cos (x-y)</math> or <math>\sin (x+y)</math> and <math>\sin(x-y)</math>, then combining them to isolate each term.<br />
<br />
== Sum-to-product identities ==<br />
Substituting <math>\alpha = x+y</math> and <math>\beta = x-y</math> into the product-to-sum identities yields the sum-to-product identities.<br />
<br />
* <math>\sin (x) + \sin (y) = 2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2})</math><br />
* <math>\sin (x) - \sin (y) = 2 \sin (\frac{x - y}{2}) \cos (\frac{x + y}{2})</math><br />
* <math>\cos (x) + \cos (y) = 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2})</math><br />
* <math>\cos (x) - \cos (y) = -2 \sin (\frac{x + y}{2}) \sin (\frac{x - y}{2})</math><br />
<br />
== Other identities ==<br />
Here are some identities that are less significant than those above, but still useful.<br />
<br />
=== Triple-angle identities ===<br />
* <math>\sin 3x = 3\sin x-4\sin^3 x</math><br />
* <math>\cos 3x = 4\cos^3 x-3\cos x</math><br />
* <math>\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}</math><br />
All of these expansions can be proved using <cmath>\sin 3x = \sin (2x+x)</cmath> trick and perform the angle addition identities. Same for <math>\cos 3x</math> and for <math>\tan 3x</math>.<br />
<br />
=== Even-odd identities ===<br />
The functions <math>\sin(x)</math>, <math>\tan(x)</math>, and <math>\csc(x)</math> are odd, while <math>cos(x)</math>, <math>\cot(x)</math>, and <math>\sec(x)</math> are even. In other words, the six trigonometric functions satisfy the following equalities:<br />
* <math>\sin (-x) = -\sin (x) </math><br />
* <math>\cos (-x) = \cos (x) </math><br />
* <math>\tan (-x) = -\tan (x) </math><br />
* <math>\cot (-x) = -\cot (x) </math><br />
* <math>\csc (-x) = -\csc (x) </math><br />
* <math>\sec (-x) = \sec (x) </math><br />
<br />
These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as reflecting it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, <math>\sin(-x) = -\sin(x)</math> and <math>\cos(-x) = \cos(x)</math>.<br />
<br />
=== Conversion identities ===<br />
The following identities are useful when converting trigonometric functions.<br />
*<math>\sin (90^{\circ} - x) = \cos (x) \textrm{ and } \cos (90^{\circ} - x) = \sin (x)</math><br />
*<math>\tan (90^{\circ} - x) = \cot (x) \textrm{ and } \cot (90^{\circ} - x) = \tan (x)</math><br />
*<math>\csc (90^{\circ} - x) = \sec (x) \textrm{ and } \sec (90^{\circ} - x) = \csc (x)</math><br />
All of these can be proven via the angle addition identities.<br />
<br />
=== Euler's identity ===<br />
[[Euler's identity]] is a formula in complex analysis that connects complex exponentiation with trigonometry. It states that for any real number <math>x</math>, <cmath>e^{ix} = \cos (x) + i \sin (x),</cmath> where <math>e</math> is Euler's constant and <math>i</math> is the imaginary unit. Euler's identity is fundamental to the study of complex numbers and is widely considered among the most beautiful formulas in math.<br />
<br />
Similar to the derivation of the product-to-sum identities, we can isolate sine and cosine by comparing <math>e^{ix}</math> and <math>e^{-ix}</math>, which yields the following identities:<br />
*<math>\cos (x) = \frac{e^{ix} + e^{-ix}}{2}</math><br />
*<math>\sin (x) = \frac{e^{ix} - e^{-ix}}{2i}</math><br />
They can also be derived by computing <math>\textrm{Re} (e^{ix})</math> and <math>\textrm{Im} (e^{ix})</math>. The definitions are occasionally used as a definition of the trigonometric functions.<br />
<br />
=== Miscellaneous ===<br />
These are the identities that are not substantial enough to warrant a section of their own.<br />
*<math>\sin (180^{\circ} - x) = \sin (x) \textrm{ and } \csc (180^{\circ} - x) = \csc (x)</math><br />
*<math>\cos (180^{\circ} - x) = -\cos (x) \textrm{ and } \sec (180^{\circ} - x) = -\sec (x)</math><br />
*<math>\tan (180^{\circ} - x) = -\tan (x) \textrm{ and } \cot (180^{\circ} - x) = -\cot (x)</math><br />
* <math>\sin (x) \cos(x) = \frac{\sin (2x)}{2}</math><br />
<br />
== Resources ==<br />
* [http://www.sosmath.com/trig/Trig5/trig5/trig5.html Table of trigonometric identities]<br />
* [https://mathwithtimmy.files.wordpress.com/2017/06/trig-identities.pdf List of Trigonometric identities]<br />
<br />
== See also ==<br />
* [[Trigonometry]]<br />
* [[Trigonometric substitution]]<br />
<br />
[[Category:Trigonometry]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=Trigonometric_identities&diff=161562Trigonometric identities2021-09-04T21:16:41Z<p>Jdong2006: /* Product-to-sum identities */ mini sin and cos error formatting</p>
<hr />
<div>In [[trigonometry]], '''trigonometric identities''' are equations involving trigonometric functions that are true for all input values. Trigonometric functions have an abundance of identities, of which only the most widely used are included in this article.<br />
<br />
== Pythagorean identities ==<br />
The Pythagorean identities state that<br />
* <math>\sin^2x + \cos^2x = 1</math><br />
* <math>1 + \cot^2x = \csc^2x</math><br />
* <math>\tan^2x + 1 = \sec^2x</math><br />
<br />
Using the unit circle definition of trigonometry, because the point <math>(\cos (x), \sin (x))</math> is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, <math>\sin^2x + \cos^2x = 1</math>. To derive the other two Pythagorean identities, divide by either <math>\sin^2 (x)</math> or <math>\cos^2 (x)</math> and substitute the respective trigonometry in place of the ratios to obtain the desired result.<br />
<br />
== Angle addition identities ==<br />
The trigonometric angle addition identities state the following identities:<br />
* <math>\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)</math><br />
* <math>\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y) </math><br />
* <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)} </math><br />
There are many proofs of these identities. For the sake of brevity, we list only one here.<br />
<br />
[[Euler's identity]] states that <math>e^{ix} = \cos (x) + i \sin(x)</math>. We have that<br />
<cmath>\begin{align*}<br />
\cos (x+y) + i \sin (x+y) &= e^{i(x+y)} \\<br />
&= e^{ix} \cdot e^{iy} \\<br />
&= (\cos (x) + i \sin (x))(\cos (y) + i \sin (y)) \\<br />
&= (\cos (x) \cos (y) - \sin (x) \sin(y)) + i(\sin (x) \cos(y) + \cos(x) \sin(y))<br />
\end{align*}</cmath><br />
By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.<br />
<br />
To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by <math>\cos (x) \cos (y)</math>, and simplify.<br />
<cmath>\begin{align*}<br />
\tan (x+y) &= \frac{\sin (x+y)}{\cos (x+y)} \\<br />
&= \frac{\sin (x) \cos(y) + \cos(x) \sin(y)}{\cos (x) \cos (y) - \sin (x) \sin(y)} \\<br />
&= \frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(x)}}{1 - \frac{\sin (x) \sin(y)}{\cos (x) \cos(y)}} \\<br />
&= \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan(y)}<br />
\end{align*}</cmath><br />
as desired.<br />
<br />
== Double-angle identities ==<br />
The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting <math>x = y </math>. Doing so yields:<br />
* <math>\sin (2x) = 2\sin (x) \cos (x)</math><br />
* <math>\cos (2x) = \cos^2 (x) - \sin^2 (x)</math><br />
* <math>\tan (2x) = \frac{2\tan (x)}{1-\tan^2 (x)}</math><br />
<br />
=== Cosine double-angle identity ===<br />
Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above.<br />
* <math>\cos (2x) = 1 - 2 \sin^2 (x)</math><br />
* <math>\cos (2x) = 2 \cos^2 (x) - 1</math><br />
<br />
In addition, the following identities are useful in [[integration]] and in deriving the half-angle identities. They are a simple rearrangement of the two above.<br />
* <math>\sin^2 (x) = \frac{1 - \cos (2x)}{2}</math><br />
* <math>\cos^2 (x) = \frac{1 + \cos (2x)}{2}</math><br />
<br />
== Half-angle identities ==<br />
The trigonometric half-angle identities state the following equalities:<br />
* <math>\sin (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{2}}</math><br />
* <math>\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}</math><br />
* <math>\tan (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{1+\cos (x)}} = \frac{\sin (x)}{1 + \cos (x)} = \frac{1-\cos (x)}{\sin (x)}</math><br />
The plus or minus does not mean that there are two answers, but that the sign of the expression depends on the quadrant in which the angle resides.<br />
<br />
Consider the two expressions listed in the cosine double-angle section for <math>\sin^2 (x)</math> and <math>\cos^2 (x)</math>, and substitute <math>\frac{1}{2} x</math> instead of <math>x</math>. Taking the square root then yields the desired half-angle identities for sine and cosine. As for the tangent identity, divide the sine and cosine half-angle identities.<br />
<br />
== Product-to-sum identities ==<br />
The product-to-sum identities are as follows:<br />
* <math>\sin (x) \sin (y) = \frac{1}{2} (\cos (x-y) - \cos (x+y))</math><br />
* <math>\sin (x) \cos (y) = \frac{1}{2} (\sin (x-y) + \sin (x+y))</math><br />
* <math>\cos (x) \cos (y) = \frac{1}{2} (\cos (x-y) + \cos (x+y))</math><br />
They can be derived by expanding out <math>\cos (x+y)</math> and <math>\cos (x-y)</math> or <math>\sin (x+y)</math> and <math>\sin(x-y)</math>, then combining them to isolate each term.<br />
<br />
== Sum-to-product identities ==<br />
Substituting <math>\alpha = x+y</math> and <math>\beta = x-y</math> into the product-to-sum identities yields the sum-to-product identities.<br />
<br />
* <math>\sin (x) + \sin (y) = 2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2})</math><br />
* <math>\sin (x) - \sin (y) = 2 \sin (\frac{x - y}{2}) \cos (\frac{x + y}{2})</math><br />
* <math>\cos (x) + \cos (y) = 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2})</math><br />
* <math>\cos (x) - \cos (y) = -2 \sin (\frac{x + y}{2}) \sin (\frac{x - y}{2})</math><br />
<br />
== Other identities ==<br />
Here are some identities that are less significant than those above, but still useful.<br />
<br />
=== Even-odd identities ===<br />
The functions <math>\sin(x)</math>, <math>\tan(x)</math>, and <math>\csc(x)</math> are odd, while <math>cos(x)</math>, <math>\cot(x)</math>, and <math>\sec(x)</math> are even. In other words, the six trigonometric functions satisfy the following equalities:<br />
* <math>\sin (-x) = -\sin (x) </math><br />
* <math>\cos (-x) = \cos (x) </math><br />
* <math>\tan (-x) = -\tan (x) </math><br />
* <math>\cot (-x) = -\cot (x) </math><br />
* <math>\csc (-x) = -\csc (x) </math><br />
* <math>\sec (-x) = \sec (x) </math><br />
<br />
These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as reflecting it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, <math>\sin(-x) = -\sin(x)</math> and <math>\cos(-x) = \cos(x)</math>.<br />
<br />
=== Conversion identities ===<br />
The following identities are useful when converting trigonometric functions.<br />
*<math>\sin (90^{\circ} - x) = \cos (x) \textrm{ and } \cos (90^{\circ} - x) = \sin (x)</math><br />
*<math>\tan (90^{\circ} - x) = \cot (x) \textrm{ and } \cot (90^{\circ} - x) = \tan (x)</math><br />
*<math>\csc (90^{\circ} - x) = \sec (x) \textrm{ and } \sec (90^{\circ} - x) = \csc (x)</math><br />
All of these can be proven via the angle addition identities.<br />
<br />
=== Euler's identity ===<br />
[[Euler's identity]] is a formula in complex analysis that connects complex exponentiation with trigonometry. It states that for any real number <math>x</math>, <cmath>e^{ix} = \cos (x) + i \sin (x),</cmath> where <math>e</math> is Euler's constant and <math>i</math> is the imaginary unit. Euler's identity is fundamental to the study of complex numbers and is widely considered among the most beautiful formulas in math.<br />
<br />
Similar to the derivation of the product-to-sum identities, we can isolate sine and cosine by comparing <math>e^{ix}</math> and <math>e^{-ix}</math>, which yields the following identities:<br />
*<math>\cos (x) = \frac{e^{ix} + e^{-ix}}{2}</math><br />
*<math>\sin (x) = \frac{e^{ix} - e^{-ix}}{2i}</math><br />
They can also be derived by computing <math>\textrm{Re} (e^{ix})</math> and <math>\textrm{Im} (e^{ix})</math>. The definitions are occasionally used as a definition of the trigonometric functions.<br />
<br />
=== Miscellaneous ===<br />
These are the identities that are not substantial enough to warrant a section of their own.<br />
*<math>\sin (180^{\circ} - x) = \sin (x) \textrm{ and } \csc (180^{\circ} - x) = \csc (x)</math><br />
*<math>\cos (180^{\circ} - x) = -\cos (x) \textrm{ and } \sec (180^{\circ} - x) = -\sec (x)</math><br />
*<math>\tan (180^{\circ} - x) = -\tan (x) \textrm{ and } \cot (180^{\circ} - x) = -\cot (x)</math><br />
* <math>\sin (x) \cos(x) = \frac{\sin (2x)}{2}</math><br />
<br />
== Resources ==<br />
* [http://www.sosmath.com/trig/Trig5/trig5/trig5.html Table of trigonometric identities]<br />
* [https://mathwithtimmy.files.wordpress.com/2017/06/trig-identities.pdf List of Trigonometric identities]<br />
<br />
== See also ==<br />
* [[Trigonometry]]<br />
* [[Trigonometric substitution]]<br />
<br />
[[Category:Trigonometry]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_9&diff=1615602016 AIME I Problems/Problem 92021-09-04T20:52:09Z<p>Jdong2006: /* Problem 9 */</p>
<hr />
<div>==Problem 9 (WARNING: CORRUPTED PROBLEM)==<br />
Triangle <math>ABC</math> has <math>AB=40,AC=31,</math> and <math>\sin{A}=\frac{1}{5}</math>. This triangle is inscribed in rectangle <math>AQRS</math> with <math>B</math> on <math>\overline{QR}</math> and <math>C</math> on <math>\overline{RS}</math>. Find the maximum possible area of <math>AQRS</math>.<br />
<br />
==Solution 1==<br />
Note that if angle <math>BAC</math> is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where <math>A</math> is obtuse. Therefore, angle A is acute. Let angle <math>CAS=n</math> and angle <math>BAQ=m</math>. Then, <math>\overline{AS}=31\cos(n)</math> and <math>\overline{AQ}=40\cos(m)</math>. Then the area of rectangle <math>AQRS</math> is <math>1240\cos(m)\cos(n)</math>. By product-to-sum, <math>\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))</math>. <math>\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}</math>. The maximum possible value of <math>\cos(m-n)</math> is 1, which occurs when <math>m=n</math>. Thus the maximum possible value of <math>\cos(m)\cos(n)</math> is <math>\frac{1}{2}(\frac{1}{5}+1)=\frac{3}{5}</math> so the maximum possible area of <math>AQRS</math> is <math>1240\times{\frac{3}{5}}=\fbox{744}</math>.<br />
<br />
==Solution 2==<br />
As above, we note that angle <math>A</math> must be acute. Therefore, let <math>A</math> be the origin, and suppose that <math>Q</math> is on the positive <math>x</math> axis and <math>S</math> is on the positive <math>y</math> axis. We approach this using complex numbers. Let <math>w=\text{cis} A</math>, and let <math>z</math> be a complex number with <math>|z|=1</math>, <math>\text{Arg}(z)\ge 0^\circ</math> and <math>\text{Arg}(zw)\le90^\circ</math>. Then we represent <math>B</math> by <math>40z</math> and <math>C</math> by <math>31zw</math>. The coordinates of <math>Q</math> and <math>S</math> depend on the real part of <math>40z</math> and the imaginary part of <math>31zw</math>. Thus<br />
<cmath>[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).</cmath><br />
We can expand this, using the fact that <math>z\overline{z}=|z|^2</math>, finding<br />
<cmath>[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).</cmath><br />
Now as <math>w=\text{cis}A</math>, we know that <math>\Im(w)=\frac15</math>. Also, <math>|z^2w|=1</math>, so the maximum possible imaginary part of <math>z^2w</math> is <math>1</math>. This is clearly achievable under our conditions on <math>z</math>. Therefore, the maximum possible area of <math>AQRS</math> is <math>620(1+\tfrac15)=\boxed{744}</math>.<br />
<br />
==Solution 3 (With Calculus)==<br />
Let <math>\theta</math> be the angle <math>\angle BAQ</math>. The height of the rectangle then can be expressed as <math>h = 31 \sin (A+\theta)</math>, and the length of the rectangle can be expressed as <math>l = 40\cos \theta</math>. The area of the rectangle can then be written as a function of <math>\theta</math>, <math>[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta</math>. For now, we will ignore the <math>1240</math> and focus on the function <math>f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta</math>.<br />
<br />
Taking the derivative, <math>f'(\theta) = \sin A \cdot -2\cos \theta \sin \theta + \cos A \cos 2\theta = \cos A \cos 2\theta - \sin A \sin 2\theta = \cos(2\theta + A)</math>. Setting this equal to <math>0</math>, we get <math>\cos(2 \theta + A) = 0 \Rightarrow 2\theta +A = 90, 270 ^\circ</math>. Since we know that <math>A+ \theta < 90</math>, the <math>270^\circ</math> solution is extraneous. Thus, we get that <math>\theta = \frac{90 - A}{2} = 45 - \frac{A}{2}</math>.<br />
<br />
Plugging this value into the original area equation, <math>a(45 - \frac{A}{2}) = 1240 \sin (45 - \frac{A}{2} + A) \cos (45 - \frac{A}{2}) = 1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2})</math>. Using a product-to-sum formula, we get that: <cmath>1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2}) = </cmath><br />
<cmath>1240\cdot \frac{1}{2}\cdot(\sin((45 + \frac{A}{2}) + (45 -\frac{A}{2}))+\sin((45 +\frac{A}{2})-(45 - \frac{A}{2})))= </cmath><br />
<cmath>620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}</cmath>.<br />
<br />
==Solution 4==<br />
Let <math>\alpha</math> be the angle <math>\angle CAS</math> and <math>\beta</math> be the angle <math>\angle BAQ</math>. Then<br />
<cmath>\alpha + \beta + \angle A = 90^\circ \Rightarrow \alpha + \beta = 90^\circ - \angle A</cmath><br />
<cmath>\cos(\alpha + \beta) = \cos(90^\circ - \angle A)</cmath><br />
<cmath>\cos(\alpha + \beta) = \sin(\angle A) = \frac{1}{5}</cmath><br />
<cmath>\cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{1}{5}</cmath><br />
<cmath>\cos\alpha\cos\beta - \sqrt{(1-\cos^2\alpha)(1-\cos^2\beta)} = \frac{1}{5}</cmath><br />
<cmath>\cos\alpha\cos\beta - \sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} = \frac{1}{5}</cmath><br />
However, by AM-GM:<br />
<cmath>\cos^2\alpha+\cos^2\beta \ge 2\cos\alpha\beta</cmath><br />
Therefore,<br />
<cmath>1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta \le 1-2\cos\alpha\beta+\cos^2\alpha\cos^2\beta = (1-\cos\alpha\cos\beta)^2</cmath><br />
<cmath>\sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} \le 1-\cos\alpha\cos\beta</cmath><br />
So,<br />
<cmath>\frac{1}{5} \ge \cos\alpha\cos\beta - (1-\cos\alpha\cos\beta) = 2\cos\alpha\cos\beta-1</cmath><br />
<cmath>\frac{3}{5} \ge \cos\alpha\cos\beta</cmath>.<br />
However, the area of the rectangle is just <math>AS \cdot AQ = 31\cos\alpha \cdot 40\cos\beta \le 31 \cdot 40 \cdot \frac{3}{5} = \boxed{744}</math>.<br />
<br />
<br />
==Note on Problem Validity==<br />
<br />
It has been noted that this answer won't actually work. Let angle <math>QAB = m</math> and angle <math>CAS = n</math> as in Solution 1. Since we know (through that solution) that <math>m = n</math>, we can call them each <math>\theta</math>. The height of the rectangle is <math>AS = 31\cos\theta</math>, and the distance <math>BQ = 40\sin\theta</math>. We know that, if the triangle is to be inscribed in a rectangle, <math>AS \geq BQ</math>.<br />
<br />
<cmath>AS \geq BQ</cmath><br />
<br />
<cmath>31\cos\theta \geq 40\sin\theta</cmath><br />
<br />
<cmath>\frac{31}{40} \geq \tan\theta</cmath><br />
<br />
However, <math>\tan\theta = \tan(\frac{90-A}{2}) = \frac{\sin(90-A)}{\cos(90-A)+1} = \frac{\cos A}{\sin A + 1} = \frac{\frac{2\sqrt6}{5}}{\frac{6}{5}} = \frac{\sqrt6}{3} > \frac{31}{40}</math>, so the triangle does not actually fit in the rectangle: specifically, B is above R and thus in the line containing segment QR but not on the actual segment or in the rectangle.<br />
<br />
<asy><br />
size(200);<br />
pair A,B,C,Q,R,S;<br />
real r = (pi/2 - asin(1/5))/2;<br />
A = (0,0);<br />
B = 40*dir(r*180/pi);<br />
C = 31*dir(90-r*180/pi);<br />
draw(A--B--C--cycle);<br />
Q = (40*cos(r),0);<br />
R = (40*cos(r),31*cos(r));<br />
S = (0, 31*cos(r));<br />
draw(A--Q--R--S--cycle);<br />
<br />
label("$A$",A,SW);<br />
label("$B$",B,NE);<br />
label("$C$",C,N);<br />
label("$Q$",Q,SE);<br />
label("$R$",R,E);<br />
label("$S$",S,NW);<br />
</asy><br />
<br />
The actual answer is a radical near <math>728</math> (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer <math>744</math> despite the invalid problem statement.<br />
<br />
==See Also==<br />
{{AIME box|year=2016|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_7&diff=1614782018 AIME II Problems/Problem 72021-09-03T04:12:23Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>==Problem 7==<br />
