https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Zeric&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T02:23:26ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_25&diff=1648572017 AMC 12A Problems/Problem 252021-11-08T05:11:17Z<p>Zeric: Added a new, simpler solution</p>
<hr />
<div>==Problem==<br />
<br />
The vertices <math>V</math> of a centrally symmetric hexagon in the complex plane are given by<br />
<cmath>V=\left\{ \sqrt{2}i,-\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.</cmath><br />
For each <math>j</math>, <math>1\leq j\leq 12</math>, an element <math>z_j</math> is chosen from <math>V</math> at random, independently of the other choices. Let <math>P={\prod}_{j=1}^{12}z_j</math> be the product of the <math>12</math> numbers selected. What is the probability that <math>P=-1</math>?<br />
<br />
<math>\textbf{(A) } \dfrac{5\cdot11}{3^{10}} \qquad \textbf{(B) } \dfrac{5^2\cdot11}{2\cdot3^{10}} \qquad \textbf{(C) } \dfrac{5\cdot11}{3^{9}} \qquad \textbf{(D) } \dfrac{5\cdot7\cdot11}{2\cdot3^{10}} \qquad \textbf{(E) } \dfrac{2^2\cdot5\cdot11}{3^{10}}</math><br />
<br />
==Solution 1==<br />
<br />
It is possible to solve this problem using elementary counting methods. This solution proceeds by a cleaner generating function.<br />
<br />
We note that <math>\pm \sqrt{2}i</math> both lie on the imaginary axis and each of the <math>\frac{1}{\sqrt{8}}(\pm 1\pm i)</math> have length <math>\frac{1}{2}</math> and angle of odd multiples of <math>\pi/4</math>, i.e. <math>\pi/4,3\pi/4,5\pi/4,7\pi/4</math>. When we draw these 6 complex numbers out on the complex plane, we get a crystal-looking thing. Note that the total number of ways to choose 12 complex numbers is <math>6^{12}</math>. Now we count the number of good combinations.<br />
<br />
We first consider the lengths. When we multiply 12 complex numbers together, their magnitudes multiply. Suppose we have <math>n</math> of the numbers <math>\pm \sqrt{2}i</math>; then we must have <math>\left(\sqrt{2}\right)^n\cdot\left(\frac{1}{2}\right)^{12-n}=1 \Longrightarrow n=8</math>. Having <math>n=8</math> will take care of the length of the product; now we need to deal with the angle.<br />
<br />
We require <math>\sum\theta\equiv\pi \mod 2\pi</math>. Letting <math>z</math> be <math>e^{i\pi/4}</math>, we see that the angles we have available are <math>\{z^1,z^2,z^3,z^5,z^6,z^7\}</math>, where we must choose exactly 8 angles from the set <math>\{z^2,z^6\}</math> and exactly 4 from the set <math>\{z^1,z^3,z^5,z^7\}</math>. If we found a good combination where we had <math>a_i</math> of each angle <math>z^i</math>, then the amount this would contribute to our count would be <math>\binom{12}{4,8}\cdot\binom{8}{a_2,a_6}\cdot\binom{4}{a_1,a_3,a_5,a_7}</math>. We want to add these all up. We proceed by generating functions.<br />
<br />
Consider <cmath>(t_2x^2+t_6x^6)^8(t_1x^1+t_3x^3+t_5x^5+t_7x^7)^4.</cmath> The expansion will be of the form <math>\sum_i\left(\sum_{\sum a=i} \binom{8}{a_2,a_6}\binom{4}{a_1,a_3,a_5,a_7}{t_1}^{a_1}{t_2}^{a_2}{t_3}^{a_3}{t_5}^{a_5}{t_6}^{a_6}{t_7}^{a_7}x^i \right)</math>. Note that if we reduced the powers of <math>x</math> mod <math>8</math> and fished out the coefficient of <math>x^4</math> and plugged in <math>t_i=1\ \forall\,i</math> (and then multiplied by <math>\binom{12}{4,8}</math>) then we would be done. Since plugging in <math>t_i=1</math> doesn't affect the <math>x</math>'s, we do that right away. The expression then becomes <cmath>x^{20}(1+x^4)^8(1+x^2+x^4+x^6)^4=x^{20}(1+x^4)^{12}(1+x^2)^4=x^4(1+x^4)^{12}(1+x^2)^4,</cmath> where the last equality is true because we are taking the powers of <math>x</math> mod <math>8</math>. Let <math>[x^n]f(x)</math> denote the coefficient of <math>x^n</math> in <math>f(x)</math>. Note <math>[x^4] x^4(1+x^4)^{12}(1+x^2)^4=[x^0](1+x^4)^{12}(1+x^2)^4</math>. We use the roots of unity filter, which states <cmath>\text{terms of }f(x)\text{ that have exponent congruent to }k\text{ mod }n=\frac{1}{n}\sum_{m=1}^n \frac{f(z^mx)}{z^{mk}},</cmath> where <math>z=e^{i\pi/n}</math>. In our case <math>k=0</math>, so we only need to find the average of the <math>f(z^mx)</math>'s.<br />
<cmath><br />
\begin{align*}<br />
z^0 &\Longrightarrow (1+x^4)^{12}(1+x^2)^4,\\<br />
z^1 &\Longrightarrow (1-x^4)^{12}(1+ix^2)^4,\\<br />
z^2 &\Longrightarrow (1+x^4)^{12}(1-x^2)^4,\\<br />
z^3 &\Longrightarrow (1-x^4)^{12}(1-ix^2)^4,\\<br />
z^4 &\Longrightarrow (1+x^4)^{12}(1+x^2)^4,\\<br />
z^5 &\Longrightarrow (1-x^4)^{12}(1+ix^2)^4,\\<br />
z^6 &\Longrightarrow (1+x^4)^{12}(1-x^2)^4,\\<br />
z^7 &\Longrightarrow (1-x^4)^{12}(1-ix^2)^4.<br />
\end{align*}<br />
</cmath><br />
We plug in <math>x=1</math> and take the average to find the sum of all coefficients of <math>x^{\text{multiple of 8}}</math>. Plugging in <math>x=1</math> makes all of the above zero except for <math>z^0</math> and <math>z^4</math>. Averaging, we get <math>2^{14}</math>. Now the answer is simply <cmath>\frac{\binom{12}{4,8}}{6^{12}}\cdot 2^{14}=\boxed{\frac{2^2\cdot 5\cdot 11}{3^{10}}}.</cmath><br />
<br />
==Solution 2==<br />
<br />
By changing <math>z_1</math> to <math>-z_1</math>, we can give a bijection between cases where <math>P=-1</math> and cases where <math>P=1</math>, so we'll just find the probability that <math>P=\pm 1</math> and divide by <math>2</math> in the end. Multiplying the hexagon's vertices by <math>i</math> doesn't change <math>P</math>, and switching any <math>z_j</math> with <math>-z_j</math> doesn't change the property <math>P=\pm 1</math>, so the probability that <math>P=\pm1</math> remains the same if we only select our <math>z_j</math>'s at random from<br />
<cmath><br />
\left\{a= \sqrt 2,\quad b=\frac1{\sqrt{8}}(1+i),\quad c=\frac1{\sqrt{8}}(1-i)\right\}.<br />
</cmath><br />
Since <math>|a|=\sqrt2</math> and <math>|b|=|c|=\frac12</math>, we must choose <math>a</math> exactly <math>8</math> times to make <math>|P|=1</math>. To ensure <math>P</math> is real, we must either choose <math>b</math> <math>4</math> times, <math>c</math> <math>4</math> times, or both <math>b</math> and <math>c</math> <math>2</math> times. This gives us a total of <br />
<cmath><br />
2\binom{12}{4}+\binom{12}{2}\binom{10}{2}=(12\cdot 11\cdot 10\cdot 9)\left(\frac1{12}+\frac14\right)=(2^3\cdot 3^3\cdot 5\cdot 11)\frac13<br />
</cmath><br />
good sequences <math>z_1,\dots,z_{12}</math>, and hence the final result is<br />
<cmath><br />
\frac12\cdot \frac{2^3\cdot 3^2\cdot 5\cdot 11}{3^{12}}=\boxed{(E)}.<br />
</cmath><br />
<br />
==Solution 3==<br />
<br />
We use generating functions and a roots of unity filter.<br />
Notice that all values in <math>V</math> are eighth roots of unity multiplied by a constant. Let <math>x</math> be a primitive eighth root of unity (<math>e^{\frac{i\pi}{4}}</math>). The numbers in <math>V</math> are then <math>\left \{ \frac{x}{2},x^2\sqrt2,\frac{x^3}{2},\frac{x^5}{2},x^6\sqrt2,\frac{x^7}{2} \right\}</math>. To have <math>P=-1</math>, we must have that <math>|P|=1</math>, so eight of the <math>(z_i)</math> must belong to <br />
<cmath>\left \{ x^2\sqrt2,x^6\sqrt2 \right\}<br />
</cmath><br />
and the other four must belong to <br />
<cmath><br />
\left \{ \frac{x}{2},\frac{x^3}{2},\frac{x^5}{2},\frac{x^7}{2} \right\}<br />
</cmath><br />
So, we write the generating function <br />
<cmath><br />
\left (x^2 +x^6 \right )^8 \left (x+x^3+x^5+x^7 \right )^4<br />
</cmath><br />
to describe the product. Note that this assumes that the <math>(z_i)</math> that belong to <math>\left \{ x^2\sqrt2,x^6\sqrt2 \right\}</math> come first, so we will need to multiply by <math>\binom{12}{4}</math> at the end. We now apply a roots of unity filter to find the sum of the coefficients of the exponents that are <math>4\pmod{8}</math>, or equivalently the coefficients of the powers that are multiples of <math>8</math> of the following function:<br />
<cmath><br />
P(x)=\left (x^2 +x^6 \right )^8 \left (x+x^3+x^5+x^7 \right )^4x^4<br />
</cmath><br />
Let <math>\zeta=e^{\frac{i\pi}{4}}</math>. We are looking for <math>\frac{P(1)+P(\zeta)+P(\zeta^2)+P(\zeta^3)+P(\zeta^4)+P(\zeta^5)+P(\zeta^6)+P(\zeta^7)}{8}</math>. <math>P(1)=P(-1)=2^{16}</math>, and all of the rest are equal to <math>0</math>. So, we get an answer of <math>\frac{2^{16}+2^{16}}{8\cdot 6^{12}}=\frac{2^{14}}{6^{12}}=\frac{2^2}{3^{10}}</math>. But wait! We need to multiply by <math>\binom{12}{4} =\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot3\cdot1}</math><math>=5\cdot9\cdot11</math>. So, the answer is <math>\boxed{\text{\textbf{(E) }} \frac{2^2\cdot5\cdot11}{3^{10}} }</math><br />
<br />
==Solution 4==<br />
We can write the points in polar form <math>(r, \theta)</math> as<br />
<cmath>V = \left\{ \left(\sqrt{2}, \frac{\pi}{2}\right), \left(\sqrt{2}, \frac{3\pi}{2}\right), \left(\frac{1}{2}, \frac{\pi}{4}\right), \left(\frac{1}{2}, \frac{3\pi}{4}\right), \left(\frac{1}{2}, \frac{5\pi}{4}\right), \left(\frac{1}{2}, \frac{7\pi}{4}\right)\right\}.</cmath>Note that when multiplying complex numbers, the <math>r</math>'s multiply and the <math>\theta</math>'s add, and since <math>-1 = (1, \pi),</math> we need <math>8</math> complex numbers with <math>r = \sqrt{2} = 2^{0.5}</math> and <math>4</math> with <math>r = \frac{1}{2} = 2^{-1}.</math> By binomial distribution, the probability of this occurring is <math>\dbinom{12}{4} \left(\frac{1}{3}\right)^8 \left(\frac{2}{3}\right)^4.</math> For the <math>\theta</math> part, note that <math>\frac{4\theta}{\pi}</math> must be congruent to <math>4 \mod 8, </math> and by using simple symmetry, the probability of the aforementioned occurring is <math>\frac{1}{4}. </math> This is since <math>2a + 6b</math> is even for all <math>a + b =8, </math> and the number of ordered quadruples <math>(a_1, a_2, a_3, a_4)</math> such that <math>a_i \in \{1, 3, 5, 7\}</math> for all <math>1 \leq i \leq 4</math> and <math>a_1 + a_2 + a_3 + a_4 \equiv 2k \mod 8</math> is the same for all <math>1 \leq k \leq 4,</math> again by using symmetry. Thus, our probability is <cmath>\left(\frac{1}{4}\right) \left(\frac{1}{3}\right)^8 \left(\frac{2}{3}\right)^4 \dbinom{12}{4} = \frac{2^2}{3^{12}} \dbinom{12}{4} = \boxed{\frac{2^2 \cdot 11 \cdot 5}{3^{10}}}.</cmath><br />