Triangle <math>ABC</math> has side lengths <math>AB = 9</math>, <math>BC =</math> <math>5\sqrt{3}</math>, and <math>AC = 12</math>. Points <math>A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B</math> are on segment <math>\overline{AB}</math> with <math>P_{k}</math> between <math>P_{k-1}</math> and <math>P_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>, and points <math>A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C</math> are on segment <math>\overline{AC}</math> with <math>Q_{k}</math> between <math>Q_{k-1}</math> and <math>Q_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>. Furthermore, each segment <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2449</math>, is parallel to <math>\overline{BC}</math>. The segments cut the triangle into <math>2450</math> regions, consisting of <math>2449</math> trapezoids and <math>1</math> triangle. Each of the <math>2450</math> regions has the same area. Find the number of segments <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2450</math>, that have rational length.<br />
<br />
== Solution 1 ==<br />
For each <math>k</math> between <math>2</math> and <math>2450</math>, the area of the trapezoid with <math>\overline{P_kQ_k}</math> as its bottom base is the difference between the areas of two triangles, both similar to <math>\triangle{ABC}</math>. Let <math>d_k</math> be the length of segment <math>\overline{P_kQ_k}</math>. The area of the trapezoid with bases <math>\overline{P_{k-1}Q_{k-1}}</math> and <math>P_kQ_k</math> is <math>\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}</math> times the area of <math>\triangle{ABC}</math>. (This logic also applies to the topmost triangle if we notice that <math>d_0 = 0</math>.) However, we also know that the area of each shape is <math>\frac{1}{2450}</math> times the area of <math>\triangle{ABC}</math>. We then have <math>\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}</math>. Simplifying, <math>d_k^2-d_{k-1}^2 = \frac{3}{98}</math>. However, we know that <math>d_0^2 = 0</math>, so <math>d_1^2 = \frac{3}{98}</math>, and in general, <math>d_k^2 = \frac{3k}{98}</math> and <math>d_k = \frac{\sqrt{\frac{3k}{2}}}{7}</math>. The smallest <math>k</math> that gives a rational <math>d_k</math> is <math>6</math>, so <math>d_k</math> is rational if and only if <math>k = 6n^2</math> for some integer <math>n</math>.The largest <math>n</math> such that <math>6n^2</math> is less than <math>2450</math> is <math>20</math>, so <math>k</math> has <math>\boxed{020}</math> possible values.<br />
<br />
Solution by zeroman<br />
<br />
==Solution 2==<br />
We have that there are <math>2449</math> trapezoids and <math>1</math> triangle of equal area, with that one triangle being <math>AP_1Q_1</math>. Notice, if we "stack" the trapezoids on top of <math>\bigtriangleup AP_1Q_1</math> the way they already are, we'd create a similar triangle, all of which are similar to <math>\bigtriangleup ABC</math>, and since the trapezoids and <math>\bigtriangleup AP_1Q_1</math> have equal area, each of these similar triangles, <math>AP_kQ_k</math> have area <math>\frac{k}{2450}\left[ ABC\right]</math>, and so <math>\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}</math>. We want the ratio of the side lengths <math>P_kQ_k:BC</math>. Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or <cmath>\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}</cmath> <cmath>\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}</cmath> <cmath>\implies k=6n^2<2450 </cmath> <cmath>\implies 0<n\leq 20</cmath> so there are <math>\boxed{020}</math> solutions. <br />
<br />
~Solution by ktong<br />
<br />
~Beautified by jdong2006<br />
<br />
==Solution 3==<br />
<br />
Let <math>T_1</math> stand for <math>AP_1Q_1</math>, and <math>T_k = AP_kQ_k</math>. All triangles <math>T</math> are similar by AA. Let the area of <math>T_1</math> be <math>x</math>. The next trapezoid will also have an area of <math>x</math>, as given. Therefore, <math>T_k</math> has an area of <math>kx</math>. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, <math>P_k Q_k=P_1 Q_1\cdot \sqrt{k}</math>, and the same if <math>Q</math> is substituted for <math>P</math> throughout. We want the side <math>P_k Q_k</math> to be rational. Setting up proportions: <cmath>5\sqrt{3} : \sqrt{2450}=35\sqrt{2}</cmath> <cmath>\sqrt{6} : 14</cmath> which shows that <math>P_1 Q_1=\frac{\sqrt{6}}{14}</math>. In order for <math>\sqrt{k} P_1 Q_1</math> to be rational, <math>\sqrt{k}</math> must be some rational multiple of <math>\sqrt{6}</math>. This is achieved at <math>\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 20\sqrt{6}</math>. We end there as <math>21\sqrt{6}=\sqrt{2646}</math>. There are 20 numbers from 1 to 20, so there are <math>\boxed{020}</math> solutions.<br />
<br />
Solution by a1b2<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=II|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_7&diff=1614772018 AIME II Problems/Problem 72021-09-03T04:11:58Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>==Problem 7==<br />
Triangle <math>ABC</math> has side lengths <math>AB = 9</math>, <math>BC =</math> <math>5\sqrt{3}</math>, and <math>AC = 12</math>. Points <math>A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B</math> are on segment <math>\overline{AB}</math> with <math>P_{k}</math> between <math>P_{k-1}</math> and <math>P_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>, and points <math>A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C</math> are on segment <math>\overline{AC}</math> with <math>Q_{k}</math> between <math>Q_{k-1}</math> and <math>Q_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>. Furthermore, each segment <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2449</math>, is parallel to <math>\overline{BC}</math>. The segments cut the triangle into <math>2450</math> regions, consisting of <math>2449</math> trapezoids and <math>1</math> triangle. Each of the <math>2450</math> regions has the same area. Find the number of segments <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2450</math>, that have rational length.<br />
<br />
== Solution 1 ==<br />
For each <math>k</math> between <math>2</math> and <math>2450</math>, the area of the trapezoid with <math>\overline{P_kQ_k}</math> as its bottom base is the difference between the areas of two triangles, both similar to <math>\triangle{ABC}</math>. Let <math>d_k</math> be the length of segment <math>\overline{P_kQ_k}</math>. The area of the trapezoid with bases <math>\overline{P_{k-1}Q_{k-1}}</math> and <math>P_kQ_k</math> is <math>\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}</math> times the area of <math>\triangle{ABC}</math>. (This logic also applies to the topmost triangle if we notice that <math>d_0 = 0</math>.) However, we also know that the area of each shape is <math>\frac{1}{2450}</math> times the area of <math>\triangle{ABC}</math>. We then have <math>\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}</math>. Simplifying, <math>d_k^2-d_{k-1}^2 = \frac{3}{98}</math>. However, we know that <math>d_0^2 = 0</math>, so <math>d_1^2 = \frac{3}{98}</math>, and in general, <math>d_k^2 = \frac{3k}{98}</math> and <math>d_k = \frac{\sqrt{\frac{3k}{2}}}{7}</math>. The smallest <math>k</math> that gives a rational <math>d_k</math> is <math>6</math>, so <math>d_k</math> is rational if and only if <math>k = 6n^2</math> for some integer <math>n</math>.The largest <math>n</math> such that <math>6n^2</math> is less than <math>2450</math> is <math>20</math>, so <math>k</math> has <math>\boxed{020}</math> possible values.<br />
<br />
Solution by zeroman<br />
<br />
==Solution 2==<br />
We have that there are <math>2449</math> trapezoids and <math>1</math> triangle of equal area, with that one triangle being <math>AP_1Q_1</math>. Notice, if we "stack" the trapezoids on top of <math>\bigtriangleup AP_1Q_1</math> the way they already are, we'd create a similar triangle, all of which are similar to <math>\bigtriangleup ABC</math>, and since the trapezoids and <math>\bigtriangleup AP_1Q_1</math> have equal area, each of these similar triangles, <math>AP_kQ_k</math> have area <math>\frac{k}{2450}\left[ ABC\right]</math>, and so <math>\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}</math>. We want the ratio of the side lengths <math>P_kQ_k:BC</math>. Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or <cmath>\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}</cmath> <cmath>\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}</cmath> <cmath>\implies k=6n^2<2450 </cmath> <cmath>\implies 0<n\leq 20</cmath> so there are <math>\boxed{020}</math> solutions. <br />
<br />
Solution by ktong<br />
Beautified by jdong2006<br />
<br />
==Solution 3==<br />
<br />
Let <math>T_1</math> stand for <math>AP_1Q_1</math>, and <math>T_k = AP_kQ_k</math>. All triangles <math>T</math> are similar by AA. Let the area of <math>T_1</math> be <math>x</math>. The next trapezoid will also have an area of <math>x</math>, as given. Therefore, <math>T_k</math> has an area of <math>kx</math>. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, <math>P_k Q_k=P_1 Q_1\cdot \sqrt{k}</math>, and the same if <math>Q</math> is substituted for <math>P</math> throughout. We want the side <math>P_k Q_k</math> to be rational. Setting up proportions: <cmath>5\sqrt{3} : \sqrt{2450}=35\sqrt{2}</cmath> <cmath>\sqrt{6} : 14</cmath> which shows that <math>P_1 Q_1=\frac{\sqrt{6}}{14}</math>. In order for <math>\sqrt{k} P_1 Q_1</math> to be rational, <math>\sqrt{k}</math> must be some rational multiple of <math>\sqrt{6}</math>. This is achieved at <math>\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 20\sqrt{6}</math>. We end there as <math>21\sqrt{6}=\sqrt{2646}</math>. There are 20 numbers from 1 to 20, so there are <math>\boxed{020}</math> solutions.<br />
<br />
Solution by a1b2<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=II|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_7&diff=1614752015 AIME I Problems/Problem 72021-09-03T02:59:56Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>==Problem ==<br />
In the diagram below, <math>ABCD</math> is a square. Point <math>E</math> is the midpoint of <math>\overline{AD}</math>. Points <math>F</math> and <math>G</math> lie on <math>\overline{CE}</math>, and <math>H</math> and <math>J</math> lie on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, so that <math>FGHJ</math> is a square. Points <math>K</math> and <math>L</math> lie on <math>\overline{GH}</math>, and <math>M</math> and <math>N</math> lie on <math>\overline{AD}</math> and <math>\overline{AB}</math>, respectively, so that <math>KLMN</math> is a square. The area of <math>KLMN</math> is 99. Find the area of <math>FGHJ</math>.<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,J,K,L,M,N;<br />
B=(0,0);<br />
real m=7*sqrt(55)/5;<br />
J=(m,0);<br />
C=(7*m/2,0);<br />
A=(0,7*m/2);<br />
D=(7*m/2,7*m/2);<br />
E=(A+D)/2;<br />
H=(0,2m);<br />
N=(0,2m+3*sqrt(55)/2);<br />
G=foot(H,E,C);<br />
F=foot(J,E,C);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E);<br />
draw(G--H--J--F);<br />
pair X=foot(N,E,C);<br />
M=extension(N,X,A,D);<br />
K=foot(N,H,G);<br />
L=foot(M,H,G);<br />
draw(K--N--M--L);<br />
label("$A$",A,NW);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",E,dir(90));<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$H$",H,W);<br />
label("$J$",J,S);<br />
label("$K$",K,SE);<br />
label("$L$",L,SE);<br />
label("$M$",M,dir(90));<br />
label("$N$",N,dir(180)); </asy><br />
<br />
==Solution 1==<br />
<br />
Let us find the proportion of the side length of <math>KLMN</math> and <math>FJGH</math>. Let the side length of <math>KLMN=y</math> and the side length of <math>FJGH=x</math>.<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;<br />
B=(0,0);<br />
real m=7*sqrt(55)/5;<br />
J=(m,0);<br />
C=(7*m/2,0);<br />
A=(0,7*m/2);<br />
D=(7*m/2,7*m/2);<br />
E=(A+D)/2;<br />
H=(0,2m);<br />
N=(0,2m+3*sqrt(55)/2);<br />
G=foot(H,E,C);<br />
F=foot(J,E,C);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E);<br />
draw(G--H--J--F);<br />
pair X=foot(N,E,C);<br />
M=extension(N,X,A,D);<br />
K=foot(N,H,G);<br />
L=foot(M,H,G);<br />
draw(K--N--M--L);<br />
label("$A$",A,NW);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",E,dir(90));<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$H$",H,W);<br />
label("$J$",J,S);<br />
label("$K$",K,SE);<br />
label("$L$",L,SE);<br />
label("$M$",M,dir(90));<br />
label("$N$",N,dir(180)); </asy><br />
<br />
<br />
Now, examine <math>BC</math>. We know <math>BC=BJ+JC</math>, and triangles <math>\Delta BHJ</math> and <math>\Delta JFC</math> are similar to <math>\Delta EDC</math> since they are <math>1-2-\sqrt{5}</math> triangles. Thus, we can rewrite <math>BC</math> in terms of the side length of <math>FJGH</math>. <br />
<cmath>BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}</cmath><br />
<br />
Now examine <math>AB</math>. We can express this length in terms of <math>x,y</math> since <math>AB=AN+NH+HB</math>. By using similar triangles as in the first part, we have <br />
<cmath>AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x</cmath><br />
<cmath>AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x</cmath><br />
<br />
Now, it is trivial to see that <math>[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.\Box</math><br />
<br />
==Solution 2==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;<br />
B=(0,0);<br />
real m=7*sqrt(55)/5;<br />
J=(m,0);<br />
C=(7*m/2,0);<br />
A=(0,7*m/2);<br />
D=(7*m/2,7*m/2);<br />
E=(A+D)/2;<br />
H=(0,2m);<br />
N=(0,2m+3*sqrt(55)/2);<br />
G=foot(H,E,C);<br />
F=foot(J,E,C);<br />
draw(A--B--C--D--cycle);<br />
draw(C--E);<br />
draw(G--H--J--F);<br />
pair X=foot(N,E,C);<br />
M=extension(N,X,A,D);<br />
K=foot(N,H,G);<br />
L=foot(M,H,G);<br />
draw(K--N--M--L);<br />
P=foot(E,M,L);<br />
draw(P--E);<br />
label("$A$",A,NW);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",E,dir(90));<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$H$",H,W);<br />
label("$J$",J,S);<br />
label("$K$",K,SE);<br />
label("$L$",L,SE);<br />
label("$M$",M,dir(90));<br />
label("$N$",N,dir(180));<br />
label("$P$",P,dir(235)); </asy><br />
<br />
We begin by denoting the length <math>ED</math> <math>a</math>, giving us <math>DC = 2a</math> and <math>EC = a\sqrt5</math>. Since angles <math>\angle DCE</math> and <math>\angle FCJ</math> are complementary, we have that <math>\triangle CDE \sim \triangle JFC</math> (and similarly the rest of the triangles are <math>1-2-\sqrt5</math> triangles). We let the sidelength of <math>FGHJ</math> be <math>b</math>, giving us: <br />
<br />
<cmath>JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}</cmath><br />
and<br />
<cmath>BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}</cmath><br />
<br />
Since <math>BC = CJ + BJ</math>, <br />
<br />
<cmath>2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}</cmath><br />
<br />
Solving for <math>b</math> in terms of <math>a</math> yields <cmath>b = \frac{4a\sqrt5}{7}</cmath><br />
<br />
We now use the given that <math>[KLMN] = 99</math>, implying that <math>KL = LM = MN = NK = 3\sqrt{11}</math>. We also draw the perpendicular from <math>E</math> to <math>ML</math> and label the point of intersection <math>P</math> as in the diagram at the top<br />
<br />
This gives that <cmath>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</cmath><br />
and <cmath>ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}</cmath><br />
<br />
Since <math>AE</math> = <math>AM + ME</math>, we get<br />
<br />
<cmath>2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a</cmath><br />
<br />
<cmath>\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a</cmath><br />
<br />
<cmath>\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a</cmath><br />
<br />
<cmath>\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}</cmath><br />
<br />
<cmath>\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a</cmath><br />
<br />
<cmath>\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}</cmath><br />
<br />
So our final answer is <math>(7\sqrt{11})^2 = \boxed{539}</math>.<br />
<br />
==Solution 3==<br />
This is a relatively quick solution but a fakesolve. We see that with a ruler, <math>KL = \frac{3}{2}</math> cm and <math>HG = \frac{7}{2}</math> cm. Thus if <math>KL</math> corresponds with an area of <math>99</math>, then <math>HG</math> (<math>FGHJ</math>'s area) would correspond with <math>99*(\frac{7}{3})^2 = \boxed{539}</math> - aops5234<br />
<br />
== See also ==<br />
{{AIME box|year=2015|n=I|num-b=6|num-a=8}}<br />
{{MAA Notice}}<br />
[[Category:Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_7&diff=1613462016 AIME II Problems/Problem 72021-09-02T04:10:35Z<p>Jdong2006: /* Solution */</p>
<hr />
<div>==Problem==<br />
Squares <math>ABCD</math> and <math>EFGH</math> have a common center and <math>\overline{AB} || \overline{EF}</math>. The area of <math>ABCD</math> is 2016, and the area of <math>EFGH</math> is a smaller positive integer. Square <math>IJKL</math> is constructed so that each of its vertices lies on a side of <math>ABCD</math> and each vertex of <math>EFGH</math> lies on a side of <math>IJKL</math>. Find the difference between the largest and smallest positive integer values for the area of <math>IJKL</math>.<br />
<br />
==Solution==<br />
Letting <math>AI=a</math> and <math>IB=b</math>, we have <cmath>IJ^{2}=a^{2}+b^{2} \geq 1008</cmath> by [[AM-GM inequality]]. Also, since <math>EFGH||ABCD</math>, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and <math>2</math> adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since <cmath>2016=12^{2} \cdot 14</cmath> we have the maximum area is <cmath>2016 \cdot \dfrac{11}{12} = 1848</cmath> (the areas of the squares from largest to smallest are <math>12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14</math> forming a geometric progression). <br />
<br />
<br />
The minimum area is <math>1008</math> (every square is half the area of the square whose sides its vertices touch), so the desired answer is <cmath>1848-1008=\boxed{840}</cmath><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,K,L;<br />
A=(0,0);<br />
B=(2016,0);<br />
C=(2016,2016);<br />
D=(0,2016);<br />
I=(1008,0);<br />
J=(2016,1008);<br />
K=(1008,2016);<br />
L=(0,1008);<br />
E=(504,504);<br />
F=(1512,504);<br />
G=(1512,1512);<br />
H=(504,1512);<br />
draw(A--B--C--D--A);<br />
draw(I--J--K--L--I);<br />
draw(E--F--G--H--E);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,SW);<br />
label("$F$",F,SE);<br />
label("$G$",G,NE);<br />
label("$H$",H,NW);<br />
label("$I$",I,S);<br />
label("$J$",J,NE);<br />
label("$K$",K,N);<br />
label("$L$",L,NW);<br />
</asy><br />
<br />
== See also ==<br />
{{AIME box|year=2016|n=II|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=User:Firebolt360&diff=161345User:Firebolt3602021-09-02T02:50:01Z<p>Jdong2006: /* User Count */</p>
<hr />
<div><br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">69</font></center><br />
</div><br />
<br />
<p>:p I don't know what else to put here, so bye!!!</p></div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_6&diff=1613442015 AIME I Problems/Problem 62021-09-02T02:46:52Z<p>Jdong2006: /* Solution */</p>
<hr />
<div>==Problem==<br />
Point <math>A,B,C,D,</math> and <math>E</math> are equally spaced on a minor arc of a circle. Points <math>E,F,G,H,I</math> and <math>A</math> are equally spaced on a minor arc of a second circle with center <math>C</math> as shown in the figure below. The angle <math>\angle ABD</math> exceeds <math>\angle AHG</math> by <math>12^\circ</math>. Find the degree measure of <math>\angle BAG</math>.<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,O;<br />
O=(0,0);<br />
C=dir(90);<br />
B=dir(70);<br />
A=dir(50);<br />
D=dir(110);<br />
E=dir(130);<br />
draw(arc(O,1,50,130));<br />
real x=2*sin(20*pi/180);<br />
F=x*dir(228)+C;<br />
G=x*dir(256)+C;<br />
H=x*dir(284)+C;<br />
I=x*dir(312)+C;<br />
draw(arc(C,x,200,340));<br />
label("$A$",A,dir(0));<br />
label("$B$",B,dir(75));<br />
label("$C$",C,dir(90));<br />
label("$D$",D,dir(105));<br />
label("$E$",E,dir(180));<br />
label("$F$",F,dir(225));<br />
label("$G$",G,dir(260));<br />
label("$H$",H,dir(280));<br />
label("$I$",I,dir(315));<br />
</asy><br />
<br />
==The Only Solution==<br />
<br />
Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it. <br />
<br />
Let <math>x</math> be the degree measurement of <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math> in circle <math>O</math> and <math>y</math> be the degree measurement of <math>\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</math> in circle <math>C</math>.<br />
<math>\angle ECA</math> is, therefore, <math>5y</math> by way of circle <math>C</math> and <cmath>\frac{360-4x}{2}=180-2x</cmath> by way of circle <math>O</math>.<br />
<math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> by way of circle <math>C</math>.<br />
<br />
This means that:<br />
<br />
<cmath>180-\frac{3x}{2}=180-\frac{3y}{2}+12</cmath><br />
<br />