<br />
-fidgetboss_4000<br />
<br />
==Solution 5 (Simple)==<br />
The absolute value of the first two complex numbers is <math>\sqrt{2}</math> while the absolute value of the latter four is <math>\frac12</math>. For the absolute value of the product to be <math>1</math>, we need <math>8</math> elements with absolute value <math>\sqrt{2}</math> and <math>4</math> elements with absolute value of <math>\frac12</math>.<br />
<br />
We pick <math>8</math> elements from <math>A = \left\{ \sqrt{2}i,-\sqrt{2}i\right\}</math> and <math>4</math> elements from <math>B = \left\{\frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}</math>. We also need to choose which <math>8</math> of <math>z_j</math> will be chosen from <math>A</math> which gives us <math>{12 \choose 8} \cdot 2^8 \cdot 4^4</math> cases. However, suppose we have chosen <math>11</math> elements and we need to choose one more element from <math>B</math>. The product of these <math>11</math> elements can have any of these values: <math>\left\{ 2\mathrm{cis}(\frac{\pi}{4}), 2\mathrm{cis}(\frac{3\pi}{4}), 2\mathrm{cis}(\frac{5\pi}{4}), 2\mathrm{cis}(\frac{7\pi}{4}), \right\}</math>. For either of these values, there is just one value of <math>z_{12} \in B</math> such that <math>P = -1</math> so we must divide by <math>4</math> which gives <math>{12 \choose 8} \cdot 2^8 \cdot 4^3</math> cases. <br />
<br />
Because there are <math>6^{12}</math> ways to pick any <math>12</math> elements, our probability is simply <math>\frac{{12 \choose 8} \cdot 2^8 \cdot 4^3}{6^{12}} = \frac{2^2}{3^{12}} {12 \choose 8} = \boxed{\text{\textbf{(E) }} \frac{2^2\cdot5\cdot11}{3^{10}} }</math><br />
<br />
~Zeric<br />
<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=A|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_25&diff=1648192017 AMC 12B Problems/Problem 252021-11-07T19:52:04Z<p>Zeric: Added some clarification and an alternative approximation for the solution</p>
<hr />
<div>==Problem==<br />
<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no two teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>9</math> people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9\leq n\leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562</math><br />
<br />
==Solution==<br />
<br />
Solution by Pieater314159<br />
<br />
Minor edits by Zeric<br />
<br />
Let there be <math>T</math> teams. For each team, there are <math>{n-5\choose 4}</math> different subsets of <math>9</math> players including that full team, so the total number of team-(group of 9) pairs is <br />
<br />
<cmath>T{n-5\choose 4}.</cmath><br />
<br />
Thus, the expected value of the number of full teams in a random set of <math>9</math> players is<br />
<br />
<cmath>\frac{T{n-5\choose 4}}{{n\choose 9}}.</cmath><br />
<br />
Similarly, the expected value of the number of full teams in a random set of <math>8</math> players is<br />
<br />
<cmath>\frac{T{n-5\choose 3}}{{n\choose 8}}.</cmath><br />
<br />
The condition is thus equivalent to the existence of a positive integer <math>T</math> such that<br />
<br />
<cmath>\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.</cmath><br />
<br />
<cmath>T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1</cmath><br />
<br />
<cmath>T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}</cmath><br />
<br />
<cmath>T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}</cmath><br />
<br />
<cmath>T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}</cmath><br />
<br />
<cmath>T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}</cmath><br />
<br />
Note that this is always less than <math>{n\choose 5}</math>, so as long as <math>T</math> is integral, <math>n</math> is a possibility. Thus, we have that this is equivalent to <br />
<br />
<cmath>2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).</cmath><br />
<br />
It is obvious that <math>5</math> divides the RHS, and that <math>7</math> does iff <math>n\equiv 0,1,2,3,4\mod 7</math>. Also, <math>3^2</math> divides it iff <math>n\not\equiv 5,8\mod 9</math>. One can also bash out that <math>2^5</math> divides it in <math>16</math> out of the <math>32</math> possible residues <math>\mod 32</math>. <br />
<br />