which when simplified yields <cmath>\frac{3x}{2}+12=\frac{3y}{2}</cmath> or <cmath>x+8=y</cmath><br />
Since:<br />
<cmath>5y=180-2x</cmath> and <cmath>5x+40=180-2x</cmath><br />
So:<br />
<cmath>7x=140\Longleftrightarrow x=20</cmath> <br />
<cmath>y=28</cmath><br />
<math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to <math>\frac{3x}{2} + y</math>.<br />
Plugging in yields <math>30+28</math>, or <math>\boxed{058}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_6&diff=1613432015 AIME I Problems/Problem 62021-09-02T02:46:30Z<p>Jdong2006: /* Solution */</p>
<hr />
<div>==Problem==<br />
Point <math>A,B,C,D,</math> and <math>E</math> are equally spaced on a minor arc of a circle. Points <math>E,F,G,H,I</math> and <math>A</math> are equally spaced on a minor arc of a second circle with center <math>C</math> as shown in the figure below. The angle <math>\angle ABD</math> exceeds <math>\angle AHG</math> by <math>12^\circ</math>. Find the degree measure of <math>\angle BAG</math>.<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,O;<br />
O=(0,0);<br />
C=dir(90);<br />
B=dir(70);<br />
A=dir(50);<br />
D=dir(110);<br />
E=dir(130);<br />
draw(arc(O,1,50,130));<br />
real x=2*sin(20*pi/180);<br />
F=x*dir(228)+C;<br />
G=x*dir(256)+C;<br />
H=x*dir(284)+C;<br />
I=x*dir(312)+C;<br />
draw(arc(C,x,200,340));<br />
label("$A$",A,dir(0));<br />
label("$B$",B,dir(75));<br />
label("$C$",C,dir(90));<br />
label("$D$",D,dir(105));<br />
label("$E$",E,dir(180));<br />
label("$F$",F,dir(225));<br />
label("$G$",G,dir(260));<br />
label("$H$",H,dir(280));<br />
label("$I$",I,dir(315));<br />
</asy><br />
<br />
==Solution==<br />
<br />
Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it. <br />
<br />
Let <math>x</math> be the degree measurement of <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math> in circle <math>O</math> and <math>y</math> be the degree measurement of <math>\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</math> in circle <math>C</math>.<br />
<math>\angle ECA</math> is, therefore, <math>5y</math> by way of circle <math>C</math> and <cmath>\frac{360-4x}{2}=180-2x</cmath> by way of circle <math>O</math>.<br />
<math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> by way of circle <math>C</math>.<br />
<br />
This means that:<br />
<br />
<cmath>180-\frac{3x}{2}=180-\frac{3y}{2}+12</cmath><br />
<br />
which when simplified yields <cmath>\frac{3x}{2}+12=\frac{3y}{2}</cmath> or <cmath>x+8=y</cmath><br />
Since:<br />
<cmath>5y=180-2x</cmath> and <cmath>5x+40=180-2x</cmath><br />
So:<br />
<cmath>7x=140\Longleftrightarrow x=20</cmath> <br />
<cmath>y=28</cmath><br />
<math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to <math>\frac{3x}{2} + y</math>.<br />
Plugging in yields <math>30+28</math>, or <math>\boxed{058}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_6&diff=1613422015 AIME I Problems/Problem 62021-09-02T02:45:56Z<p>Jdong2006: </p>
<hr />
<div>==Problem==<br />
Point <math>A,B,C,D,</math> and <math>E</math> are equally spaced on a minor arc of a circle. Points <math>E,F,G,H,I</math> and <math>A</math> are equally spaced on a minor arc of a second circle with center <math>C</math> as shown in the figure below. The angle <math>\angle ABD</math> exceeds <math>\angle AHG</math> by <math>12^\circ</math>. Find the degree measure of <math>\angle BAG</math>.<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,O;<br />
O=(0,0);<br />
C=dir(90);<br />
B=dir(70);<br />
A=dir(50);<br />
D=dir(110);<br />
E=dir(130);<br />
draw(arc(O,1,50,130));<br />
real x=2*sin(20*pi/180);<br />
F=x*dir(228)+C;<br />
G=x*dir(256)+C;<br />
H=x*dir(284)+C;<br />
I=x*dir(312)+C;<br />
draw(arc(C,x,200,340));<br />
label("$A$",A,dir(0));<br />
label("$B$",B,dir(75));<br />
label("$C$",C,dir(90));<br />
label("$D$",D,dir(105));<br />
label("$E$",E,dir(180));<br />
label("$F$",F,dir(225));<br />
label("$G$",G,dir(260));<br />
label("$H$",H,dir(280));<br />
label("$I$",I,dir(315));<br />
</asy><br />
<br />
==Solution==<br />
<br />
Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it. <br />
<br />
Let <math>x</math> be the degree measurement of <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math> in circle <math>O</math> and <math>y</math> be the degree measurement of <math>\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</math> in circle <math>C</math>.<br />
<math>\angle ECA</math> is, therefore, <math>5y</math> by way of circle <math>C</math> and <cmath>\frac{360-4x}{2}=180-2x</cmath> by way of circle <math>O</math>.<br />
<math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> by way of circle <math>C</math>.<br />
<br />
This means that:<br />
<br />
<cmath>180-\frac{3x}{2}=180-\frac{3y}{2}+12</cmath><br />
<br />
which when simplified yields <cmath>3x/2+12=3y/2</cmath> or <cmath>x+8=y</cmath><br />
Since:<br />
<cmath>5y=180-2x</cmath> and <cmath>5x+40=180-2x</cmath><br />
So:<br />
<cmath>7x=140\Longleftrightarrow x=20</cmath> <br />
<cmath>y=28</cmath><br />
<math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to <math>\frac{3x}{2} + y</math>.<br />
Plugging in yields <math>30+28</math>, or <math>\boxed{058}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_6&diff=1613412015 AIME I Problems/Problem 62021-09-02T02:44:51Z<p>Jdong2006: /* Solution */</p>
<hr />
<div>==Problem==<br />
Point <math>A,B,C,D,</math> and <math>E</math> are equally spaced on a minor arc of a circle. Points <math>E,F,G,H,I</math> and <math>A</math> are equally spaced on a minor arc of a second circle with center <math>C</math> as shown in the figure below. The angle <math>\angle ABD</math> exceeds <math>\angle AHG</math> by <math>12^\circ</math>. Find the degree measure of <math>\angle BAG</math>.<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,O;<br />
O=(0,0);<br />
C=dir(90);<br />
B=dir(70);<br />
A=dir(50);<br />
D=dir(110);<br />
E=dir(130);<br />
draw(arc(O,1,50,130));<br />
real x=2*sin(20*pi/180);<br />
F=x*dir(228)+C;<br />
G=x*dir(256)+C;<br />
H=x*dir(284)+C;<br />
I=x*dir(312)+C;<br />
draw(arc(C,x,200,340));<br />
label("$A$",A,dir(0));<br />
label("$B$",B,dir(75));<br />
label("$C$",C,dir(90));<br />
label("$D$",D,dir(105));<br />
label("$E$",E,dir(180));<br />
label("$F$",F,dir(225));<br />
label("$G$",G,dir(260));<br />
label("$H$",H,dir(280));<br />
label("$I$",I,dir(315));<br />
</asy><br />
<br />
==Solution==<br />
<br />
Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it. <br />
<br />
Let <math>x</math> be the degree measurement of <cmath>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</cmath> in circle <math>O</math> and <math>y</math> be the degree measurement of <cmath>\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</cmath> in circle <math>C</math>.<br />
<math>\angle ECA</math> is, therefore, <math>5y</math> by way of circle <math>C</math> and <cmath>\frac{360-4x}{2}=180-2x</cmath> by way of circle <math>O</math>.<br />
<math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> by way of circle <math>C</math>.<br />
<br />
This means that:<br />
<br />
<cmath>180-\frac{3x}{2}=180-\frac{3y}{2}+12</cmath><br />
<br />
which when simplified yields <cmath>3x/2+12=3y/2</cmath> or <cmath>x+8=y</cmath><br />
Since:<br />
<cmath>5y=180-2x</cmath> and <cmath>5x+40=180-2x</cmath><br />
So:<br />
<cmath>7x=140\Longleftrightarrow x=20</cmath> <br />
<cmath>y=28</cmath><br />
<math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to <math>\frac{3x}{2} + y</math>.<br />
Plugging in yields <math>30+28</math>, or <math>\boxed{058}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1613402018 AIME I Problems/Problem 42021-09-01T23:48:30Z<p>Jdong2006: /* Solution 1 */</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 0==<br />
By the Law of Cosines, we have: <cmath>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</cmath> Let <math>M</math> be midpoint of <math>AE</math>, then <cmath>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</cmath> So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
== Solution 12 (Double Angle Identity)==<br />
<br />
We let <math>AD=x</math>. Then, angle <math>A</math> is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2\sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DG=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7000}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} \implies \fbox{289}.</math><br />
<br />
~john0512<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_4&diff=1612962015 AIME I Problems/Problem 42021-09-01T03:51:34Z<p>Jdong2006: /* Diagram */</p>
<hr />
<div>==Problem==<br />
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.<br />
<br />
==Diagram==<br />
<asy><br />
pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3));<br />
draw(A--B--D--cycle);<br />
draw(B--C--EE--cycle);<br />
draw(A--EE);<br />
draw(C--D);<br />
draw(B--M--NN--cycle);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(EE);<br />
dot(M);<br />
dot(NN);<br />
label("A", A, SW);<br />
label("B", B, S);<br />
label("C", C, SE);<br />
label("D", D, N);<br />
label("E", EE, N);<br />
label("M", M, NW);<br />
label("N", NN, NE);<br />
</asy><br />
<br />
Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FFFFFF">talk</font>]]) 18:52, 15 February 2021 (EST)<br />
<br />
==Solution 1 (fastest)==<br />
Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula <math>\textbf{OR}</math> by Shoelace Theorem, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>.<br />
<br />
==Solution 2==<br />
Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS.<br />
<br />
From this, it is clear that a <math>60^{\circ}</math> rotation about <math>B</math> will map <math>\triangle ABE</math> to <math>\triangle DBC</math>.<br />
This rotation also maps <math>M</math> to <math>N</math>. Thus, <math>BM=BN</math> and <math>\angle MBN=60^{\circ}</math>. Thus, <math>\triangle BMN</math> is equilateral.<br />
<br />
Using the Law of Cosines on <math>\triangle ABE</math>,<br />
<cmath>AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)</cmath><br />
<cmath>AE = 4\sqrt{21}</cmath><br />
Thus, <math>AM=ME=2\sqrt{21}</math>.<br />
<br />
Using Stewart's Theorem on <math>\triangle ABE</math>,<br />
<cmath>AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME</cmath><br />
<cmath>BM = 2\sqrt{13}</cmath><br />
<br />
Calculating the area of <math>\triangle BMN</math>,<br />
<cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2</cmath><br />
<cmath>[BMN] = 13\sqrt{3}</cmath><br />
Thus, <math>x=13\sqrt{3}</math>, so <math>x^2 = 507</math>. Our final answer is <math>\boxed{507}</math>.<br />
<br />
Admittedly, this is much more tedious than the coordinate solutions.<br />
<br />
I also noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral:<br />
<br />
One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>.<br />
<br />
The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. The weights are the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_4&diff=1612952015 AIME I Problems/Problem 42021-09-01T03:50:53Z<p>Jdong2006: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.<br />
<br />
==Diagram==<br />
<asy><br />
pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3));<br />
draw(A--B--D--cycle);<br />
draw(B--C--EE--cycle);<br />
draw(A--EE);<br />
draw(C--D);<br />
draw(B--M--NN--cycle);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(EE);<br />
dot(M);<br />
dot(NN);<br />
label("A", A, SW);<br />
label("B", B, S);<br />
label("C", C, SE);<br />
label("D", D, N);<br />
label("E", EE, N);<br />
label("M", M, NW);<br />
label("N", NN, NE);<br />
</asy><br />
<br />
Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 18:52, 15 February 2021 (EST)<br />
<br />
==Solution 1 (fastest)==<br />
Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula <math>\textbf{OR}</math> by Shoelace Theorem, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>.<br />
<br />
==Solution 2==<br />
Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS.<br />
<br />
From this, it is clear that a <math>60^{\circ}</math> rotation about <math>B</math> will map <math>\triangle ABE</math> to <math>\triangle DBC</math>.<br />
This rotation also maps <math>M</math> to <math>N</math>. Thus, <math>BM=BN</math> and <math>\angle MBN=60^{\circ}</math>. Thus, <math>\triangle BMN</math> is equilateral.<br />
<br />
Using the Law of Cosines on <math>\triangle ABE</math>,<br />
<cmath>AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)</cmath><br />
<cmath>AE = 4\sqrt{21}</cmath><br />
Thus, <math>AM=ME=2\sqrt{21}</math>.<br />
<br />
Using Stewart's Theorem on <math>\triangle ABE</math>,<br />
<cmath>AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME</cmath><br />
<cmath>BM = 2\sqrt{13}</cmath><br />
<br />
Calculating the area of <math>\triangle BMN</math>,<br />
<cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2</cmath><br />
<cmath>[BMN] = 13\sqrt{3}</cmath><br />
Thus, <math>x=13\sqrt{3}</math>, so <math>x^2 = 507</math>. Our final answer is <math>\boxed{507}</math>.<br />
<br />
Admittedly, this is much more tedious than the coordinate solutions.<br />
<br />
I also noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral:<br />
<br />
One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>.<br />
<br />
The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. The weights are the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_4&diff=1612932015 AIME I Problems/Problem 42021-09-01T03:48:30Z<p>Jdong2006: /* Solution */</p>
<hr />
<div>==Problem==<br />
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.<br />
<br />
==Diagram==<br />
<asy><br />
pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3));<br />
draw(A--B--D--cycle);<br />
draw(B--C--EE--cycle);<br />
draw(A--EE);<br />
draw(C--D);<br />
draw(B--M--NN--cycle);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(EE);<br />
dot(M);<br />
dot(NN);<br />
label("A", A, SW);<br />
label("B", B, S);<br />
label("C", C, SE);<br />
label("D", D, N);<br />
label("E", EE, N);<br />
label("M", M, NW);<br />
label("N", NN, NE);<br />
</asy><br />
<br />
Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 18:52, 15 February 2021 (EST)<br />
<br />
==Solution 1==<br />
Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula <math>\textbf{OR}</math> by Shoelace Theorem, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>.<br />
<br />
==Solution 2==<br />
Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS.<br />
<br />
From this, it is clear that a <math>60^{\circ}</math> rotation about <math>B</math> will map <math>\triangle ABE</math> to <math>\triangle DBC</math>.<br />
This rotation also maps <math>M</math> to <math>N</math>. Thus, <math>BM=BN</math> and <math>\angle MBN=60^{\circ}</math>. Thus, <math>\triangle BMN</math> is equilateral.<br />
<br />
Using the Law of Cosines on <math>\triangle ABE</math>,<br />
<cmath>AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)</cmath><br />
<cmath>AE = 4\sqrt{21}</cmath><br />
Thus, <math>AM=ME=2\sqrt{21}</math>.<br />
<br />
Using Stewart's Theorem on <math>\triangle ABE</math>,<br />
<cmath>AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME</cmath><br />
<cmath>BM = 2\sqrt{13}</cmath><br />
<br />
Calculating the area of <math>\triangle BMN</math>,<br />
<cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2</cmath><br />
<cmath>[BMN] = 13\sqrt{3}</cmath><br />
Thus, <math>x=13\sqrt{3}</math>, so <math>x^2 = 507</math>. Our final answer is <math>\boxed{507}</math>.<br />
<br />
Admittedly, this is much more tedious than the coordinate solutions.<br />
<br />
I also noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral:<br />
<br />
One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>.<br />
<br />
The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. The weights are the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_4&diff=1612922015 AIME I Problems/Problem 42021-09-01T03:47:49Z<p>Jdong2006: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.<br />
<br />
==Diagram==<br />
<asy><br />
pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3));<br />
draw(A--B--D--cycle);<br />
draw(B--C--EE--cycle);<br />
draw(A--EE);<br />
draw(C--D);<br />
draw(B--M--NN--cycle);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(EE);<br />
dot(M);<br />
dot(NN);<br />
label("A", A, SW);<br />
label("B", B, S);<br />
label("C", C, SE);<br />
label("D", D, N);<br />
label("E", EE, N);<br />
label("M", M, NW);<br />
label("N", NN, NE);<br />
</asy><br />
<br />
Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 18:52, 15 February 2021 (EST)<br />
<br />
==Solution==<br />
Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>.<br />
==Solution 2==<br />
Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS.<br />
<br />
From this, it is clear that a <math>60^{\circ}</math> rotation about <math>B</math> will map <math>\triangle ABE</math> to <math>\triangle DBC</math>.<br />
This rotation also maps <math>M</math> to <math>N</math>. Thus, <math>BM=BN</math> and <math>\angle MBN=60^{\circ}</math>. Thus, <math>\triangle BMN</math> is equilateral.<br />
<br />
Using the Law of Cosines on <math>\triangle ABE</math>,<br />
<cmath>AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)</cmath><br />
<cmath>AE = 4\sqrt{21}</cmath><br />
Thus, <math>AM=ME=2\sqrt{21}</math>.<br />
<br />
Using Stewart's Theorem on <math>\triangle ABE</math>,<br />
<cmath>AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME</cmath><br />
<cmath>BM = 2\sqrt{13}</cmath><br />
<br />
Calculating the area of <math>\triangle BMN</math>,<br />
<cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2</cmath><br />
<cmath>[BMN] = 13\sqrt{3}</cmath><br />
Thus, <math>x=13\sqrt{3}</math>, so <math>x^2 = 507</math>. Our final answer is <math>\boxed{507}</math>.<br />
<br />
Admittedly, this is much more tedious than the coordinate solutions.<br />
<br />
I also noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral:<br />
<br />
One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>.<br />
<br />
The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. The weights are the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_4&diff=1612912015 AIME I Problems/Problem 42021-09-01T03:47:32Z<p>Jdong2006: /* Solution 2 */ unecessary solution</p>
<hr />
<div>==Problem==<br />
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.<br />
<br />
==Diagram==<br />
<asy><br />
pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3));<br />
draw(A--B--D--cycle);<br />
draw(B--C--EE--cycle);<br />
draw(A--EE);<br />
draw(C--D);<br />
draw(B--M--NN--cycle);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(EE);<br />
dot(M);<br />
dot(NN);<br />
label("A", A, SW);<br />
label("B", B, S);<br />
label("C", C, SE);<br />
label("D", D, N);<br />
label("E", EE, N);<br />
label("M", M, NW);<br />
label("N", NN, NE);<br />
</asy><br />
<br />
Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 18:52, 15 February 2021 (EST)<br />
<br />
==Solution==<br />
Let point <math>A</math> be at <math>(0,0)</math>. Then, <math>B</math> is at <math>(16,0)</math>, and <math>C</math> is at <math>(20,0)</math>. Due to symmetry, it is allowed to assume <math>D</math> and <math>E</math> are in quadrant 1. By equilateral triangle calculations, Point <math>D</math> is at <math>(8,8\sqrt{3})</math>, and Point <math>E</math> is at <math>(18,2\sqrt{3})</math>. By Midpoint Formula, <math>M</math> is at <math>(9,\sqrt{3})</math>, and <math>N</math> is at <math>(14,4\sqrt{3})</math>. The distance formula shows that <math>BM=BN=MN=2\sqrt{13}</math>. Therefore, by equilateral triangle area formula, <math>x=13\sqrt{3}</math>, so <math>x^2</math> is <math>\boxed{507}</math>.<br />
==Solution 3==<br />
Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS.<br />
<br />
From this, it is clear that a <math>60^{\circ}</math> rotation about <math>B</math> will map <math>\triangle ABE</math> to <math>\triangle DBC</math>.<br />
This rotation also maps <math>M</math> to <math>N</math>. Thus, <math>BM=BN</math> and <math>\angle MBN=60^{\circ}</math>. Thus, <math>\triangle BMN</math> is equilateral.<br />
<br />
Using the Law of Cosines on <math>\triangle ABE</math>,<br />
<cmath>AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)</cmath><br />
<cmath>AE = 4\sqrt{21}</cmath><br />
Thus, <math>AM=ME=2\sqrt{21}</math>.<br />
<br />
Using Stewart's Theorem on <math>\triangle ABE</math>,<br />
<cmath>AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME</cmath><br />
<cmath>BM = 2\sqrt{13}</cmath><br />
<br />
Calculating the area of <math>\triangle BMN</math>,<br />
<cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2</cmath><br />
<cmath>[BMN] = 13\sqrt{3}</cmath><br />
Thus, <math>x=13\sqrt{3}</math>, so <math>x^2 = 507</math>. Our final answer is <math>\boxed{507}</math>.<br />
<br />
Admittedly, this is much more tedious than the coordinate solutions.<br />
<br />
I also noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral:<br />
<br />
One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>.<br />
<br />
The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. The weights are the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_14&diff=1612822013 AIME I Problems/Problem 142021-09-01T01:57:35Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
For <math>\pi \le \theta < 2\pi</math>, let<br />
<br />
<cmath>\begin{align*}<br />
P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
and<br />
<br />
<cmath>\begin{align*}<br />
Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
Noticing the <math>\sin</math> and <math>\cos</math> in both <math>P</math> and <math>Q,</math> we think of the angle addition identities: <br />
<br />
<cmath>\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b + \sin a \sin b</cmath><br />
<br />
With this in mind, we multiply <math>P</math> by <math>\sin \theta</math> and <math>Q</math> by <math>\cos \theta</math> to try and use some angle addition identities. Indeed, we get <br />
<cmath>\begin{align*}<br />
P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\<br />
&= \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots \\<br />
&= \cos \theta - \dfrac{1}{2}P<br />
\end{align*}</cmath><br />
after adding term-by-term. Similar term-by-term adding yields <cmath>P \cos \theta + Q \sin \theta = -2(Q - 1).</cmath> <br />
This is a system of equations; rearrange and rewrite to get <cmath>P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta</cmath> and <cmath>P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.</cmath> Subtract the two and rearrange to get <cmath>\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.</cmath> Then, square both sides and use Pythagorean Identity to get a quadratic in <math>\sin \theta.</math> Factor that quadratic and solve for <math>\sin \theta = -17/19, 1/3.</math> The answer format tells us it's the negative solution, and our desired answer is <math>17 + 19 = \boxed{036}.</math><br />
<br />
==Solution 2==<br />
<br />
Use sum to product formulas to rewrite <math>P</math> and <math>Q</math><br />
<br />
<br />
<cmath>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </cmath><br />
<br />
Therefore, <cmath>P \sin \theta - Q \cos \theta = -2P</cmath><br />
<br />
Using <cmath>\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P</cmath><br />
<br />
Plug in to the previous equation and cancel out the "P" terms to get: <cmath>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</cmath><br />
<br />
Then use the pythagorean identity to solve for <math>\sin\theta</math>, <cmath>\sin\theta = -\frac{17}{19} \implies \boxed{036}</cmath><br />
<br />
==Solution 3==<br />
<br />
Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath><br />
<br />
Thus, the following identities follow immediately:<br />
<cmath>ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)</cmath><br />
<cmath>i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)</cmath><br />
<cmath>i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)</cmath><br />
<br />
Consider, now, the sum <math>Q+iP</math>. It follows fairly immediately that:<br />
<br />
<cmath>Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}</cmath><br />
<cmath>Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}</cmath><br />
<br />
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:<br />
<br />
<cmath>Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)</cmath><br />
<cmath>Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}</cmath><br />
<br />
Comparing real and imaginary parts, we find:<br />
<cmath>\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}</cmath><br />
<br />
Squaring this equation and letting <math>\sin^2(\theta)=x</math>:<br />
<br />
<math>\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}</math><br />
<br />
Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>.<br />
<br />
<math>017+019=\boxed{036}</math><br />
<br />
==Solution 4==<br />
A bit similar to Solution 3. We use <math>\phi = \theta+90^\circ</math> because the progression cycles in <math>P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)</math>. So we could rewrite that as <math>P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)</math>.<br />
<br />
Similarly, <math>Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)</math>.<br />
<br />
Setting complex <math>z=q_1+p_1i</math>, we get <math>z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)</math><br />
<br />
<math>(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}</math>.<br />
<br />
The important part is the ratio of the imaginary part <math>i</math> to the real part. To cancel out the imaginary part from the denominator, we must add <math>0.5i\sin\phi</math> to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would cancel out.<br />
<br />
<math>\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}</math>.<br />
<br />
Setting <math>\sin\theta=y</math>, we obtain<br />
<cmath>\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}</cmath><br />
<cmath>7\sqrt{1-y^2}=2\sqrt{2}(2+y)</cmath><br />
<cmath>49-49y^2=8y^2+32y+32</cmath><br />
<cmath>57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}</cmath><br />
<cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>.<br />
<br />
Since <math>y<0</math> because <math>\pi<\theta<2\pi</math>, <math>y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}</math>. Adding up, <math>17+19=\boxed{036}</math>.<br />
<br />
==Solution 5 (lots of room for sillies, I wouldn't recommend it)==<br />
<br />
We notice <math>\sin\theta=\frac{-i}{2}(e^{i\theta}-e^{-i\theta})</math> and <math>\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})</math><br />
<br />
With these, we just quickly find the sum of the infinite geometric series' in <math>P</math> and <math>Q</math>. <math>P</math> has 2 parts, the <math>\cos</math> and the <math>\sin</math> parts. The <math>\cos</math> part is: <math>\frac12\cos\theta-\frac18\cos3\theta+\cdots</math>, which can be turned into: <math>\frac14(e^{i\theta}(1-\frac{e^{i2\theta}}{4}+\cdots)+e^{-i\theta}(1-\frac{e^{-i2\theta}}{4}+\cdots))</math>, which is <math>\frac{1}{4}(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}})</math>. This turns into <math>\frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
Following the same process as above, we find that the <math>\sin</math> part of <math>P</math> is <math>\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, the <math>\cos</math> part of <math>Q</math> is <math>\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, and finally, the <math>\sin</math> part of <math>Q</math> is <math>\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
We convert all 4 of these equations into trig, and we end up getting <math>\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}</math>, we divide by <math>2</math> on both numerator and denominator, and we get <math>\frac{2\sqrt{2}}{7}=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}</math>. We use some trig identities and we get <math>\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}=\frac{2\sqrt2}{7}</math>. We factor the denominator into <math>(5-4\sin\theta)(2+\sin\theta)</math>. We cancel out <math>5-4\sin\theta</math> on both numerator and denominator to get <math>\frac{\cos\theta}{2+\sin\theta}=\frac{2\sqrt2}{7}</math>. We set <math>\sin\theta</math> as <math>x</math>, and we just solve a quadratic in terms of <math>x</math>, <math>\frac{1-x^2}{x^2+4x+4}=\frac{8}{49}</math>, cross multiply and simplify, and we get <math>57x^2+32x-17=0</math>. We can actually factor this to get <math>(19x+17)(3x-1)=0</math>, which yields the 2 solutions <math>x=-\frac{17}{19}, x=\frac{1}{3}</math>. Since <math>\pi\leq\theta<2\pi</math>, the latter solution is deemed invalid, and we are left with <math>\sin\theta=-\frac{17}{19}</math>. Our final answer is <math>17+19=\boxed{036}</math>.<br />
<br />
~ASAB<br />
<br />
<cmath>\textbf{omg this was definitely the hardest problem on the AIME I}</cmath><br />
<br />
==Solution 6==<br />
Follow solution 3, up to the point of using the geometric series formula<br />
<cmath>Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}</cmath><br />
<br />
Moving everything to the other side, and considering only the imaginary part, we get<br />
<cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath><br />
<br />
We can then write <math>P = 2 \sqrt{2} k</math>, and <math>Q = 7k</math>, (<math>k \neq 0</math>). Thus, we can substitute and divide out by k.<br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)</cmath><br />
<cmath>8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta</cmath><br />
<cmath>\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0</cmath><br />
<cmath>57\sin^{2}\theta+32\sin\theta-17 = 0</cmath><br />
<cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath><br />
<br />
Since <math>\pi \le \theta < 2\pi</math>, we get <math>\sin \theta < 0</math>, and thus, <math>\sin\theta = \frac{-19}{17} \implies \boxed{036}</math><br />
<br />
-Alexlikemath<br />
== See also ==<br />
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_14&diff=1612812013 AIME I Problems/Problem 142021-09-01T01:57:09Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
For <math>\pi \le \theta < 2\pi</math>, let<br />
<br />
<cmath>\begin{align*}<br />
P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
and<br />
<br />
<cmath>\begin{align*}<br />
Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
Noticing the <math>\sin</math> and <math>\cos</math> in both <math>P</math> and <math>Q,</math> we think of the angle addition identities: <br />
<br />
<cmath>\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b + \sin a \sin b</cmath><br />
<br />
With this in mind, we multiply <math>P</math> by <math>\sin \theta</math> and <math>Q</math> by <math>\cos \theta</math> to try and use some angle addition identities. Indeed, we get <br />
<cmath>\begin{align*}<br />
P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\<br />
&= \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots \\<br />
&= \cos \theta - \dfrac{1}{2}P<br />
\end{align*}</cmath><br />
after adding term-by-term. Similar term-by-term adding yields <cmath>P \cos \theta + Q \sin \theta = -2(Q - 1).</cmath> <br />
This is a system of equations; rearrange and rewrite to get <cmath>P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta</cmath> and <cmath>P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.</cmath> Subtract the two and rearrange to get <cmath>\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.</cmath> Then, square both sides and use Pythagorean Identity to get a quadratic in <math>\sin \theta.</math> Factor that quadratic and solve for <math>\sin \theta = -17/19, 1/3.</math> The answer format tells us it's the negative solution, and our desired answer is <math>17 + 19 = \boxed{036}.</math><br />
<br />
==Solution 2==<br />
<br />
Use sum to product formulas to rewrite <math>P</math> and <math>Q</math><br />
<br />
<br />
<cmath>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </cmath><br />
<br />
Therefore, <cmath>P \sin \theta - Q \cos \theta = -2P</cmath><br />
<br />
Using <cmath>\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P</cmath><br />
<br />
Plug in to the previous equation and cancel out the "P" terms to get: <cmath>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</cmath><br />
<br />
Then use the pythagorean identity to solve for <math></math>\sin\theta<math>, </math>\sin\theta = -\frac{17}{19} \implies \boxed{036}<math></math><br />
<br />
==Solution 3==<br />
<br />
Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath><br />
<br />
Thus, the following identities follow immediately:<br />
<cmath>ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)</cmath><br />
<cmath>i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)</cmath><br />
<cmath>i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)</cmath><br />
<br />
Consider, now, the sum <math>Q+iP</math>. It follows fairly immediately that:<br />
<br />
<cmath>Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}</cmath><br />
<cmath>Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}</cmath><br />
<br />
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:<br />
<br />
<cmath>Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)</cmath><br />
<cmath>Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}</cmath><br />
<br />
Comparing real and imaginary parts, we find:<br />
<cmath>\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}</cmath><br />
<br />
Squaring this equation and letting <math>\sin^2(\theta)=x</math>:<br />
<br />
<math>\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}</math><br />
<br />
Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>.<br />
<br />
<math>017+019=\boxed{036}</math><br />
<br />
==Solution 4==<br />
A bit similar to Solution 3. We use <math>\phi = \theta+90^\circ</math> because the progression cycles in <math>P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)</math>. So we could rewrite that as <math>P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)</math>.<br />
<br />
Similarly, <math>Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)</math>.<br />
<br />
Setting complex <math>z=q_1+p_1i</math>, we get <math>z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)</math><br />
<br />
<math>(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}</math>.<br />
<br />
The important part is the ratio of the imaginary part <math>i</math> to the real part. To cancel out the imaginary part from the denominator, we must add <math>0.5i\sin\phi</math> to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would cancel out.<br />
<br />
<math>\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}</math>.<br />
<br />
Setting <math>\sin\theta=y</math>, we obtain<br />
<cmath>\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}</cmath><br />
<cmath>7\sqrt{1-y^2}=2\sqrt{2}(2+y)</cmath><br />
<cmath>49-49y^2=8y^2+32y+32</cmath><br />
<cmath>57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}</cmath><br />
<cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>.<br />
<br />
Since <math>y<0</math> because <math>\pi<\theta<2\pi</math>, <math>y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}</math>. Adding up, <math>17+19=\boxed{036}</math>.<br />
<br />
==Solution 5 (lots of room for sillies, I wouldn't recommend it)==<br />
<br />
We notice <math>\sin\theta=\frac{-i}{2}(e^{i\theta}-e^{-i\theta})</math> and <math>\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})</math><br />
<br />
With these, we just quickly find the sum of the infinite geometric series' in <math>P</math> and <math>Q</math>. <math>P</math> has 2 parts, the <math>\cos</math> and the <math>\sin</math> parts. The <math>\cos</math> part is: <math>\frac12\cos\theta-\frac18\cos3\theta+\cdots</math>, which can be turned into: <math>\frac14(e^{i\theta}(1-\frac{e^{i2\theta}}{4}+\cdots)+e^{-i\theta}(1-\frac{e^{-i2\theta}}{4}+\cdots))</math>, which is <math>\frac{1}{4}(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}})</math>. This turns into <math>\frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
Following the same process as above, we find that the <math>\sin</math> part of <math>P</math> is <math>\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, the <math>\cos</math> part of <math>Q</math> is <math>\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, and finally, the <math>\sin</math> part of <math>Q</math> is <math>\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
We convert all 4 of these equations into trig, and we end up getting <math>\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}</math>, we divide by <math>2</math> on both numerator and denominator, and we get <math>\frac{2\sqrt{2}}{7}=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}</math>. We use some trig identities and we get <math>\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}=\frac{2\sqrt2}{7}</math>. We factor the denominator into <math>(5-4\sin\theta)(2+\sin\theta)</math>. We cancel out <math>5-4\sin\theta</math> on both numerator and denominator to get <math>\frac{\cos\theta}{2+\sin\theta}=\frac{2\sqrt2}{7}</math>. We set <math>\sin\theta</math> as <math>x</math>, and we just solve a quadratic in terms of <math>x</math>, <math>\frac{1-x^2}{x^2+4x+4}=\frac{8}{49}</math>, cross multiply and simplify, and we get <math>57x^2+32x-17=0</math>. We can actually factor this to get <math>(19x+17)(3x-1)=0</math>, which yields the 2 solutions <math>x=-\frac{17}{19}, x=\frac{1}{3}</math>. Since <math>\pi\leq\theta<2\pi</math>, the latter solution is deemed invalid, and we are left with <math>\sin\theta=-\frac{17}{19}</math>. Our final answer is <math>17+19=\boxed{036}</math>.<br />
<br />
~ASAB<br />
<br />
<cmath>\textbf{omg this was definitely the hardest problem on the AIME I}</cmath><br />
<br />
==Solution 6==<br />
Follow solution 3, up to the point of using the geometric series formula<br />
<cmath>Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}</cmath><br />
<br />
Moving everything to the other side, and considering only the imaginary part, we get<br />
<cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath><br />
<br />
We can then write <math>P = 2 \sqrt{2} k</math>, and <math>Q = 7k</math>, (<math>k \neq 0</math>). Thus, we can substitute and divide out by k.<br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)</cmath><br />
<cmath>8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta</cmath><br />
<cmath>\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0</cmath><br />
<cmath>57\sin^{2}\theta+32\sin\theta-17 = 0</cmath><br />
<cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath><br />
<br />
Since <math>\pi \le \theta < 2\pi</math>, we get <math>\sin \theta < 0</math>, and thus, <math>\sin\theta = \frac{-19}{17} \implies \boxed{036}</math><br />
<br />
-Alexlikemath<br />
== See also ==<br />
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_14&diff=1612802013 AIME I Problems/Problem 142021-09-01T01:56:35Z<p>Jdong2006: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
For <math>\pi \le \theta < 2\pi</math>, let<br />
<br />
<cmath>\begin{align*}<br />
P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
and<br />
<br />
<cmath>\begin{align*}<br />
Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
Noticing the <math>\sin</math> and <math>\cos</math> in both <math>P</math> and <math>Q,</math> we think of the angle addition identities: <br />
<br />
<cmath>\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b + \sin a \sin b</cmath><br />
<br />
With this in mind, we multiply <math>P</math> by <math>\sin \theta</math> and <math>Q</math> by <math>\cos \theta</math> to try and use some angle addition identities. Indeed, we get <br />
<cmath>\begin{align*}<br />
P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\<br />
&= \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots \\<br />
&= \cos \theta - \dfrac{1}{2}P<br />
\end{align*}</cmath><br />
after adding term-by-term. Similar term-by-term adding yields <cmath>P \cos \theta + Q \sin \theta = -2(Q - 1).</cmath> <br />
This is a system of equations; rearrange and rewrite to get <cmath>P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta</cmath> and <cmath>P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.</cmath> Subtract the two and rearrange to get <cmath>\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.</cmath> Then, square both sides and use Pythagorean Identity to get a quadratic in <math>\sin \theta.</math> Factor that quadratic and solve for <math>\sin \theta = -17/19, 1/3.</math> The answer format tells us it's the negative solution, and our desired answer is <math>17 + 19 = \boxed{036}.</math><br />
<br />
==Solution 2==<br />
<br />
Use sum to product formulas to rewrite <math>P</math> and <math>Q</math><br />
<br />
<br />
<math>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math><br />
<br />
Therefore, <math>P \sin \theta - Q \cos \theta = -2P</math><br />
<br />
Using <math>\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P</math><br />
<br />
Plug in to the previous equation and cancel out the "P" terms to get: <math>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</math>.<br />
<br />
Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = -\frac{17}{19} \implies \boxed{036}</math><br />
<br />
==Solution 3==<br />
<br />
Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath><br />
<br />
Thus, the following identities follow immediately:<br />
<cmath>ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)</cmath><br />
<cmath>i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)</cmath><br />
<cmath>i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)</cmath><br />
<br />
Consider, now, the sum <math>Q+iP</math>. It follows fairly immediately that:<br />
<br />
<cmath>Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}</cmath><br />