Note that <math>2016 = 7*9*32</math> so by using all numbers from <math>2</math> to <math>2017</math>, inclusive, it is clear that each possible residue <math>\mod 7,9,32</math> is reached an equal number of times, so the total number of working <math>n</math> in that range is <math>5\cdot 7\cdot 16 = 560</math>. However, we must subtract the number of "working" <math>2\leq n\leq 8</math>, which is <math>3</math>. Thus, the answer is <math>\boxed{\textbf{(D) } 557}</math>.<br />
<br />
Alternatively, it is enough to approximate by finding the floor of <math>2017 \cdot \frac57 \cdot \frac79 \cdot \frac12 - 3</math> to get <math>\boxed{\textbf{(D) } 557}</math>.<br />
<br />
Video Solution by Dr. Nal:<br />
<br />
https://www.youtube.com/watch?v=2p2qYRWbvV4&feature=emb_logo<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_25&diff=1648182017 AMC 12B Problems/Problem 252021-11-07T19:51:08Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
<br />
A set of <math>n</math> people participate in an online video basketball tournament. Each person may be a member of any number of <math>5</math>-player teams, but no two teams may have exactly the same <math>5</math> members. The site statistics show a curious fact: The average, over all subsets of size <math>9</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>9</math> people is equal to the reciprocal of the average, over all subsets of size <math>8</math> of the set of <math>n</math> participants, of the number of complete teams whose members are among those <math>8</math> people. How many values <math>n</math>, <math>9\leq n\leq 2017</math>, can be the number of participants?<br />
<br />
<math>\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562</math><br />
<br />
==Solution==<br />
<br />
Solution by Pieater314159<br />
Minor edits by Zeric<br />
<br />
Let there be <math>T</math> teams. For each team, there are <math>{n-5\choose 4}</math> different subsets of <math>9</math> players including that full team, so the total number of team-(group of 9) pairs is <br />
<br />
<cmath>T{n-5\choose 4}.</cmath><br />
<br />
Thus, the expected value of the number of full teams in a random set of <math>9</math> players is<br />
<br />
<cmath>\frac{T{n-5\choose 4}}{{n\choose 9}}.</cmath><br />
<br />
Similarly, the expected value of the number of full teams in a random set of <math>8</math> players is<br />
<br />
<cmath>\frac{T{n-5\choose 3}}{{n\choose 8}}.</cmath><br />
<br />
The condition is thus equivalent to the existence of a positive integer <math>T</math> such that<br />
<br />
<cmath>\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.</cmath><br />
<br />
<cmath>T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1</cmath><br />
<br />
<cmath>T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}</cmath><br />
<br />
<cmath>T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}</cmath><br />
<br />
<cmath>T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}</cmath><br />
<br />
<cmath>T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}</cmath><br />
<br />
Note that this is always less than <math>{n\choose 5}</math>, so as long as <math>T</math> is integral, <math>n</math> is a possibility. Thus, we have that this is equivalent to <br />
<br />
<cmath>2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).</cmath><br />
<br />
It is obvious that <math>5</math> divides the RHS, and that <math>7</math> does iff <math>n\equiv 0,1,2,3,4\mod 7</math>. Also, <math>3^2</math> divides it iff <math>n\not\equiv 5,8\mod 9</math>. One can also bash out that <math>2^5</math> divides it in <math>16</math> out of the <math>32</math> possible residues <math>\mod 32</math>. <br />
<br />
Note that <math>2016 = 7*9*32</math> so by using all numbers from <math>2</math> to <math>2017</math>, inclusive, it is clear that each possible residue <math>\mod 7,9,32</math> is reached an equal number of times, so the total number of working <math>n</math> in that range is <math>5\cdot 7\cdot 16 = 560</math>. However, we must subtract the number of "working" <math>2\leq n\leq 8</math>, which is <math>3</math>. Thus, the answer is <math>\boxed{\textbf{(D) } 557}</math>.<br />
<br />
Alternatively, it is enough to approximate by finding the floor of <math>2017 \cdot \frac57 \cdot \frac79 \cdot \frac12 - 3</math> to get <math>\boxed{\textbf{(D) } 557}</math>.<br />
<br />
Video Solution by Dr. Nal:<br />
<br />
https://www.youtube.com/watch?v=2p2qYRWbvV4&feature=emb_logo<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_24&diff=1647802017 AMC 12B Problems/Problem 242021-11-07T06:17:02Z<p>Zeric: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\tfrac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields:<br />
<cmath>[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))</cmath><br />
<cmath>[ABCD]=[AED]+[DEC]+[CEB]+[BEA]</cmath><br />
<cmath>(AB+CD)(BC)/2= 17*[CEB]+ [CEB] + [CEB] + [CEB]</cmath><br />
<cmath>(x^3+x)/2=(20x^3)/(2(x^2+1))</cmath><br />
<cmath>(x)(x^2+1)=20x^3/(x^2+1)</cmath><br />
<cmath>(x^2+1)^2=20x^2</cmath><br />