<cmath>Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}</cmath><br />
<br />
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:<br />
<br />
<cmath>Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)</cmath><br />
<cmath>Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}</cmath><br />
<br />
Comparing real and imaginary parts, we find:<br />
<cmath>\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}</cmath><br />
<br />
Squaring this equation and letting <math>\sin^2(\theta)=x</math>:<br />
<br />
<math>\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}</math><br />
<br />
Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>.<br />
<br />
<math>017+019=\boxed{036}</math><br />
<br />
==Solution 4==<br />
A bit similar to Solution 3. We use <math>\phi = \theta+90^\circ</math> because the progression cycles in <math>P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)</math>. So we could rewrite that as <math>P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)</math>.<br />
<br />
Similarly, <math>Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)</math>.<br />
<br />
Setting complex <math>z=q_1+p_1i</math>, we get <math>z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)</math><br />
<br />
<math>(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}</math>.<br />
<br />
The important part is the ratio of the imaginary part <math>i</math> to the real part. To cancel out the imaginary part from the denominator, we must add <math>0.5i\sin\phi</math> to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would cancel out.<br />
<br />
<math>\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}</math>.<br />
<br />
Setting <math>\sin\theta=y</math>, we obtain<br />
<cmath>\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}</cmath><br />
<cmath>7\sqrt{1-y^2}=2\sqrt{2}(2+y)</cmath><br />
<cmath>49-49y^2=8y^2+32y+32</cmath><br />
<cmath>57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}</cmath><br />
<cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>.<br />
<br />
Since <math>y<0</math> because <math>\pi<\theta<2\pi</math>, <math>y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}</math>. Adding up, <math>17+19=\boxed{036}</math>.<br />
<br />
==Solution 5 (lots of room for sillies, I wouldn't recommend it)==<br />
<br />
We notice <math>\sin\theta=\frac{-i}{2}(e^{i\theta}-e^{-i\theta})</math> and <math>\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})</math><br />
<br />
With these, we just quickly find the sum of the infinite geometric series' in <math>P</math> and <math>Q</math>. <math>P</math> has 2 parts, the <math>\cos</math> and the <math>\sin</math> parts. The <math>\cos</math> part is: <math>\frac12\cos\theta-\frac18\cos3\theta+\cdots</math>, which can be turned into: <math>\frac14(e^{i\theta}(1-\frac{e^{i2\theta}}{4}+\cdots)+e^{-i\theta}(1-\frac{e^{-i2\theta}}{4}+\cdots))</math>, which is <math>\frac{1}{4}(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}})</math>. This turns into <math>\frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
Following the same process as above, we find that the <math>\sin</math> part of <math>P</math> is <math>\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, the <math>\cos</math> part of <math>Q</math> is <math>\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, and finally, the <math>\sin</math> part of <math>Q</math> is <math>\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
We convert all 4 of these equations into trig, and we end up getting <math>\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}</math>, we divide by <math>2</math> on both numerator and denominator, and we get <math>\frac{2\sqrt{2}}{7}=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}</math>. We use some trig identities and we get <math>\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}=\frac{2\sqrt2}{7}</math>. We factor the denominator into <math>(5-4\sin\theta)(2+\sin\theta)</math>. We cancel out <math>5-4\sin\theta</math> on both numerator and denominator to get <math>\frac{\cos\theta}{2+\sin\theta}=\frac{2\sqrt2}{7}</math>. We set <math>\sin\theta</math> as <math>x</math>, and we just solve a quadratic in terms of <math>x</math>, <math>\frac{1-x^2}{x^2+4x+4}=\frac{8}{49}</math>, cross multiply and simplify, and we get <math>57x^2+32x-17=0</math>. We can actually factor this to get <math>(19x+17)(3x-1)=0</math>, which yields the 2 solutions <math>x=-\frac{17}{19}, x=\frac{1}{3}</math>. Since <math>\pi\leq\theta<2\pi</math>, the latter solution is deemed invalid, and we are left with <math>\sin\theta=-\frac{17}{19}</math>. Our final answer is <math>17+19=\boxed{036}</math>.<br />
<br />
~ASAB<br />
<br />
<cmath>\textbf{omg this was definitely the hardest problem on the AIME I}</cmath><br />
<br />
==Solution 6==<br />
Follow solution 3, up to the point of using the geometric series formula<br />
<cmath>Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}</cmath><br />
<br />
Moving everything to the other side, and considering only the imaginary part, we get<br />
<cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath><br />
<br />
We can then write <math>P = 2 \sqrt{2} k</math>, and <math>Q = 7k</math>, (<math>k \neq 0</math>). Thus, we can substitute and divide out by k.<br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)</cmath><br />
<cmath>8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta</cmath><br />
<cmath>\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0</cmath><br />
<cmath>57\sin^{2}\theta+32\sin\theta-17 = 0</cmath><br />
<cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath><br />
<br />
Since <math>\pi \le \theta < 2\pi</math>, we get <math>\sin \theta < 0</math>, and thus, <math>\sin\theta = \frac{-19}{17} \implies \boxed{036}</math><br />
<br />
-Alexlikemath<br />
== See also ==<br />
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_14&diff=1612762013 AIME I Problems/Problem 142021-09-01T01:01:10Z<p>Jdong2006: /* Solution 5 (lots of room for sillies, I wouldn't recommend it) */</p>
<hr />
<div>== Problem ==<br />
For <math>\pi \le \theta < 2\pi</math>, let<br />
<br />
<cmath>\begin{align*}<br />
P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
and<br />
<br />
<cmath>\begin{align*}<br />
Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots<br />
\end{align*}</cmath><br />
<br />
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
Noticing the <math>\sin</math> and <math>\cos</math> in both <math>P</math> and <math>Q,</math> we think of the angle addition identities: <cmath>\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b + \sin a \sin b.</cmath> With this in mind, we multiply <math>P</math> by <math>\sin \theta</math> and <math>Q</math> by <math>\cos \theta</math> to try and use some angle addition identities. Indeed, we get <br />
<cmath>\begin{align*}<br />
P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\<br />
&= \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots \\<br />
&= \cos \theta - \dfrac{1}{2}P<br />
\end{align*}</cmath><br />
after adding term-by-term. Similar term-by-term adding yields <cmath>P \cos \theta + Q \sin \theta = -2(Q - 1).</cmath> <br />
This is a system of equations; rearrange and rewrite to get <cmath>P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta</cmath> and <cmath>P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.</cmath> Subtract the two and rearrange to get <cmath>\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.</cmath> Then, square both sides and use Pythagorean Identity to get a quadratic in <math>\sin \theta.</math> Factor that quadratic and solve for <math>\sin \theta = -17/19, 1/3.</math> The answer format tells us it's the negative solution, and our desired answer is <math>17 + 19 = \boxed{036}.</math><br />
<br />
==Solution 2==<br />
<br />
Use sum to product formulas to rewrite <math>P</math> and <math>Q</math><br />
<br />
<br />
<math>P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ... </math><br />
<br />
Therefore, <math>P \sin \theta - Q \cos \theta = -2P</math><br />
<br />
Using <math>\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P</math><br />
<br />
Plug in to the previous equation and cancel out the "P" terms to get: <math>\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2</math>.<br />
<br />
Then use the pythagorean identity to solve for <math>\sin\theta</math>, <math>\sin\theta = -\frac{17}{19} \implies \boxed{036}</math><br />
<br />
==Solution 3==<br />
<br />
Note that <cmath>e^{i\theta}=\cos(\theta)+i\sin(\theta)</cmath><br />
<br />
Thus, the following identities follow immediately:<br />
<cmath>ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)</cmath><br />
<cmath>i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)</cmath><br />
<cmath>i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)</cmath><br />
<br />
Consider, now, the sum <math>Q+iP</math>. It follows fairly immediately that:<br />
<br />
<cmath>Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}</cmath><br />
<cmath>Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}</cmath><br />
<br />
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:<br />
<br />
<cmath>Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)</cmath><br />
<cmath>Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}</cmath><br />
<br />
Comparing real and imaginary parts, we find:<br />
<cmath>\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}</cmath><br />
<br />
Squaring this equation and letting <math>\sin^2(\theta)=x</math>:<br />
<br />
<math>\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}</math><br />
<br />
Clearing denominators and solving for <math>x</math> gives sine as <math>x=-\frac{17}{19}</math>.<br />
<br />
<math>017+019=\boxed{036}</math><br />
<br />
==Solution 4==<br />
A bit similar to Solution 3. We use <math>\phi = \theta+90^\circ</math> because the progression cycles in <math>P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)</math>. So we could rewrite that as <math>P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)</math>.<br />
<br />
Similarly, <math>Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)</math>.<br />
<br />
Setting complex <math>z=q_1+p_1i</math>, we get <math>z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)</math><br />
<br />
<math>(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}</math>.<br />
<br />
The important part is the ratio of the imaginary part <math>i</math> to the real part. To cancel out the imaginary part from the denominator, we must add <math>0.5i\sin\phi</math> to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find <math>\frac{P}{Q}=\tan\text{arg}(\Sigma)</math> a PROPORTION of values. So denominators would cancel out.<br />
<br />
<math>\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}</math>.<br />
<br />
Setting <math>\sin\theta=y</math>, we obtain<br />
<cmath>\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}</cmath><br />
<cmath>7\sqrt{1-y^2}=2\sqrt{2}(2+y)</cmath><br />
<cmath>49-49y^2=8y^2+32y+32</cmath><br />
<cmath>57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}</cmath><br />
<cmath>y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}</cmath>.<br />
<br />
Since <math>y<0</math> because <math>\pi<\theta<2\pi</math>, <math>y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}</math>. Adding up, <math>17+19=\boxed{036}</math>.<br />
<br />
==Solution 5 (lots of room for sillies, I wouldn't recommend it)==<br />
<br />
We notice <math>\sin\theta=\frac{-i}{2}(e^{i\theta}-e^{-i\theta})</math> and <math>\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})</math><br />
<br />
With these, we just quickly find the sum of the infinite geometric series' in <math>P</math> and <math>Q</math>. <math>P</math> has 2 parts, the <math>\cos</math> and the <math>\sin</math> parts. The <math>\cos</math> part is: <math>\frac12\cos\theta-\frac18\cos3\theta+\cdots</math>, which can be turned into: <math>\frac14(e^{i\theta}(1-\frac{e^{i2\theta}}{4}+\cdots)+e^{-i\theta}(1-\frac{e^{-i2\theta}}{4}+\cdots))</math>, which is <math>\frac{1}{4}(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}})</math>. This turns into <math>\frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
Following the same process as above, we find that the <math>\sin</math> part of <math>P</math> is <math>\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, the <math>\cos</math> part of <math>Q</math> is <math>\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>, and finally, the <math>\sin</math> part of <math>Q</math> is <math>\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}</math>.<br />
<br />
We convert all 4 of these equations into trig, and we end up getting <math>\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}</math>, we divide by <math>2</math> on both numerator and denominator, and we get <math>\frac{2\sqrt{2}}{7}=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}</math>. We use some trig identities and we get <math>\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}=\frac{2\sqrt2}{7}</math>. We factor the denominator into <math>(5-4\sin\theta)(2+\sin\theta)</math>. We cancel out <math>5-4\sin\theta</math> on both numerator and denominator to get <math>\frac{\cos\theta}{2+\sin\theta}=\frac{2\sqrt2}{7}</math>. We set <math>\sin\theta</math> as <math>x</math>, and we just solve a quadratic in terms of <math>x</math>, <math>\frac{1-x^2}{x^2+4x+4}=\frac{8}{49}</math>, cross multiply and simplify, and we get <math>57x^2+32x-17=0</math>. We can actually factor this to get <math>(19x+17)(3x-1)=0</math>, which yields the 2 solutions <math>x=-\frac{17}{19}, x=\frac{1}{3}</math>. Since <math>\pi\leq\theta<2\pi</math>, the latter solution is deemed invalid, and we are left with <math>\sin\theta=-\frac{17}{19}</math>. Our final answer is <math>17+19=\boxed{036}</math>.<br />
<br />
~ASAB<br />
<br />
<cmath>\textbf{omg this was definitely the hardest problem on the AIME I}</cmath><br />
<br />
==Solution 6==<br />
Follow solution 3, up to the point of using the geometric series formula<br />
<cmath>Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}</cmath><br />
<br />
Moving everything to the other side, and considering only the imaginary part, we get<br />
<cmath>Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0</cmath><br />
<br />
We can then write <math>P = 2 \sqrt{2} k</math>, and <math>Q = 7k</math>, (<math>k \neq 0</math>). Thus, we can substitute and divide out by k.<br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0</cmath><br />
<cmath>2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)</cmath><br />
<cmath>8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta</cmath><br />
<cmath>\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0</cmath><br />
<cmath>57\sin^{2}\theta+32\sin\theta-17 = 0</cmath><br />
<cmath>\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0</cmath><br />
<br />
Since <math>\pi \le \theta < 2\pi</math>, we get <math>\sin \theta < 0</math>, and thus, <math>\sin\theta = \frac{-19}{17} \implies \boxed{036}</math><br />
<br />
-Alexlikemath<br />
== See also ==<br />
{{AIME box|year=2013|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=1612742020 AIME I Problems/Problem 142021-09-01T00:16:28Z<p>Jdong2006: /* Solution 5 (Official MAA) */</p>
<hr />
<div><br />
== Problem ==<br />
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math><br />
<br />
== Solution 1 ==<br />
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333<br />
<br />
<b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>.<br />
<br />
== Solution 2 ==<br />
Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <cmath>P(x)=x^2-(m+n)x+mn</cmath> The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now casework: <br />
If <math>P(3)=P(4)=m</math>, then <br />
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath><br />
<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath><br />
so this gives <math>(a+b)^2=7^2=49</math>. <br />
Next, if <math>P(3)=P(a)=m</math>, then <br />
<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath><br />
<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath><br />
Subtracting the first part of the first equation from the first part of the second equation gives <br />
<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath><br />
Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. <br />
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong<br />
<br />
== Solution 3 ==<br />
Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.<br />
<br />
Case 1: We have <math>P(3) \ne P(4)</math>. We must have <br />
<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath><br />
Rearrange and divide through by <math>8</math> to obtain<br />
<cmath>w = \frac{-25-2z}{8}.</cmath><br />
Now, note that<br />
<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath><br />
<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath><br />
Now, rearrange to get<br />
<cmath>\frac{z}{8} = -\frac{21}{16}</cmath><br />
and thus<br />
<cmath>z = -\frac{21}{2}.</cmath><br />
Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.<br />
<br />
Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.<br />
<br />
Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton<br />
<br />
==Solution 4==<br />
Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>.<br />
<br />
In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>.<br />
<br />
Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>.<br />
<br />
The requested sum is <math>36+49=85</math>.~TheUltimate123<br />
<br />
==Solution 5 (Official MAA)==<br />
Note that because <math>P\big(P(3)\big)=P\big(P(4)\big)= 0</math>, <math>P(3)</math> and <math>P(4)</math> are roots of <math>P(x)</math>. There are two cases.<br />
CASE 1: <math>P(3) = P(4)</math>. Then <math>P(x)</math> is symmetric about <math>x=\tfrac72</math>; that is to say, <math>P(r) = P(7-r)</math> for all <math>r</math>. Thus the remaining two roots must sum to <math>7</math>. Indeed, the polynomials <math>P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3</math> satisfy the conditions.<br />
CASE 2: <math>P(3)\neq P(4)</math>. Then <math>P(3)</math> and <math>P(4)</math> are the two distinct roots of <math>P(x)</math>, so<cmath>P(x) = \big(x-P(3)\big)\big(x-P(4)\big)</cmath>for all <math>x</math>. Note that any solution to <math>P\big(P(x)\big) = 0</math> must satisfy either <math>P(x) = P(3)</math> or <math>P(x) = P(4)</math>. Because <math>P(x)</math> is quadratic, the polynomials <math>P(x) - P(3)</math> and <math>P(x) - P(4)</math> each have the same sum of roots as the polynomial <math>P(x)</math>, which is <math>P(3) + P(4)</math>. Thus the answer in this case is <math>2\big(P(3) + P(4)\big)-7</math>, and so it suffices to compute the value of <math>P(3)+P(4)</math>.<br />
<br />
Let <math>P(3)=u</math> and <math>P(4) = v</math>. Substituting <math>x=3</math> and <math>x=4</math> into the above quadratic polynomial yields the system of equations<br />
<cmath>\begin{align*}<br />
u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\<br />
v &= (4-u)(4-v) = 16 - 4u - 4v + uv.<br />
\end{align*}</cmath>Subtracting the first equation from the second gives <math>v - u = 7 - u - v</math>, yielding <math>v = \frac72.</math> Substituting this value into the second equation gives<cmath>\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),</cmath>yielding <math>u = -3.</math> The sum of the two solutions is <math>2\left(\tfrac72-3\right)-7 = -6</math>. In this case, <math>P(x)= (x+3)\left(x-\frac72\right)</math>.<br />
<br />
The requested sum of squares is <math>7^2+(-6)^2 = \boxed{085}</math>.<br />
<br />
==Solution 6==<br />
<br />
Let <math>P(x) = (x-c)^2 - d</math> for some <math>c</math>, <math>d</math>.<br />
<br />
Then, we can write <math>P(P(x)) = ((x-c)^2 - d - c)^2 - d</math>. Setting the expression equal to <math>0</math> and solving for <math>x</math> gives:<br />
<br />
<math>x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c</math><br />
<br />
Therefore, we have that <math>x</math> takes on the four values <math>\sqrt{\sqrt{d} + d + c} + c</math>, <math>-\sqrt{\sqrt{d} + d + c} + c</math>, <math>\sqrt{-\sqrt{d} + d + c} + c</math>, and <math>-\sqrt{-\sqrt{d} + d + c} + c</math>. Two of these values are <math>3</math> and <math>4</math>, and the other two are <math>a</math> and <math>b</math>.<br />
<br />
We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.<br />
<br />
<math>\textbf{Case 1}</math>: Both the 3 and 4 values are from the same group.<br />
<br />
In this case, the <math>a</math> and <math>b</math> values are both from the other group. The sum of this is just <math>2c</math> because the radical cancels out. Because of this, we can see that <math>c</math> is just the average of <math>3</math> and <math>4</math>, so we have <math>2c = 3 + 4 = 7</math>, so <math>(a+b)^2 = 7^2 = 49</math>.<br />
<br />