<cmath>x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5</cmath><br />
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math><br />
<br />
<br />
==Solution 2== <br />
Draw line <math>FG</math> through <math>E</math>, with <math>F</math> on <math>BC</math> and <math>G</math> on <math>AD</math>, <math>FG \parallel AB</math>. WLOG let <math>CD=1</math>, <math>CB=x</math>, <math>AB=x^2</math>. By weighted average <math>FG=\frac{1+x^4}{1+x^2}</math>. <br />
<br />
Meanwhile, <math>FE:EG=[\triangle CBE]:[\triangle ADE]=1:17</math>. This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio <math>EG:FE</math>.<br />
<br />
<math>FE=\frac{x^2}{1+x^2}</math>. We obtain <math>\frac{1+x^4}{1+x^2}=\frac{18x^2}{1+x^2}</math>,<br />
namely <math>x^4-18x^2+1=0</math>.<br />
<br />
The rest is the same as Solution 1.<br />
<br />
== Solution 3==<br />
Let <math>A=(-1,4a), B=(-1,0), C=(1,0), D=\bigg(1,\frac{1}{a}\bigg)</math>. Then from the similar triangles condition, we compute <math>CE=\frac{4a}{\sqrt{4a^2+1}}</math> and <math>BE=\frac{2}{\sqrt{4a^2+1}}</math>. Hence, the <math>y</math>-coordinate of <math>E</math> is just <math>\frac{BE\cdot CE}{BC}=\frac{4a}{4a^2+1}</math>. Since <math>E</math> lies on the unit circle, we can compute the <math>x</math> coordinate as <math>\frac{1-4a^2}{4a^2+1}</math>. By Shoelace, we want <cmath>\frac{1}{2}\det\begin{bmatrix}<br />
-1 & 4a & 1\\ <br />
\frac{1-4a^2}{4a^2+1} & \frac{4a}{4a^2+1} & 1\\ <br />
1 & \frac{1}{a} & 1<br />
\end{bmatrix}=17\cdot\frac{1}{2}\cdot 2 \cdot \frac{4a}{4a^2+1}</cmath>Factoring out denominators and expanding by minors, this is equivalent to<br />
<cmath>\frac{32a^4-8a^2+2}{2a(4a^2+1)}=\frac{68a}{4a^2+1} \Longrightarrow 16a^4-72a^2+1=0</cmath>This factors as <math>(4a^2-8a-1)(4a^2+8a-1)=0</math>, so <math>a=1+\frac{\sqrt{5}}{2}</math> and so the answer is <math> \textbf{(D) \ }</math>.<br />
<br />
==Solution 4==<br />
<br />
Let <math>C = (0,0), D=(\frac1a, 0), B = (0,1), A = (a,1)</math> where <math>a>1</math>. Because <math>BC = 1, a = \frac{AB}{BC}</math>. Notice that the diagonals are perpendicular with slopes of <math>\frac1a</math> and <math>-a</math>. Let the intersection of <math>AC</math> and <math>BD</math> be <math>F</math>, then <math>\triangle BFC \sim \triangle ABC</math>. However, because <math>ABCD</math> is a trapezoid, <math>\triangle</math><math>BCF</math> and <math>\triangle ADF</math> share the same area, therefore <math>\triangle</math><math>BCE</math> is the reflection of <math>\triangle</math><math>BCF</math> over the perpendicular bisector of <math>BC</math>, which is <math>y=\frac12</math>. We use the linear equations of the diagonals, <math>y = -ax + 1, y = \frac1a x</math>, to find the coordinates of <math>F</math>. <cmath>-ax+1 = \frac1ax \Longrightarrow x = \frac{1}{a+\frac1a} = \frac{a}{a^2+1}</cmath> <cmath>y = \frac1ax = \frac{1}{a^2+1}</cmath><br />
The y-coordinate of <math>E</math> is simply <math>1-\frac{1}{a^2+1} = \frac{a^2}{a^2+1}</math><br />
The area of <math>\triangle BCE</math> is <math>\frac12 \frac{a}{a^2+1}</math>. We apply shoelace theorem to solve for the area of <math>\triangle ADE</math>. The coordinates of the triangle are <math>\{(\frac{a}{a^2+1}, \frac{a^2}{a^2+1}), (a,1), (\frac1a, 0)\}</math>, so the area is <br />
<br />
<cmath>\frac12 |\frac{a^3}{a^2+1} + \frac1a - \frac{a}{a^2+1} - \frac{a}{a^2+1}| = \frac12 |\frac{a^3-2a}{a^2+1} + \frac1a|</cmath><br />
<cmath>= \frac12 |\frac{a^4-2a^2}{a(a^2+1)} + \frac{a^2+1}{a(a^2+1)}| = \frac12 \frac{a^4-a^2+1}{a(a^2+1)}</cmath><br />
Finally, we use the property that the ratio of areas equals <math>17</math> <cmath>\frac{\frac12}{\frac12} \frac{\frac{a^4-a^2+1}{a(a^2+1)}}{\frac{a}{a^2+1}} = 17 \Rightarrow \frac{a^4-a^2+1}{a^2} = 17 \Rightarrow a^4-18a^2+1 = 0</cmath><br />
<cmath>a^2 = 9+4\sqrt{5} = (2+\sqrt{5})^2 \Longrightarrow a = \boxed{\textbf{(D) } 2+\sqrt{5}}</cmath><br />
<br />
~Zeric<br />
<br />
== Notes==<br />
1) <math>\sqrt{17}</math> is the most relevant answer choice because it shares numbers with the givens of the problem.<br />
<br />
2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice).<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_24&diff=1647792017 AMC 12B Problems/Problem 242021-11-07T06:14:59Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
<br />
Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\tfrac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields:<br />
<cmath>[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))</cmath><br />
<cmath>[ABCD]=[AED]+[DEC]+[CEB]+[BEA]</cmath><br />
<cmath>(AB+CD)(BC)/2= 17*[CEB]+ [CEB] + [CEB] + [CEB]</cmath><br />
<cmath>(x^3+x)/2=(20x^3)/(2(x^2+1))</cmath><br />
<cmath>(x)(x^2+1)=20x^3/(x^2+1)</cmath><br />
<cmath>(x^2+1)^2=20x^2</cmath><br />
<cmath>x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5</cmath><br />