<math>\textbf{Case 2}</math>: The 3 and 4 values come from different groups.<br />
<br />
It is easy to see that all possibilities in this case are basically symmetric and yield the same value for <math>(a+b)^2</math>. Without loss of generality, assume that <math>\sqrt{\sqrt{d} + d + c} + c = 4</math> and <math>\sqrt{-\sqrt{d} + d + c} + c = 3</math>. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.<br />
<br />
We can write <math>\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c</math>.<br />
<br />
Moving most terms to the left side and simplifying gives <math>\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1</math>.<br />
<br />
We can square both sides and simplify:<br />
<br />
<math>\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}</math><br />
<br />
Squaring both sides again gives the following:<br />
<br />
<math>d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}</math><br />
<br />
Nearly all terms cancel out, yielding <math>c = \frac{1}{4}</math>.<br />
<br />
By substituting this back in, we obtain <math>\sqrt{\sqrt{d} + d + c} = \frac{15}{4}</math> and <math>\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}</math>.<br />
<br />
The sum of <math>a</math> and <math>b</math> is equal to <math>-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6</math>, so <math>(a+b)^2 = 36</math>.<br />
<br />
Adding up both values gives <math>49 + 36 = \boxed{085}</math> as our final answer.<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=1612732020 AIME I Problems/Problem 142021-09-01T00:13:51Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div><br />
== Problem ==<br />
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math><br />
<br />
== Solution 1 ==<br />
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333<br />
<br />
<b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>.<br />
<br />
== Solution 2 ==<br />
Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <cmath>P(x)=x^2-(m+n)x+mn</cmath> The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now casework: <br />
If <math>P(3)=P(4)=m</math>, then <br />
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath><br />
<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath><br />
so this gives <math>(a+b)^2=7^2=49</math>. <br />
Next, if <math>P(3)=P(a)=m</math>, then <br />
<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath><br />
<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath><br />
Subtracting the first part of the first equation from the first part of the second equation gives <br />
<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath><br />
Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. <br />
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong<br />
<br />
== Solution 3 ==<br />
Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.<br />
<br />
Case 1: We have <math>P(3) \ne P(4)</math>. We must have <br />
<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath><br />
Rearrange and divide through by <math>8</math> to obtain<br />
<cmath>w = \frac{-25-2z}{8}.</cmath><br />
Now, note that<br />
<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath><br />
<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath><br />
Now, rearrange to get<br />
<cmath>\frac{z}{8} = -\frac{21}{16}</cmath><br />
and thus<br />
<cmath>z = -\frac{21}{2}.</cmath><br />
Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.<br />
<br />
Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.<br />
<br />
Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton<br />
<br />
==Solution 4==<br />
Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>.<br />
<br />
In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>.<br />
<br />
Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>.<br />
<br />
The requested sum is <math>36+49=85</math>.~TheUltimate123<br />
<br />
==Solution 5 (Official MAA)==<br />
Note that because <math>P\big(P(3)\big)=P\big(P(4)\big)= 0</math>, <math>P(3)</math> and <math>P(4)</math> are roots of <math>P(x)</math>. There are two cases.<br />
CASE 1: <math>P(3) = P(4)</math>. Then <math>P(x)</math> is symmetric about <math>x=\tfrac72</math>; that is to say, <math>P(r) = P(7-r)</math> for all <math>r</math>. Thus the remaining two roots must sum to <math>7</math>. Indeed, the polynomials <math>P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3</math> satisfy the conditions.<br />
CASE 2: <math>P(3)\neq P(4)</math>. Then <math>P(3)</math> and <math>P(4)</math> are the two distinct roots of <math>P(x)</math>, so<cmath>P(x) = \big(x-P(3)\big)\big(x-P(4)\big)</cmath>for all <math>x</math>. Note that any solution to <math>P\big(P(x)\big) = 0</math> must satisfy either <math>P(x) = P(3)</math> or <math>P(x) = P(4)</math>. Because <math>P(x)</math> is quadratic, the polynomials <math>P(x) - P(3)</math> and <math>P(x) - P(4)</math> each have the same sum of roots as the polynomial <math>P(x)</math>, which is <math>P(3) + P(4)</math>. Thus the answer in this case is <math>2\big(P(3) + P(4)\big)-7</math>, and so it suffices to compute the value of <math>P(3)+P(4)</math>.<br />
<br />
Let <math>P(3)=u</math> and <math>P(4) = v</math>. Substituting <math>x=3</math> and <math>x=4</math> into the above quadratic polynomial yields the system of equations<br />
<cmath>\begin{align*}<br />
u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\<br />
v &= (4-u)(4-v) = 16 - 4u - 4v + uv.<br />
\end{align*}</cmath>Subtracting the first equation from the second gives <math>v - u = 7 - u - v</math>, yielding <math>v = \frac72.</math> Substituting this value into the second equation gives<cmath>\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),</cmath>yielding <math>u = -3.</math> The sum of the two solutions is <math>2\left(\tfrac72-3\right)-7 = -6</math>. In this case, <math>P(x)= (x+3)\left(x-\frac72\right)</math>.<br />
<br />
The requested sum of squares is <math>7^2+(-6)^2 = {85}</math>.<br />
<br />
==Solution 6==<br />
<br />
Let <math>P(x) = (x-c)^2 - d</math> for some <math>c</math>, <math>d</math>.<br />
<br />
Then, we can write <math>P(P(x)) = ((x-c)^2 - d - c)^2 - d</math>. Setting the expression equal to <math>0</math> and solving for <math>x</math> gives:<br />
<br />
<math>x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c</math><br />
<br />
Therefore, we have that <math>x</math> takes on the four values <math>\sqrt{\sqrt{d} + d + c} + c</math>, <math>-\sqrt{\sqrt{d} + d + c} + c</math>, <math>\sqrt{-\sqrt{d} + d + c} + c</math>, and <math>-\sqrt{-\sqrt{d} + d + c} + c</math>. Two of these values are <math>3</math> and <math>4</math>, and the other two are <math>a</math> and <math>b</math>.<br />
<br />
We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.<br />
<br />
<math>\textbf{Case 1}</math>: Both the 3 and 4 values are from the same group.<br />
<br />
In this case, the <math>a</math> and <math>b</math> values are both from the other group. The sum of this is just <math>2c</math> because the radical cancels out. Because of this, we can see that <math>c</math> is just the average of <math>3</math> and <math>4</math>, so we have <math>2c = 3 + 4 = 7</math>, so <math>(a+b)^2 = 7^2 = 49</math>.<br />
<br />
<math>\textbf{Case 2}</math>: The 3 and 4 values come from different groups.<br />
<br />
It is easy to see that all possibilities in this case are basically symmetric and yield the same value for <math>(a+b)^2</math>. Without loss of generality, assume that <math>\sqrt{\sqrt{d} + d + c} + c = 4</math> and <math>\sqrt{-\sqrt{d} + d + c} + c = 3</math>. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.<br />
<br />
We can write <math>\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c</math>.<br />
<br />
Moving most terms to the left side and simplifying gives <math>\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1</math>.<br />
<br />
We can square both sides and simplify:<br />
<br />
<math>\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}</math><br />
<br />
Squaring both sides again gives the following:<br />
<br />
<math>d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}</math><br />
<br />
Nearly all terms cancel out, yielding <math>c = \frac{1}{4}</math>.<br />
<br />
By substituting this back in, we obtain <math>\sqrt{\sqrt{d} + d + c} = \frac{15}{4}</math> and <math>\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}</math>.<br />
<br />
The sum of <math>a</math> and <math>b</math> is equal to <math>-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6</math>, so <math>(a+b)^2 = 36</math>.<br />
<br />
Adding up both values gives <math>49 + 36 = \boxed{085}</math> as our final answer.<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_12&diff=1612142013 AIME II Problems/Problem 122021-08-30T03:49:55Z<p>Jdong2006: </p>
<hr />
<div>==Problem 12==<br />
<br />
Let <math>S</math> be the set of all polynomials of the form <math>z^3 + az^2 + bz + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers. Find the number of polynomials in <math>S</math> such that each of its roots <math>z</math> satisfies either <math>|z| = 20</math> or <math>|z| = 13</math>.<br />
<br />
==Solution 1==<br />
<br />
Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from [[Vieta's formulas]]. <br />
<br />
*Case 1: <math>f(z)=(z-r)(z-\omega)(z-\omega^*)</math>, where <math>r\in \mathbb{R}</math>, <math>\omega</math> is nonreal, and <math>\omega^*</math> is the complex conjugate of omega (note that we may assume that <math>\Im(\omega)>0</math>).<br />
<br />
The real root <math>r</math> must be one of <math>-20</math>, <math>20</math>, <math>-13</math>, or <math>13</math>. By Viète's formulas, <math>a=-(r+\omega+\omega^*)</math>, <math>b=|\omega|^2+r(\omega+\omega^*)</math>, and <math>c=-r|\omega|^2</math>. But <math>\omega+\omega^*=2\Re{(\omega)}</math> (i.e., adding the conjugates cancels the imaginary part). Therefore, to make <math>a</math> an integer, <math>2\Re{(\omega)}</math> must be an integer. Conversely, if <math>\omega+\omega^*=2\Re{(\omega)}</math> is an integer, then <math>a,b,</math> and <math>c</math> are clearly integers. Therefore <math>2\Re{(\omega)}\in \mathbb{Z}</math> is equivalent to the desired property. Let <math>\omega=\alpha+i\beta</math>.<br />
<br />
*Subcase 1.1: <math>|\omega|=20</math>.<br />
In this case, <math>\omega</math> lies on a circle of radius <math>20</math> in the complex plane. As <math>\omega</math> is nonreal, we see that <math>\beta\ne 0</math>. Hence <math>-20<\Re{(\omega)}< 20</math>, or rather <math>-40<2\Re{(\omega)}< 40</math>. We count <math>79</math> integers in this interval, each of which corresponds to a unique complex number on the circle of radius <math>20</math> with positive imaginary part.<br />
<br />
*Subcase 1.2: <math>|\omega|=13</math>.<br />
In this case, <math>\omega</math> lies on a circle of radius <math>13</math> in the complex plane. As <math>\omega</math> is nonreal, we see that <math>\beta\ne 0</math>. Hence <math>-13<\Re{(\omega)}< 13</math>, or rather <math>-26<2\Re{(\omega)}< 26</math>. We count <math>51</math> integers in this interval, each of which corresponds to a unique complex number on the circle of radius <math>13</math> with positive imaginary part.<br />
<br />
Therefore, there are <math>79+51=130</math> choices for <math>\omega</math>. We also have <math>4</math> choices for <math>r</math>, hence there are <math>4\cdot 130=520</math> total polynomials in this case.<br />
<br />
*Case 2: <math>f(z)=(z-r_1)(z-r_2)(z-r_3)</math>, where <math>r_1,r_2,r_3</math> are all real.<br />
In this case, there are four possible real roots, namely <math>\pm 13, \pm20</math>. Let <math>p</math> be the number of times that <math>13</math> appears among <math>r_1,r_2,r_3</math>, and define <math>q,r,s</math> similarly for <math>-13,20</math>, and <math>-20</math>, respectively. Then <math>p+q+r+s=3</math> because there are three roots. We wish to find the number of ways to choose nonnegative integers <math>p,q,r,s</math> that satisfy that equation. By balls and urns, these can be chosen in <math>\binom{6}{3}=20</math> ways.<br />
<br />
Therefore, there are a total of <math>520+20=\boxed{540}</math> polynomials with the desired property.<br />
<br />
==Solution 2 (Systematics)==<br />
This combinatorics problem involves counting, and casework is most appropriate.<br />
There are two cases: either all three roots are real, or one is real and there are two imaginary roots.<br />
<br />
Case 1: Three roots are of the set <math>{13, -13, 20, -20}</math>. By stars and bars, there is <math>\binom{6}{3}=20</math> ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves).<br />
<br />
Case 2: One real root: one of <math>13, -13, 20, -20</math>. Then two imaginary roots left; it is well known that because coefficients of the polynomial are integral (and thus not imaginary), these roots are conjugates. Therefore, either both roots have a norm (also called magnitude) of <math>20</math> or <math>13</math>. Call the root <math>a+bi</math>, where <math>a</math> is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of <math>(x-(a+bi))(x-(a-bi))=-2ax+x^2+(a^2+b^2)</math> tells us that we just need <math>2a</math> to be integral, because <math>a^2+b^2</math> IS the norm of the root! (Note that it is not necessary to multiply by the real root. That won't affect whether or not a coefficient is imaginary.) <br />
Therefore, when the norm is <math>20</math>, the <math>a</math> term can range from <math>-19.5, -19, ...., 0, 0.5, ..., 19.5</math> or <math>79</math> solutions. When the norm is <math>13</math>, the <math>a</math> term has <math>51</math> possibilities from <math>-12.5, -12, ..., 12.5</math>. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, <math>4</math>, and you get <math>520</math> for this case.<br />
<br />
And <math>520+20=540</math> and we are done.<br />
<br />
==Solution 3 (Comments)==<br />
If the polynomial has one real root and two complex roots, then it can be factored as <math>(z-r)(z^2+pz+q), </math> where <math>r</math> is real with <math>|r|=13,20</math> and <math>p,q</math> are integers with <math>p^2 <4q.</math> The roots <math>z_1</math> and <math>z_2</math> are conjugates. We have <math>|z_1|^2=|z_2|^2=z_1z_2=q.</math> So <math>q</math> is either <math>20^2</math> or <math>13^2</math>. The only requirement for <math>p</math> is <math>p<\sqrt{4q^2}=2\sqrt{q}.</math> All such quadratic equations are listed as follows:<br />
<br />
<math>z^2+pz+20^2,</math> where <math>p=0,\pm1,\pm2,\cdots,\pm 39,</math><br />
<br />
<math>z^2+pz+13^2,</math> where <math>p=0,\pm1,\pm2,\cdots,\pm 25</math>.<br />
<br />
Total of 130 equations, multiplied by 4 (the number of cases for real <math>r</math>, we have 520 equations, as indicated in the solution.<br />
<br />
-JZ<br />
<br />
==See Also==<br />
{{AIME box|year=2013|n=II|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_12&diff=1612062019 AIME I Problems/Problem 122021-08-30T00:34:23Z<p>Jdong2006: /* Solution 5 (Official MAA) */</p>
<hr />
<div>==Problem==<br />
Given <math>f(z) = z^2-19z</math>, there are complex numbers <math>z</math> with the property that <math>z</math>, <math>f(z)</math>, and <math>f(f(z))</math> are the vertices of a right triangle in the complex plane with a right angle at <math>f(z)</math>. There are positive integers <math>m</math> and <math>n</math> such that one such value of <math>z</math> is <math>m+\sqrt{n}+11i</math>. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
Notice that we must have <cmath>\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .</cmath>However, <math>f(t)-t=t(t-20)</math>, so<br />
<cmath><br />
\begin{align*}<br />
\frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\<br />
&=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\<br />
&=(z-19)(z+1)\\<br />
&=(z-9)^2-100.<br />
\end{align*}<br />
</cmath><br />
Then, the real part of <math>(z-9)^2</math> is <math>100</math>. Since <math>\text{Im}(z-9)=\text{Im}(z)=11</math>, let <math>z-9=a+11i</math>. Then, <cmath>100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.</cmath>It follows that <math>z=9+\sqrt{221}+11i</math>, and the requested sum is <math>9+221=\boxed{230}</math>.<br />
<br />
(Solution by TheUltimate123)<br />
<br />
==Solution 2==<br />
<br />
We will use the fact that segments <math>AB</math> and <math>BC</math> are perpendicular in the complex plane if and only if <math>\frac{a-b}{b-c}\in i\mathbb{R}</math>. To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part. <br />
<br />
Now to apply this: <br />
<cmath>\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}</cmath><br />
<cmath>\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}</cmath><br />
<cmath>\frac{z^2-20z}{z^4-38z^3+341z^2+380z}</cmath><br />
<cmath>\frac{z(z-20)}{z(z+1)(z-19)(z-20)}</cmath><br />
<cmath>\frac{1}{(z+1)(z-19)}\in i\mathbb{R}</cmath><br />
<br />
The factorization of the nasty denominator above is made easier with the intuition that <math>(z-20)</math> must be a divisor for the problem to lead anywhere. Now we know <math>(z+1)(z-19)\in i\mathbb{R}</math> so using the fact that the imaginary part of <math>z</math> is <math>11i</math> and calling the real part r, <br />
<br />
<cmath>(r+1+11i)(r-19+11i)\in i\mathbb{R}</cmath><br />
<cmath>r^2-18r-140=0</cmath><br />
<br />
solving the above quadratic yields <math>r=9+\sqrt{221}</math> so our answer is <math>9+221=\boxed{230}</math><br />
<br />
==Solution 3==<br />
I would like to use a famous method, namely the coni method. <br />
<br />
Statement .If we consider there complex number <math>A,B,C</math> in argand plane then <math>\angle ABC =\arg{\frac{B-A}{B-C}}</math>.<br />
<br />
According to the question given, we can assume ,<math>A= f(f(z)),B=f(z),C= x</math> respectively.<br />
<br />
WLOG,<math>Z_1= \frac{f(f(z))-f(z)}{z-f(z)}</math>.<br />
According to the question <math>\arg{Z_1}=\frac{\pi}{2}</math>.<br />
<br />
So,<math>\Re (Z_1)=0</math>.<br />
<br />
Now, <math>Z_1=\frac{z(z-19)(z+1)(z-20)}{-z(z-20)}</math>.<br />
<br />
<math>\implies Z_1= -(z^2-18z-19)</math>.<br />
WLOG,<math>z=a+11i</math> .where <math>a=m+\sqrt{n}</math>.<br />
<br />
So,<math>\Re (Z_1)= -(a^2-18a-140)</math>.<br />
Solving,<math>a^2-18a-140=0</math> .get ,<br />
<br />
<math>a=9</math>±<math>\sqrt{221}</math>.<br />
So, possible value of <math>a=9+\sqrt{221}</math>.<br />
<br />
<math>m+n=\boxed{230}</math>.<br />
~ftheftics.<br />
<br />
==Proof of this method==<br />
Note that if we translate a triangle, the measures of all of its angles stay the same. So we can translate <math>ABC</math> on the complex plane so that <math>B=0</math>. Let the images of <math>A, B, C</math> be <math>A', B', C'</math> respectively. Then, we can use the formula:<br />
<br />
<cmath>re^{i\theta}=r\cos(\theta)+i \cdot r\sin(\theta)</cmath> (This is known as Euler's Theorem.)<br />
<br />
Using Euler's theorem (represent each complex number in polar form, then use exponent identities), we can show that <cmath>\text{arg}(\frac{A'}{C'})=\text{arg}(A')-\text{arg}(C')=\angle(A'B'C')=\angle(ABC)</cmath> So this method is valid.<br />
~Math4Life2020<br />
<br />
==Solution 4==<br />
It is well known that <math>AB</math> is perpendicular to <math>CD</math> iff <math>\frac{d-c}{b-a}</math> is a pure imaginary number. Here, we have that <math>A=z</math>, <math>B,C=f(z)</math>, and <math>D=f(f(z))</math>. This means that this is equivalent to <math>\frac{f(f(z))-f(z)}{f(z)-z}</math> being a pure imaginary number. Plugging in <math>f(z)=z^2-19z</math>, we have that <math>\frac{(z^2-19z)-19(z^2-19z)-(z^2-19z)}{z^2-19z-z}</math> being pure imaginary. Factoring and simplifying, we find that this is simply equivalent to <math>(z-19)(z+1)</math> being pure imaginary. We let <math>z=a+bi</math>, so this is equivalent to <math>(a+bi-19)(a+bi+1)</math> being pure imaginary. Expanding the product, this is equivalent to <math>a^2+abi+a+abi-b^2+bi-19a-19bi-19</math> being pure imaginary. Taking the real part of this, and setting this equal to <math>0</math>, we have that <math>a^2-18a-b^2-19=0</math>. Since <math>b=11</math>, we have that <math>a^2-18a-140=0</math>. By the quadratic formula, <math>a=9 \pm \sqrt{221}</math>, and taking the positive root gives that <math>a=9+ \sqrt{221}</math>, so the answer is <math>9+221=230</math><br />
<br />
~smartninja2000<br />
==Solution 5 (Official MAA)==<br />
The arguments of the two complex numbers differ by <math>90^\circ</math> if the ratio of the numbers is a pure imaginary number. Thus three distinct complex numbers <math>A,\,B,</math> and <math>C</math> form a right triangle at <math>B</math> if and only if <math>\tfrac{C-B}{B-A}</math> has real part equal to <math>0.</math> Hence <cmath>\begin{align*}<br />
\frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}{(z^2-19z)-z}\\<br />
&=\frac{(z^2-19z)(z^2-19z-19-1)}{z^2-20z}\\<br />
&=\frac{z(z-19)(z+1)(z-20)}{z(z-20)} \\<br />
&=z^2-18z-19<br />
\end{align*}</cmath> must have real part equal to <math>0.</math> If <math>z=x+11i,</math> the real part of <math>z^2-18z-19</math> is <math>x^2-11^2-18x-19,</math> which is <math>0</math> when <math>x=9\pm\sqrt{221}.</math> The requested sum is <math>9+221=\boxed{230}.</math><br />
<br />
==See also==<br />
{{AIME box|year=2019|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_12&diff=1612052019 AIME I Problems/Problem 122021-08-30T00:27:31Z<p>Jdong2006: /* Proof of this method */</p>
<hr />
<div>==Problem==<br />
Given <math>f(z) = z^2-19z</math>, there are complex numbers <math>z</math> with the property that <math>z</math>, <math>f(z)</math>, and <math>f(f(z))</math> are the vertices of a right triangle in the complex plane with a right angle at <math>f(z)</math>. There are positive integers <math>m</math> and <math>n</math> such that one such value of <math>z</math> is <math>m+\sqrt{n}+11i</math>. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
Notice that we must have <cmath>\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .</cmath>However, <math>f(t)-t=t(t-20)</math>, so<br />
<cmath><br />
\begin{align*}<br />
\frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\<br />
&=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\<br />
&=(z-19)(z+1)\\<br />
&=(z-9)^2-100.<br />
\end{align*}<br />
</cmath><br />
Then, the real part of <math>(z-9)^2</math> is <math>100</math>. Since <math>\text{Im}(z-9)=\text{Im}(z)=11</math>, let <math>z-9=a+11i</math>. Then, <cmath>100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.</cmath>It follows that <math>z=9+\sqrt{221}+11i</math>, and the requested sum is <math>9+221=\boxed{230}</math>.<br />
<br />
(Solution by TheUltimate123)<br />
<br />
==Solution 2==<br />
<br />
We will use the fact that segments <math>AB</math> and <math>BC</math> are perpendicular in the complex plane if and only if <math>\frac{a-b}{b-c}\in i\mathbb{R}</math>. To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part. <br />
<br />
Now to apply this: <br />
<cmath>\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}</cmath><br />
<cmath>\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}</cmath><br />
<cmath>\frac{z^2-20z}{z^4-38z^3+341z^2+380z}</cmath><br />
<cmath>\frac{z(z-20)}{z(z+1)(z-19)(z-20)}</cmath><br />
<cmath>\frac{1}{(z+1)(z-19)}\in i\mathbb{R}</cmath><br />
<br />
The factorization of the nasty denominator above is made easier with the intuition that <math>(z-20)</math> must be a divisor for the problem to lead anywhere. Now we know <math>(z+1)(z-19)\in i\mathbb{R}</math> so using the fact that the imaginary part of <math>z</math> is <math>11i</math> and calling the real part r, <br />
<br />
<cmath>(r+1+11i)(r-19+11i)\in i\mathbb{R}</cmath><br />
<cmath>r^2-18r-140=0</cmath><br />
<br />
solving the above quadratic yields <math>r=9+\sqrt{221}</math> so our answer is <math>9+221=\boxed{230}</math><br />
<br />
==Solution 3==<br />
I would like to use a famous method, namely the coni method. <br />
<br />
Statement .If we consider there complex number <math>A,B,C</math> in argand plane then <math>\angle ABC =\arg{\frac{B-A}{B-C}}</math>.<br />
<br />
According to the question given, we can assume ,<math>A= f(f(z)),B=f(z),C= x</math> respectively.<br />
<br />
WLOG,<math>Z_1= \frac{f(f(z))-f(z)}{z-f(z)}</math>.<br />
According to the question <math>\arg{Z_1}=\frac{\pi}{2}</math>.<br />
<br />
So,<math>\Re (Z_1)=0</math>.<br />
<br />
Now, <math>Z_1=\frac{z(z-19)(z+1)(z-20)}{-z(z-20)}</math>.<br />
<br />
<math>\implies Z_1= -(z^2-18z-19)</math>.<br />
WLOG,<math>z=a+11i</math> .where <math>a=m+\sqrt{n}</math>.<br />
<br />
So,<math>\Re (Z_1)= -(a^2-18a-140)</math>.<br />
Solving,<math>a^2-18a-140=0</math> .get ,<br />
<br />
<math>a=9</math>±<math>\sqrt{221}</math>.<br />
So, possible value of <math>a=9+\sqrt{221}</math>.<br />
<br />
<math>m+n=\boxed{230}</math>.<br />
~ftheftics.<br />
<br />
==Proof of this method==<br />
Note that if we translate a triangle, the measures of all of its angles stay the same. So we can translate <math>ABC</math> on the complex plane so that <math>B=0</math>. Let the images of <math>A, B, C</math> be <math>A', B', C'</math> respectively. Then, we can use the formula:<br />
<br />
<cmath>re^{i\theta}=r\cos(\theta)+i \cdot r\sin(\theta)</cmath> (This is known as Euler's Theorem.)<br />
<br />
Using Euler's theorem (represent each complex number in polar form, then use exponent identities), we can show that <cmath>\text{arg}(\frac{A'}{C'})=\text{arg}(A')-\text{arg}(C')=\angle(A'B'C')=\angle(ABC)</cmath> So this method is valid.<br />
~Math4Life2020<br />
<br />
==Solution 4==<br />
It is well known that <math>AB</math> is perpendicular to <math>CD</math> iff <math>\frac{d-c}{b-a}</math> is a pure imaginary number. Here, we have that <math>A=z</math>, <math>B,C=f(z)</math>, and <math>D=f(f(z))</math>. This means that this is equivalent to <math>\frac{f(f(z))-f(z)}{f(z)-z}</math> being a pure imaginary number. Plugging in <math>f(z)=z^2-19z</math>, we have that <math>\frac{(z^2-19z)-19(z^2-19z)-(z^2-19z)}{z^2-19z-z}</math> being pure imaginary. Factoring and simplifying, we find that this is simply equivalent to <math>(z-19)(z+1)</math> being pure imaginary. We let <math>z=a+bi</math>, so this is equivalent to <math>(a+bi-19)(a+bi+1)</math> being pure imaginary. Expanding the product, this is equivalent to <math>a^2+abi+a+abi-b^2+bi-19a-19bi-19</math> being pure imaginary. Taking the real part of this, and setting this equal to <math>0</math>, we have that <math>a^2-18a-b^2-19=0</math>. Since <math>b=11</math>, we have that <math>a^2-18a-140=0</math>. By the quadratic formula, <math>a=9 \pm \sqrt{221}</math>, and taking the positive root gives that <math>a=9+ \sqrt{221}</math>, so the answer is <math>9+221=230</math><br />
<br />
~smartninja2000<br />
==Solution 5 (Official MAA)==<br />
The arguments of the two complex numbers differ by <math>90^\circ</math> if the ratio of the numbers is a pure imaginary number. Thus three distinct complex numbers <math>A,\,B,</math> and <math>C</math> form a right triangle at <math>B</math> if and only if <math>\tfrac{C-B}{B-A}</math> has real part equal to <math>0.</math> Hence <cmath>\begin{align*}<br />
\frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}{(z^2-19z)-z}\\<br />
&=\frac{(z^2-19z)(z^2-19z-19-1)}{z^2-20z}\\<br />
&=\frac{z(z-19)(z+1)(z-20)}{z(z-20)} \\<br />
&=z^2-18z-19<br />
\end{align*}</cmath> must have real part equal to <math>0.</math> If <math>z=x+11i,</math> the real part of <math>z^2-18z-19</math> is <math>x^2-11^2-18x-19,</math> which is <math>0</math> when <math>x=9\pm\sqrt{221}.</math> The requested sum is <math>9+221=230.</math><br />
==See also==<br />
{{AIME box|year=2019|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_9&diff=1611712014 AIME I Problems/Problem 92021-08-29T16:33:19Z<p>Jdong2006: /* Solution 3 */</p>
<hr />
<div>== Problem 9 ==<br />
<br />
Let <math>x_1<x_2<x_3</math> be the three real roots of the equation <math>\sqrt{2014}x^3-4029x^2+2=0</math>. Find <math>x_2(x_1+x_3)</math>.<br />
<br />
== Solution 1==<br />
<br />
Substituting <math>n</math> for <math>2014</math>, we get <cmath>\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2</cmath> <cmath>= x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0</cmath> Noting that <math>nx^2 - 1</math> factors as a difference of squares to <cmath>(\sqrt{n}x - 1)(\sqrt{n}x+1)</cmath> we can factor the left side as <cmath>(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))</cmath> This means that <math>\frac{1}{\sqrt{n}}</math> is a root, and the other two roots are the roots of <math>x^2 - 2\sqrt{n}x - 2</math>. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to <math>2\sqrt{n}</math>, so the positive root must be greater than <math>2\sqrt{n}</math> in order to produce this sum when added to a negative value. Since <math>0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}</math> is clearly true, <math>x_2 = \frac{1}{\sqrt{2014}}</math> and <math>x_1 + x_3 = 2\sqrt{2014}</math>. Multiplying these values together, we find that <math>x_2(x_1+x_3) = \boxed{002}</math>.<br />
<br />
==Solution 2==<br />
From Vieta's formulae, we know that <cmath>x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}</cmath> <cmath>x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}}</cmath> and <cmath>x_1x_2 + x_2x_3 + x_1x_3 = 0</cmath> Thus, we know that <cmath>x_2(x_1 + x_3) = -x_1x_3</cmath><br />
<br />
Now consider the polynomial with roots <math>x_1x_2, x_2x_3,</math> and <math>x_1x_3</math>. Expanding the polynomial <cmath>(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)</cmath>we get the polynomial <cmath>x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2</cmath> Substituting the values obtained from Vieta's formulae, we find that this polynomial is <cmath>x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}</cmath> We know <math>x_1x_3</math> is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to <cmath>1007x^3 - 4029x - 2 = 0</cmath><br />
<br />
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the <math>x^3</math> term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that <math>x = -2</math> is a solution. Factoring it out, we get that <cmath>1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)</cmath> Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, <cmath>x_1x_3 = -2</cmath> so <cmath>-x_1x_3 = \boxed{002}</cmath>and we're done.<br />
<br />
==Solution 3==<br />
Observing the equation, we notice that the coefficient for the middle term <math>-4029</math> is equal to <cmath>-2{\sqrt{2014}}^2-1</cmath>. <br />
<br />
Also notice that the coefficient for the <math>{x^3}</math> term is <math>\sqrt{2014}</math>. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the <math>x</math> term of the binomial would have a coefficient of <math>\sqrt{2014}</math>. Similarly, the <math>x</math> term of the trinomial would also have a coefficient of <math>\sqrt{2014}</math>. The factored form of the expression would look something like the following:<br />
<cmath>({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)</cmath> where <math>{a, b,c}</math> are all positive integers (because the <math>{x^2}</math> term of the original expression is negative, and the constant term is positive), and <cmath>{ab=2}</cmath><br />
<br />
Multiplying this expression out gives <cmath>{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}</cmath> Equating this with the original expression gives <cmath>{2014n+a}=-4029</cmath> The only positive integer solutions of this expression is <math>(n, a)=(1, 2015)</math> or <math>(2, 1)</math>. If <math>(n, a)=(1, 2015)</math> then setting <math>{an{\sqrt{2014}}-b{\sqrt{2014}}}=0</math> yields <math>{b=2015}</math> and therefore <math>{ab=2015^2}</math> which clearly isn't equal to <math>2</math> as the constant term. Therefore, <math>(n, a)=(2, 1)</math> and the factored form of the expression is:<br />
<cmath>({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)</cmath> Therefore, one of the three roots of the original expression is <cmath>{x=\dfrac{1}{\sqrt{2014}}}</cmath><br />
Using the quadratic formula yields the other two roots as <cmath>{x={\sqrt{2014}}+{\sqrt{2016}}}</cmath> and <cmath>{x={\sqrt{2014}}-{\sqrt{2016}}}</cmath> Arranging the roots in ascending order (in the order <math>x_1<x_2<x_3</math>), <cmath>{\sqrt{2014}}-{\sqrt{2016}}<\dfrac{1}{\sqrt{2014}}<{\sqrt{2014}}+{\sqrt{2016}}</cmath><br />
Therefore, <cmath>x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}</cmath><br />
<br />
==Solution 4==<br />
By Vieta's, we are seeking to find <math>x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}</math>. Substitute <math>n=-x_1x_3</math> and <math>x_2=\frac{2}{\sqrt{2014}n}</math>. Substituting this back into the original equation, we have <math>\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0</math>, so <math>2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0</math>. Hence, <math>8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}</math>, and <math>n\equiv 2\pmod{1007}</math>. But since <math>n\le 999</math> because it is our desired answer, the only possible value for <math>n</math> is <math>\boxed{002}</math><br />
BEST PROOOFFFF<br />
Stormersyle & mathleticguyyy<br />
<br />
<br />
== Solution 5 ==<br />
<br />
Let <math>y =\frac{x}{\sqrt{2014}}.</math> The original equation simplifies to <math>\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029y + 4028=0.</math> Here we clearly see that <math>y=1</math> is a root. Divideing <math>y-1</math> from the sum we find that <math>(y-1)(y^2-4028y-4028)=0.</math> From simple bounding we see that <math>y=1</math> is the middle root. Therefore <math>x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.</math><br />
<br />
== Solution 6 ==<br />
<br />
<math>\sqrt{2014}</math> occurs multiple times, so let k = <math>\sqrt{2014}</math>.<br />
<br />
The equation becomes <math>0 = kx^3 - (2k^2 + 1)x^2 + 2</math>. Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula.<br />
<br />
We get the roots are:<br />
<br />
<math>y = \frac{1}{x}</math>, and <math>y = \frac{x}{2} - \frac{1}{x}</math>.<br />
<br />
In the first case, <math>x = \frac{1}{y} = \frac{1}{\sqrt{2014}}</math>.<br />
<br />
In the second case, <math>x^2 - 2\sqrt{2014} - 2 = 0</math>. The solutions are <math>\sqrt{2014} \pm \sqrt{2016}</math>. The sum of these 2 solutions is <math>2 \sqrt{2014}</math>, and <math>\frac{1}{\sqrt{2014}}</math> is the middle solution, and thus, <math>(x_1 + x_3) \cdot x_2 = 2</math><br />
<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_9&diff=1611702014 AIME I Problems/Problem 92021-08-29T16:31:01Z<p>Jdong2006: /* Solution 1 */</p>
<hr />
<div>== Problem 9 ==<br />
<br />
Let <math>x_1<x_2<x_3</math> be the three real roots of the equation <math>\sqrt{2014}x^3-4029x^2+2=0</math>. Find <math>x_2(x_1+x_3)</math>.<br />
<br />
== Solution 1==<br />
<br />
Substituting <math>n</math> for <math>2014</math>, we get <cmath>\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2</cmath> <cmath>= x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0</cmath> Noting that <math>nx^2 - 1</math> factors as a difference of squares to <cmath>(\sqrt{n}x - 1)(\sqrt{n}x+1)</cmath> we can factor the left side as <cmath>(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))</cmath> This means that <math>\frac{1}{\sqrt{n}}</math> is a root, and the other two roots are the roots of <math>x^2 - 2\sqrt{n}x - 2</math>. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to <math>2\sqrt{n}</math>, so the positive root must be greater than <math>2\sqrt{n}</math> in order to produce this sum when added to a negative value. Since <math>0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}</math> is clearly true, <math>x_2 = \frac{1}{\sqrt{2014}}</math> and <math>x_1 + x_3 = 2\sqrt{2014}</math>. Multiplying these values together, we find that <math>x_2(x_1+x_3) = \boxed{002}</math>.<br />
<br />
==Solution 2==<br />
From Vieta's formulae, we know that <cmath>x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}</cmath> <cmath>x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}}</cmath> and <cmath>x_1x_2 + x_2x_3 + x_1x_3 = 0</cmath> Thus, we know that <cmath>x_2(x_1 + x_3) = -x_1x_3</cmath><br />
<br />
Now consider the polynomial with roots <math>x_1x_2, x_2x_3,</math> and <math>x_1x_3</math>. Expanding the polynomial <cmath>(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)</cmath>we get the polynomial <cmath>x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2</cmath> Substituting the values obtained from Vieta's formulae, we find that this polynomial is <cmath>x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}</cmath> We know <math>x_1x_3</math> is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to <cmath>1007x^3 - 4029x - 2 = 0</cmath><br />
<br />
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the <math>x^3</math> term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that <math>x = -2</math> is a solution. Factoring it out, we get that <cmath>1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)</cmath> Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, <cmath>x_1x_3 = -2</cmath> so <cmath>-x_1x_3 = \boxed{002}</cmath>and we're done.<br />
<br />
==Solution 3==<br />
Observing the equation, we notice that the coefficient for the middle term <math>-4029</math> is equal to <math>-2{\sqrt{2014}}^2-1</math>. Also notice that the coefficient for the <math>{x^3}</math> term is <math>\sqrt{2014}</math>. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the <math>x</math> term of the binomial would have a coefficient of <math>\sqrt{2014}</math>. Similarly, the <math>x</math> term of the trinomial would also have a coefficient of <math>\sqrt{2014}</math>. The factored form of the expression would look something like the following:<br />
<math>({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)</math> where <math>{a, b,c}</math> are all positive integers (because the <math>{x^2}</math> term of the original expression is negative, and the constant term is positive), and <math>{ab=2}</math>.<br />
<br />
Multiplying this expression out gives <math>{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}</math>. Equating this with the original expression gives <math>{2014n+a}=-4029</math>. The only positive integer solutions of this expression is <math>(n, a)=(1, 2015)</math> or <math>(2, 1)</math>. If <math>(n, a)=(1, 2015)</math> then setting <math>{an{\sqrt{2014}}-b{\sqrt{2014}}}=0</math> yields <math>{b=2015}</math> and therefore <math>{ab=2015^2}</math> which clearly isn't equal to <math>2</math> as the constant term. Therefore, <math>(n, a)=(2, 1)</math> and the factored form of the expression is:<br />
<math>({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)</math>. Therefore, one of the three roots of the original expression is <math>{x=\dfrac{1}{\sqrt{2014}}}</math>.<br />
Using the quadratic formula yields the other two roots as <math>{x={\sqrt{2014}}+{\sqrt{2016}}}</math> and <math>{x={\sqrt{2014}}-{\sqrt{2016}}}</math>. Arranging the roots in ascending order (in the order <math>x_1<x_2<x_3</math>), <math>{\sqrt{2014}}-{\sqrt{2016}}<\dfrac{1}{\sqrt{2014}}<{\sqrt{2014}}+{\sqrt{2016}}</math>.<br />
Therefore, <math>x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}</math>.<br />
<br />
==Solution 4==<br />
By Vieta's, we are seeking to find <math>x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}</math>. Substitute <math>n=-x_1x_3</math> and <math>x_2=\frac{2}{\sqrt{2014}n}</math>. Substituting this back into the original equation, we have <math>\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0</math>, so <math>2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0</math>. Hence, <math>8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}</math>, and <math>n\equiv 2\pmod{1007}</math>. But since <math>n\le 999</math> because it is our desired answer, the only possible value for <math>n</math> is <math>\boxed{002}</math><br />
BEST PROOOFFFF<br />
Stormersyle & mathleticguyyy<br />
<br />
<br />
== Solution 5 ==<br />
<br />
Let <math>y =\frac{x}{\sqrt{2014}}.</math> The original equation simplifies to <math>\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029y + 4028=0.</math> Here we clearly see that <math>y=1</math> is a root. Divideing <math>y-1</math> from the sum we find that <math>(y-1)(y^2-4028y-4028)=0.</math> From simple bounding we see that <math>y=1</math> is the middle root. Therefore <math>x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.</math><br />
<br />
== Solution 6 ==<br />
<br />
<math>\sqrt{2014}</math> occurs multiple times, so let k = <math>\sqrt{2014}</math>.<br />
<br />
The equation becomes <math>0 = kx^3 - (2k^2 + 1)x^2 + 2</math>. Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula.<br />
<br />
We get the roots are:<br />
<br />
<math>y = \frac{1}{x}</math>, and <math>y = \frac{x}{2} - \frac{1}{x}</math>.<br />
<br />
In the first case, <math>x = \frac{1}{y} = \frac{1}{\sqrt{2014}}</math>.<br />
<br />
In the second case, <math>x^2 - 2\sqrt{2014} - 2 = 0</math>. The solutions are <math>\sqrt{2014} \pm \sqrt{2016}</math>. The sum of these 2 solutions is <math>2 \sqrt{2014}</math>, and <math>\frac{1}{\sqrt{2014}}</math> is the middle solution, and thus, <math>(x_1 + x_3) \cdot x_2 = 2</math><br />
<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_9&diff=1611692014 AIME I Problems/Problem 92021-08-29T16:29:59Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>== Problem 9 ==<br />
<br />
Let <math>x_1<x_2<x_3</math> be the three real roots of the equation <math>\sqrt{2014}x^3-4029x^2+2=0</math>. Find <math>x_2(x_1+x_3)</math>.<br />
<br />
== Solution 1==<br />
<br />