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math><br />
<br />
<br />
==Solution 2== <br />
Draw line <math>FG</math> through <math>E</math>, with <math>F</math> on <math>BC</math> and <math>G</math> on <math>AD</math>, <math>FG \parallel AB</math>. WLOG let <math>CD=1</math>, <math>CB=x</math>, <math>AB=x^2</math>. By weighted average <math>FG=\frac{1+x^4}{1+x^2}</math>. <br />
<br />
Meanwhile, <math>FE:EG=[\triangle CBE]:[\triangle ADE]=1:17</math>. This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio <math>EG:FE</math>.<br />
<br />
<math>FE=\frac{x^2}{1+x^2}</math>. We obtain <math>\frac{1+x^4}{1+x^2}=\frac{18x^2}{1+x^2}</math>,<br />
namely <math>x^4-18x^2+1=0</math>.<br />
<br />
The rest is the same as Solution 1.<br />
<br />
== Solution 3==<br />
Let <math>A=(-1,4a), B=(-1,0), C=(1,0), D=\bigg(1,\frac{1}{a}\bigg)</math>. Then from the similar triangles condition, we compute <math>CE=\frac{4a}{\sqrt{4a^2+1}}</math> and <math>BE=\frac{2}{\sqrt{4a^2+1}}</math>. Hence, the <math>y</math>-coordinate of <math>E</math> is just <math>\frac{BE\cdot CE}{BC}=\frac{4a}{4a^2+1}</math>. Since <math>E</math> lies on the unit circle, we can compute the <math>x</math> coordinate as <math>\frac{1-4a^2}{4a^2+1}</math>. By Shoelace, we want <cmath>\frac{1}{2}\det\begin{bmatrix}<br />
-1 & 4a & 1\\ <br />
\frac{1-4a^2}{4a^2+1} & \frac{4a}{4a^2+1} & 1\\ <br />
1 & \frac{1}{a} & 1<br />
\end{bmatrix}=17\cdot\frac{1}{2}\cdot 2 \cdot \frac{4a}{4a^2+1}</cmath>Factoring out denominators and expanding by minors, this is equivalent to<br />
<cmath>\frac{32a^4-8a^2+2}{2a(4a^2+1)}=\frac{68a}{4a^2+1} \Longrightarrow 16a^4-72a^2+1=0</cmath>This factors as <math>(4a^2-8a-1)(4a^2+8a-1)=0</math>, so <math>a=1+\frac{\sqrt{5}}{2}</math> and so the answer is <math> \textbf{(D) \ }</math>.<br />
<br />
==Solution 4==<br />
<br />
Let <math>C = (0,0), D=(\frac1a, 0), B = (0,1), A = (a,1)</math> where <math>a>1</math>. Because <math>BC = 1, a = \frac{AB}{BC}</math>. Notice that the diagonals are perpendicular with slopes of <math>\frac1a</math> and <math>-a</math>. Let the intersection of <math>AC</math> and <math>BD</math> be <math>F</math>, then <math>\triangle BFC \sim \triangle ABC</math>. However, because <math>ABCD</math> is a trapezoid, <math>\triangle</math><math>BCF</math> and <math>\triangle ADF</math> share the same area, therefore <math>\triangle</math><math>BCE</math> is the reflection of <math>\triangle</math><math>BCF</math> over the perpendicular bisector of <math>BC</math>, which is <math>y=\frac12</math>. We use the linear equations of the diagonals, <math>y = -ax + 1, y = \frac1a x</math>, to find the coordinates of <math>F</math>. <cmath>-ax+1 = \frac1ax \Longrightarrow x = \frac{1}{a+\frac1a} = \frac{a}{a^2+1}</cmath> <cmath>y = \frac1ax = \frac{1}{a^2+1}</cmath><br />
The y-coordinate of <math>E</math> is simply <math>1-\frac{1}{a^2+1} = \frac{a^2}{a^2+1}</math><br />
The area of <math>\triangle BCE</math> is <math>\frac12 \frac{a}{a^2+1}</math>. We apply shoelace theorem to solve for the area of <math>\triangle ADE</math>. The coordinates of the triangle are <math>(\frac{a}{a^2+1}, \frac{a^2}{a^2+1}), (a,1), (\frac1a, 0)</math>, so the area is <br />
<br />
<cmath>\frac12 |\frac{a^3}{a^2+1} + \frac1a - \frac{a}{a^2+1} - \frac{a}{a^2+1}| = \frac12 |\frac{a^3-2a}{a^2+1} + \frac1a|</cmath><br />
<cmath>= \frac12 |\frac{a^4-2a^2}{a(a^2+1)} + \frac{a^2+1}{a(a^2+1)}| = \frac12 \frac{a^4-a^2+1}{a(a^2+1)}</cmath><br />
Finally, we use the property that the ratio of areas equals <math>17</math> <cmath>\frac{\frac12}{\frac12} \frac{\frac{a^4-a^2+1}{a(a^2+1)}}{\frac{a}{a^2+1}} = 17 \Rightarrow \frac{a^4-a^2+1}{a^2} = 17 \Rightarrow a^4-18a^2+1 = 0</cmath><br />
<cmath>a^2 = 9+4\sqrt{5} = (2+\sqrt{5})^2 \Longrightarrow a = \boxed{\textbf{(D) } 2+\sqrt{5}}</cmath><br />
<br />
~Zeric<br />
<br />
== Notes==<br />
1) <math>\sqrt{17}</math> is the most relevant answer choice because it shares numbers with the givens of the problem.<br />
<br />
2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice).<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=1647692020 AMC 10A Problems/Problem 202021-11-07T02:53:51Z<p>Zeric: </p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #18]] and [[2020 AMC 10A Problems|2020 AMC 10A #20]]}}<br />
<br />
== Problem ==<br />
Quadrilateral <math>ABCD</math> satisfies <math>\angle ABC = \angle ACD = 90^{\circ}, AC=20,</math> and <math>CD=30.</math> Diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at point <math>E,</math> and <math>AE=5.</math> What is the area of quadrilateral <math>ABCD?</math><br />
<br />
<math>\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370</math><br />
<br />
== Solution 1 (Just Drop An Altitude)==<br />
<br />
<asy><br />
size(15cm,0);<br />
import olympiad;<br />
draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);<br />
label("A", (0,2), NW);<br />
label("B", (0,0), SW);<br />
label("C", (4,0), SE);<br />
label("D", (6,4), NE);<br />
label("E", (1.714,1.143), N);<br />
label("F", (1,1.5), N);<br />
draw((0,2)--(4,0), dashed);<br />
draw((0,0)--(6,4), dashed);<br />
draw((0,0)--(1,1.5), dashed);<br />
label("20", (0,2)--(4,0), SW);<br />
label("30", (4,0)--(6,4), SE);<br />
label("$x$", (1,1.5)--(1.714,1.143), NE);<br />
draw(rightanglemark((0,2),(0,0),(4,0)));<br />
draw(rightanglemark((0,2),(4,0),(6,4)));<br />
draw(rightanglemark((0,0),(1,1.5),(0,2)));<br />
</asy><br />
<br />
It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. <br />
<br />
By dropping this altitude, we can also see two similar triangles, <math>\triangle BFE \sim \triangle DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. <br />
<br />
Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math> because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into. <br />
<br />
Expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, that means the altitude <math>BF=2\cdot3=6</math>, or <math>[ABC]=60</math>. Thus <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}</math><br />
<br />
~ Solution by Ultraman<br />
<br />
~ Diagram by ciceronii<br />
<br />
~ Formatting by BakedPotato66<br />
<br />
==Solution 2 (Coordinates)==<br />
<asy><br />
size(10cm,0);<br />
draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br />
draw((-20,0)--(20,0));<br />
draw((0,-15)--(0,35));<br />
draw((10,30)--(-8,-6));<br />
draw(circle((0,0),10));<br />
label("E",(-4.05,-.25),S);<br />
label("D",(10,30),NE);<br />
label("C",(10,0),NE);<br />
label("B",(-8,-6),SW);<br />
label("A",(-10,0),NW);<br />
label("5",(-10,0)--(-5,0), NE);<br />
label("15",(-5,0)--(10,0), N);<br />
label("30",(10,0)--(10,30), E);<br />
dot((-5,0));<br />
dot((-10,0));<br />
dot((-8,-6));<br />
dot((10,0));<br />
dot((10,30));<br />
</asy><br />
Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>.<br />
<br />
==Solution 3 (Trigonometry)==<br />
Let <math>\angle C = \angle{ACB}</math> and <math>\angle{B} = \angle{CBE}.</math> Using Law of Sines on <math>\triangle{BCE}</math> we get <cmath>\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}</cmath> and LoS on <math>\triangle{ABE}</math> yields <cmath>\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.</cmath> Divide the two to get <math>\tan{B} = 3 \tan{C}.</math> Now, <cmath>\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}</cmath> and solve the quadratic, taking the positive solution (C is acute) to get <math>\tan{C} = \frac{1}{3}.</math> So if <math>AB = a,</math> then <math>BC = 3a</math> and <math>[ABC] = \frac{3a^2}{2}.</math> By Pythagorean Theorem, <math>10a^2 = 400 \iff \frac{3a^2}{2} = 60</math> and the answer is <math>300 + 60 \iff \boxed{\textbf{(D)}}.</math><br />
<br />
(This solution is incomplete, can someone complete it please-Lingjun) ok<br />
Latex edited by kc5170<br />
<br />
We could use the famous m-n rule in trigonometry in <math>\triangle ABC</math> with Point <math>E</math> <br />
[Unable to write it here.Could anybody write the expression]<br />
. We will find that <math>\overrightarrow{BD}</math> is an angle bisector of <math>\triangle ABC</math> (because we will get <math>\tan(x) = 1</math>). <br />
Therefore by converse of angle bisector theorem <math>AB:BC = 1:3</math>. By using Pythagorean theorem, we have values of <math>AB</math> and <math>AC</math>. <br />
Computing <math>AB \cdot AC = 120</math>. Adding the areas of <math>ABC</math> and <math>ACD</math>, hence the answer is <math>\boxed{\textbf{(D)}\:360}</math>.<br />
<br />
By: Math-Amaze<br />
Latex: Catoptrics.<br />
<br />
==Solution 4 (Answer Choices)==<br />
We know that the big triangle has area 300. Using the answer choices would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. We guess that the legs are equal to <math>\sqrt{20a}</math> and <math>\sqrt{20b}</math>, and because the hypotenuse is 20 we get <math>a+b=20</math>. Testing small numbers, we get that when <math>a=2</math> and <math>b=18</math>, <math>ab</math> is indeed a square. The area of the triangle is thus <math>60</math>, so the answer is <math>\boxed {\textbf{(D) }360}</math>.<br />
<br />
~tigershark22<br />
<br />
==Solution 5 (Law of Cosines)==<br />
<br />
<asy><br />
import olympiad;<br />
pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798);<br />
dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);<br />
draw(A--B--C--D--A);<br />
draw(A--C, dotted); draw(B--D, dotted);<br />
</asy><br />
<br />
Denote <math>EB</math> as <math>x</math>. By the Law of Cosine:<br />
<cmath>AB^2 = 25 + x^2 - 10x\cos(\angle DEC)</cmath><br />
<cmath>BC^2 = 225 + x^2 + 30x\cos(\angle DEC)</cmath><br />
<br />
Adding these up yields:<br />
<cmath>400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0</cmath><br />