Substituting <math>n</math> for <math>2014</math>, we get <math>\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0</math>. Noting that <math>nx^2 - 1</math> factors as a difference of squares to <math>(\sqrt{n}x - 1)(\sqrt{n}x+1)</math>, we can factor the left side as <math>(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))</math>. This means that <math>\frac{1}{\sqrt{n}}</math> is a root, and the other two roots are the roots of <math>x^2 - 2\sqrt{n}x - 2</math>. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to <math>2\sqrt{n}</math>, so the positive root must be greater than <math>2\sqrt{n}</math> in order to produce this sum when added to a negative value. Since <math>0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}</math> is clearly true, <math>x_2 = \frac{1}{\sqrt{2014}}</math> and <math>x_1 + x_3 = 2\sqrt{2014}</math>. Multiplying these values together, we find that <math>x_2(x_1+x_3) = \boxed{002}</math>.<br />
<br />
<br />
==Solution 2==<br />
From Vieta's formulae, we know that <cmath>x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}</cmath> <cmath>x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}}</cmath> and <cmath>x_1x_2 + x_2x_3 + x_1x_3 = 0</cmath> Thus, we know that <cmath>x_2(x_1 + x_3) = -x_1x_3</cmath><br />
<br />
Now consider the polynomial with roots <math>x_1x_2, x_2x_3,</math> and <math>x_1x_3</math>. Expanding the polynomial <cmath>(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)</cmath>we get the polynomial <cmath>x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2</cmath> Substituting the values obtained from Vieta's formulae, we find that this polynomial is <cmath>x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}</cmath> We know <math>x_1x_3</math> is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to <cmath>1007x^3 - 4029x - 2 = 0</cmath><br />
<br />
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the <math>x^3</math> term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that <math>x = -2</math> is a solution. Factoring it out, we get that <cmath>1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)</cmath> Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, <cmath>x_1x_3 = -2</cmath> so <cmath>-x_1x_3 = \boxed{002}</cmath>and we're done.<br />
<br />
==Solution 3==<br />
Observing the equation, we notice that the coefficient for the middle term <math>-4029</math> is equal to <math>-2{\sqrt{2014}}^2-1</math>. Also notice that the coefficient for the <math>{x^3}</math> term is <math>\sqrt{2014}</math>. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the <math>x</math> term of the binomial would have a coefficient of <math>\sqrt{2014}</math>. Similarly, the <math>x</math> term of the trinomial would also have a coefficient of <math>\sqrt{2014}</math>. The factored form of the expression would look something like the following:<br />
<math>({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)</math> where <math>{a, b,c}</math> are all positive integers (because the <math>{x^2}</math> term of the original expression is negative, and the constant term is positive), and <math>{ab=2}</math>.<br />
<br />
Multiplying this expression out gives <math>{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}</math>. Equating this with the original expression gives <math>{2014n+a}=-4029</math>. The only positive integer solutions of this expression is <math>(n, a)=(1, 2015)</math> or <math>(2, 1)</math>. If <math>(n, a)=(1, 2015)</math> then setting <math>{an{\sqrt{2014}}-b{\sqrt{2014}}}=0</math> yields <math>{b=2015}</math> and therefore <math>{ab=2015^2}</math> which clearly isn't equal to <math>2</math> as the constant term. Therefore, <math>(n, a)=(2, 1)</math> and the factored form of the expression is:<br />
<math>({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)</math>. Therefore, one of the three roots of the original expression is <math>{x=\dfrac{1}{\sqrt{2014}}}</math>.<br />
Using the quadratic formula yields the other two roots as <math>{x={\sqrt{2014}}+{\sqrt{2016}}}</math> and <math>{x={\sqrt{2014}}-{\sqrt{2016}}}</math>. Arranging the roots in ascending order (in the order <math>x_1<x_2<x_3</math>), <math>{\sqrt{2014}}-{\sqrt{2016}}<\dfrac{1}{\sqrt{2014}}<{\sqrt{2014}}+{\sqrt{2016}}</math>.<br />
Therefore, <math>x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}</math>.<br />
<br />
==Solution 4==<br />
By Vieta's, we are seeking to find <math>x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}</math>. Substitute <math>n=-x_1x_3</math> and <math>x_2=\frac{2}{\sqrt{2014}n}</math>. Substituting this back into the original equation, we have <math>\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0</math>, so <math>2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0</math>. Hence, <math>8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}</math>, and <math>n\equiv 2\pmod{1007}</math>. But since <math>n\le 999</math> because it is our desired answer, the only possible value for <math>n</math> is <math>\boxed{002}</math><br />
BEST PROOOFFFF<br />
Stormersyle & mathleticguyyy<br />
<br />
<br />
== Solution 5 ==<br />
<br />
Let <math>y =\frac{x}{\sqrt{2014}}.</math> The original equation simplifies to <math>\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029y + 4028=0.</math> Here we clearly see that <math>y=1</math> is a root. Divideing <math>y-1</math> from the sum we find that <math>(y-1)(y^2-4028y-4028)=0.</math> From simple bounding we see that <math>y=1</math> is the middle root. Therefore <math>x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.</math><br />
<br />
== Solution 6 ==<br />
<br />
<math>\sqrt{2014}</math> occurs multiple times, so let k = <math>\sqrt{2014}</math>.<br />
<br />
The equation becomes <math>0 = kx^3 - (2k^2 + 1)x^2 + 2</math>. Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula.<br />
<br />
We get the roots are:<br />
<br />
<math>y = \frac{1}{x}</math>, and <math>y = \frac{x}{2} - \frac{1}{x}</math>.<br />
<br />
In the first case, <math>x = \frac{1}{y} = \frac{1}{\sqrt{2014}}</math>.<br />
<br />
In the second case, <math>x^2 - 2\sqrt{2014} - 2 = 0</math>. The solutions are <math>\sqrt{2014} \pm \sqrt{2016}</math>. The sum of these 2 solutions is <math>2 \sqrt{2014}</math>, and <math>\frac{1}{\sqrt{2014}}</math> is the middle solution, and thus, <math>(x_1 + x_3) \cdot x_2 = 2</math><br />
<br />
<br />
== See also ==<br />
{{AIME box|year=2014|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_9&diff=1611662011 AIME II Problems/Problem 92021-08-29T14:13:53Z<p>Jdong2006: /* Solution 2 (Not legit) */</p>
<hr />
<div>==Problem 9==<br />
Let <math>x_1, x_2, ... , x_6</math> be non-negative real numbers such that <math>x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1</math>, and <math>x_1 x_3 x_5 +x_2 x_4 x_6 \ge {\frac{1}{540}}</math>. Let <math>p</math> and <math>q</math> be positive relatively prime integers such that <math>\frac{p}{q}</math> is the maximum possible value of<br />
<math>x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2</math>. Find <math>p+q</math>.<br />
<br />
==Solution==<br />
Note that neither the constraint nor the expression we need to maximize involves products <math>x_i x_j</math> with <math>i \equiv j \pmod 3</math>. Factoring out say <math>x_1</math> and <math>x_4</math> we see that the constraint is <math>x_1(x_3x_5) + x_4(x_2x_6) \ge {\frac1{540}}</math>, while the expression we want to maximize is <math>x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)</math>. Adding the left side of the constraint to the expression, we get: <math>(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)</math>. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most <math>\frac1{27}</math>. Since we have added at least <math>\frac{1}{540}</math> the desired maximum is at most <math>\frac1{27} - \frac1{540} =\frac{19}{540}</math>. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to <math>\frac1{540}</math> with <math>x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =\frac13</math>&mdash;for example, by choosing <math>x_1</math> and <math>x_2</math> small enough&mdash;so our answer is <math>540 + 19 = \fbox{559}.</math><br />
<br />
An example is:<br />
<cmath><br />
\begin{align*}<br />
x_3 &= x_6 = \frac16 \\<br />
x_1 &= x_2 = \frac{5 - \sqrt{20}}{30} \\<br />
x_5 &= x_4 = \frac{5 + \sqrt{20}}{30}<br />
\end{align*}<br />
</cmath><br />
<br />
Another example is <br />
<cmath><br />
\begin{align*}<br />
x_1 = x_3 = \frac{1}{3} \\<br />
x_2 = \frac{19}{60}, \ x_5 = \frac{1}{60} \\<br />
x_4 &= x_6 = 0<br />
\end{align*}<br />
</cmath><br />
<br />
==Solution 2 (Not legit)==<br />
There's a symmetry between <math>x_1, x_3, x_5</math> and <math>x_2,x_4,x_6</math>. Therefore, a good guess is that <math>a = x_1 = x_3 = x_5</math> and <math>b = x_2 = x_4 = x_6</math>, at which point we know that <math>a+b = 1/3</math>, <math>a^3+b^3 \geq 1/540</math>, and we are trying to maximize <math>3a^2b+3ab^2</math>. Then,<br />
<br />
<cmath>3a^2b+3ab^2 = (a+b)^3-a^3-b^3 \leq \frac{1}{27} - \frac{1}{540} = \boxed{\frac{19}{540}}</cmath> which is the answer.<br />
<br />
This solution is extremely lucky; if you attempt to solve for <math>a</math> and <math>b</math> you receive complex answers (which contradict the problem statement), but the final answer is correct.<br />
<br />
==See also==<br />
{{AIME box|year=2011|n=II|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_9&diff=1611652010 AIME I Problems/Problem 92021-08-29T13:40:08Z<p>Jdong2006: /* Remark */</p>
<hr />
<div>== Problem ==<br />
Let <math>(a,b,c)</math> be the [[real number|real]] solution of the system of equations <math>x^3 - xyz = 2</math>, <math>y^3 - xyz = 6</math>, <math>z^3 - xyz = 20</math>. The greatest possible value of <math>a^3 + b^3 + c^3</math> can be written in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.<br />
<br />
== Solution ==<br />
<br />
===Solution 1===<br />
Add the three equations to get <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Now, let <math>abc = p</math>. <math>a = \sqrt [3]{p + 2}</math>, <math>b = \sqrt [3]{p + 6}</math> and <math>c = \sqrt [3]{p + 20}</math>, so <math>p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})</math>. Now [[cube]] both sides; the <math>p^3</math> terms cancel out. Solve the remaining [[quadratic]] to get <math>p = - 4, - \frac {15}{7}</math>. To maximize <math>a^3 + b^3 + c^3</math> choose <math>p = - \frac {15}{7}</math> and so the sum is <math>28 - \frac {45}{7} = \frac {196 - 45}{7}</math> giving <math>151 + 7 = \fbox{158}</math>.<br />
<br />
===Solution 2===<br />
This is almost the same as solution 1. Note <math>a^3 + b^3 + c^3 = 28 + 3abc</math>. Next, let <math>k = a^3</math>. Note that <math>b = \sqrt [3]{k + 4}</math> and <math>c = \sqrt [3]{k + 18}</math>, so we have <math>28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22</math>. Move 28 over, divide both sides by 3, then cube to get <math>k^3-6k^2+12k-8 = k^3+22k^2+18k</math>. The <math>k^3</math> terms cancel out, so solve the quadratic to get <math>k = -2, -\frac{1}{7}</math>. We maximize <math>abc</math> by choosing <math>k = -\frac{1}{7}</math>, which gives us <math>a^3+b^3+c^3 = 3k + 22 = \frac{151}{7}</math>. Thus, our answer is <math>151+7=\boxed{158}</math>.<br />
<br />
===Solution 3===<br />
We have that <math>x^3 = 2 + xyz</math>, <math>y^3 = 6 + xyz</math>, and <math>z^3 = 20 + xyz</math>. Multiplying the three equations, and letting <math>m = xyz</math>, we have that <math>m^3 = (2+m)(6+m)(20+m)</math>, and reducing, that <math>7m^2 + 43m + 60 = 0</math>, which has solutions <math>m = -\frac{15}{7}, -4</math>. Adding the three equations and testing both solutions, we find the answer of <math>\frac{151}{7}</math>, so the desired quantity is <math>151 + 7 = \fbox{158}</math>.<br />
<br />
== Remark ==<br />
It is tempting to add the equations and then use the well-known factorization <math>x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-xz-yz)</math>. Unfortunately such a factorization is just a red herring: it doesn't give much information on <math>a^3+b^3+c^3</math>.<br />
== Another Remark ==<br />
The real problem with adding the equations is that <math>x, y, z</math> are real numbers based on the problem, but the adding trick only works when <math>x, y, z</math> are integers.<br />
<br />
==Video Solution==<br />
https://youtu.be/LXct4j_rYfw<br />
<br />
~Shreyas S<br />
<br />
== See Also ==<br />
{{AIME box|year=2010|num-b=8|num-a=10|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=1611282011 AIME I Problems/Problem 72021-08-29T00:11:11Z<p>Jdong2006: /* Solution 2 */</p>
<hr />
<div>== Problem 7 ==<br />
Find the number of positive integers <math>m</math> for which there exist nonnegative integers <math>x_0</math>, <math>x_1</math> , <math>\dots</math> , <math>x_{2011}</math> such that<br />
<cmath>m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.</cmath><br />
<br />
==Solution 1==<br />
<math> m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{2011}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>. Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{2010}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.<br />
<br />
==Solution 2==<br />
Let <cmath>P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}</cmath>. The problem then becomes finding the number of positive integer roots <math>m</math> for which <math>P(m) = 0</math> and <math>x_0, x_1, ..., x_{2011}</math> are nonnegative integers. We plug in <math>m = 1</math> and see that <cmath>P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010</cmath> Now, we can say that <cmath>P(m) = (m-1)Q(m) - 2010</cmath> for some polynomial <math>Q(m)</math> with integer coefficients. Then if <math>P(m) = 0</math>, <math>(m-1)Q(m) = 2010</math>. Thus, if <math>P(m) = 0</math>, then <math>m-1 | 2010</math> .<br />
Now, we need to show that <cmath>m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}\forall m-1|2011 </cmath> We try with the first few <math>m</math> that satisfy this.<br />
For <math>m = 2</math>, we see we can satisfy this if <math>x_0 = 2010</math>, <math>x_1 = 2009</math>, <math>x_2 = 2008</math>, <math>\cdots</math> , <math>x_{2008} = 2</math>, <math>x_{2009} = 1</math>, <math> x_{2010} = 0</math>, <math>x_{2011} = 0</math>, because <cmath>2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots</cmath> (based on the idea <math>2^n + 2^n = 2^{n+1}</math>, leading to a chain of substitutions of this kind) <cmath>= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}</cmath>. Thus <math>2</math> is a possible value of <math>m</math>. For other values, for example <math>m = 3</math>, we can use the same strategy, with <cmath>x_{2011} = x_{2010} = x_{2009} = 0</cmath>, <cmath>x_{2008} = x_{2007} = 1</cmath>, <cmath>x_{2006} = x_{2005} = 2</cmath>, <cmath>\cdots</cmath>, <cmath>x_2 = x_1 = 1004</cmath> and <cmath>x_0 = 1005</cmath>, because <br />
<cmath>3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004}</cmath><br />
<cmath>=\cdots = 3^{1004} +3^{1004}+3^{1004} = 3^{1005}</cmath><br />
It's clearly seen we can use the same strategy for all <math>m-1 |2010</math>. We count all positive <math>m</math> satisfying <math>m-1 |2010</math>, and see there are <math>\boxed{016}</math><br />
<br />
==Solution 3==<br />
One notices that <math>m-1 \mid 2010</math> if and only if there exist non-negative integers <math>x_0,x_1,\ldots,x_{2011}</math> such that <math>m^{x_0} = \sum_{k=1}^{2011}m^{x_k}</math>.<br />
<br />
To prove the forward case, we proceed by directly finding <math>x_0,x_1,\ldots,x_{2011}</math>. Suppose <math>m</math> is an integer such that <math>m^{x_0} = \sum_{k=1}^{2011}m^{x_k}</math>. We will count how many <math>x_k = 0</math>, how many <math>x_k = 1</math>, etc. Suppose the number of <math>x_k = 0</math> is non-zero. Then, there must be at least <math>m</math> such <math>x_k</math> since <math>m</math> divides all the remaining terms, so <math>m</math> must also divide the sum of all the <math>m^0</math> terms. Thus, if we let <math>x_k = 0</math> for <math>k = 1,2,\ldots,m</math>, we have,<br />
<cmath>m^{x_0} = m + \sum_{k=m+1}^{2011}m^{x_k}.</cmath><br />
Well clearly, <math>m^{x_0}</math> is greater than <math>m</math>, so <math>m^2 \mid m^{x_0}</math>. <math>m^2</math> will also divide every term, <math>m^{x_k}</math>, where <math>x_k \geq 2</math>. So, all the terms, <math>m^{x_k}</math>, where <math>x_k < 2</math> must sum to a multiple of <math>m^2</math>. If there are exactly <math>m</math> terms where <math>x_k = 0</math>, then we must have at least <math>m-1</math> terms where <math>x_k = 1</math>. Suppose there are exactly <math>m-1</math> such terms and <math>x_k = 1</math> for <math>k = m+1,m+2,2m-1</math>. Now, we have,<br />
<cmath>m^{x_0} = m^2 + \sum_{k=2m}^{2011}m^{x_k}.</cmath><br />
One can repeat this process for successive powers of <math>m</math> until the number of terms reaches 2011. Since there are <math>m + j(m-1)</math> terms after the <math>j</math>th power, we will only hit exactly 2011 terms if <math>m-1</math> is a factor of 2010. To see this, <br />
<cmath>m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) = 2010 \Rightarrow (m-1)(j+1) = 2010.</cmath><br />
Thus, when <math>j = 2010/(m-1) - 1</math> (which is an integer since <math>m-1 \mid 2010</math> by assumption, there are exactly 2011 terms. To see that these terms sum to a power of <math>m</math>, we realize that the sum is a geometric series:<br />
<cmath>1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j = 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.</cmath><br />
Thus, we have found a solution for the case <math>m-1 \mid 2010</math>.<br />
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Now, for the reverse case, we use the formula <cmath>x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\cdots+1).</cmath> Suppose <math>m^{x_0} = \sum_{k=1}^{2011}m^{x^k}</math> has a solution. Subtract 2011 from both sides to get <cmath>m^{x_0}-1-2010 = \sum_{k=1}^{2011}(m^{x^k}-1).</cmath> Now apply the formula to get <cmath>(m-1)a_0-2010 = \sum_{k=1}^{2011}[(m-1)a_k],</cmath> where <math>a_k</math> are some integers. Rearranging this equation, we find <cmath>(m-1)A = 2010,</cmath> where <math>A = a_0 - \sum_{k=1}^{2011}a_k</math>. Thus, if <math>m</math> is a solution, then <math>m-1 \mid 2010</math>. <br />
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So, there is one positive integer solution corresponding to each factor of 2010. Since <math>2010 = 2\cdot 3\cdot 5\cdot 67</math>, the number of solutions is <math>2^4 = \boxed{016}</math>.<br />
<br />
==Solution 4 (for noobs like me)==<br />
The problem is basically asking how many integers <math>m</math> have a power that can be expressed as the sum of 2011 other powers of <math>m</math> (not necessarily distinct). Notice that <math>2+2+4+8+16=32</math>, <math>3+3+3+9+9+27+27+81+81=243</math>, and <math>4+4+4+4+16+16+16+64+64+64+256+256+256=1024</math>. Thus, we can safely assume that the equation <math>2011 = (m-1)x + m</math> must have an integer solution <math>x</math>. To find the number of <math>m</math>-values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant <math>m</math> to make the equation equal a friendlier number, <math>2010</math>, instead of the ugly prime number <math>2011</math>: <math>2010 = (m-1)x+(m-1)</math>. Factor the equation and we get <math>2010 = (m-1)(x+1)</math>. The number of values of <math>m-1</math> that allow <math>x+1</math> to be an integer is quite obviously the number of factors of <math>2010</math>. Factoring <math>2010</math>, we obtain <math>2010 = 2 \times 3 \times 5 \times 67</math>, so the number of positive integers <math>m</math> that satisfy the required condition is <math>2^4 = \boxed{016}</math>.<br />
<br />
-fidgetboss_4000<br />
<br />
==Solution 5==<br />
<br />
First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of <math>m</math> we can multiply to get to non-negative powers.<br />
Now we see that our problem is just a matter of m-chopping blocks.<br />
What is meant by <math>m</math>-chopping is taking an existing block of say <math>m^{k}</math> and turning it into <math>m</math> blocks of <math>m^{k-1}</math>. This process increases the total number of blocks by <math>m-1</math> per chop.<br />
The problem wants us to find the number of positive integers <math>m</math> where some number of chops will turn <math>1</math> block into <math>2011</math> such blocks, thus increasing the total amount by <math>2010= 2 \cdot 3 \cdot 5 \cdot 67</math>. Thus <math>m-1 | 2010</math>, and a cursory check on extreme cases will confirm that there are indeed <math>\boxed{016}</math> possible <math>m</math>s.<br />
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==An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)==<br />
https://artofproblemsolving.com/community/c6h84155p486903<br />
2001 Austrian-Polish Math Individual Competition #1<br />
~MSC<br />
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== See also ==<br />
{{AIME box|year=2011|n=I|num-b=6|num-a=8}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Jdong2006