By the quadratic formula, <math>x = 3\sqrt5</math>.<br />
<br />
Observe:<br />
<cmath>[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60</cmath>.<br />
<br />
Thus the desired area is <math>\frac{1}{2}(30)(20) + 60 = \boxed{\textbf{(D) } 360}</math><br />
<br />
~qwertysri987<br />
<br />
==Solution 6 (Basic Vectors / Coordinates)==<br />
<br />
Let <math>C = (0, 0)</math> and <math>D = (0, 30)</math>. Then <math>E = (-15, 0), A = (-20, 0),</math> and <math>B</math> lies on the line <math>y=2x+30.</math> So the coordinates of <math>B</math> are <cmath>(x, 2x+30).</cmath><br />
<br />
We can make this a vector problem.<br />
<math>\overrightarrow{\mathbf{B}} = \begin{pmatrix}<br />
x \\<br />
2x+30<br />
\end{pmatrix}.</math> We notice that point <math>B</math> forms a right angle, meaning vectors <math>\overrightarrow{\mathbf{BC}}</math> and <math>\overrightarrow{\mathbf{BA}}</math> are orthogonal, and their dot-product is <math>0</math>.<br />
<br />
We determine <math>\overrightarrow{\mathbf{BC}}</math> and <math>\overrightarrow{\mathbf{BA}}</math> to be <math>\begin{pmatrix}<br />
-x \\<br />
-2x-30<br />
\end{pmatrix}</math> and <math>\begin{pmatrix}<br />
-20-x \\<br />
-2x-30<br />
\end{pmatrix}</math> , respectively. (To get this, we use the fact that <math>\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}</math> and similarly, <math>\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.</math> )<br />
<br />
Equating the cross-product to <math>0</math> gets us the quadratic <math>-x(-20-x)+(-2x-30)(-2x-30)=0.</math> The solutions are <math>x=-18, -10.</math> Since <math>B</math> clearly has a more negative x-coordinate than <math>E</math>, we take <math>x=-18</math>. So <math>B = (-18, -6).</math><br />
<br />
From here, there are multiple ways to get the area of <math>\Delta{ABC}</math> to be <math>60</math>, and since the area of <math>\Delta{ACD}</math> is <math>300</math>, we get our final answer to be <cmath>60 + 300 = \boxed{\text{(D) } 360}.</cmath><br />
<br />
-PureSwag<br />
<br />
== Solution 7 (Power of a Point/No quadratics)==<br />
<br />
<asy><br />
import olympiad;<br />
pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798),F=(285,94.5),G=(361.2,361.2);<br />
dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);dot("$F$",F,N);dot("$G$",G,N);<br />
draw(A--B--C--D--A);<br />
draw(A--C, dotted); draw(B--D, dotted); draw(F--G, dotted);<br />
</asy><br />
<br />
Let <math>F</math> be the midpoint of <math>AC</math>, and draw <math>FG // CD</math> where <math>G</math> is on <math>BD</math>. We have <math>EF=5,FC=10</math>.<br />
<br />
<math>\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC</math>. Therefore <math>ABCG</math> is a cyclic quadrilateral.<br />
<br />
Notice that <math>\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}</math> via Power of a Point.<br />
<br />
The altitude from <math>B</math> to <math>AC</math> is then equal to <math>GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6</math>. <br />
<br />
Finally, the total area of <math>ABCD</math> is equal to <math>\frac 12 \cdot 20 \left(30+6 \right) =\boxed{\text{(D) } 360}.</math><br />
<br />
~asops<br />
<br />
==Solution 8==<br />
This diagram is simple to draw. Use a ruler to draw right triangle <math>ACD</math> with side ratio <math>2:3</math>. Measure out point <math>E</math> then draw and extend line <math>DE</math>. Because <math>\angle ABC = 90</math>, we find a point <math>B</math> on <math>DE</math> such that the distance of <math>B</math> to the midpoint of <math>AC = 10</math>. Simply find the height from point <math>B</math> to get the answer <math>\boxed{\text{(D) } 360}.</math><br />
<br />
==Video Solutions==<br />
===Video Solution 1===<br />
Education, The Study of Everything<br />
https://youtu.be/5lb8kk1qbaA<br />
<br />
<br />
===Video Solution 2===<br />
On The Spot STEM<br />
https://www.youtube.com/watch?v=hIdNde2Vln4<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=sHrjx968ZaM&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=2 ~ MathEx<br />
<br />
<br />
===Video Solution 4===<br />
The Beauty Of Math https://www.youtube.com/watch?v=RKlG6oZq9so&ab_channel=TheBeautyofMath<br />
<br />
===Video Solution 5===<br />
https://youtu.be/R220vbM_my8?t=658<br />
<br />
(amritvignesh0719062.0)<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_15&diff=1646252018 AMC 12B Problems/Problem 152021-11-05T03:26:30Z<p>Zeric: </p>
<hr />
<div>== Problem ==<br />
How many odd positive <math>3</math>-digit integers are divisible by <math>3</math> but do not contain the digit <math>3</math>?<br />
<br />
<math>\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 </math><br />
<br />
== Solution 1 ==<br />
Let <math>\underline{ABC}</math> be one such odd positive <math>3</math>-digit integer with hundreds digit <math>A,</math> tens digit <math>B,</math> and ones digit <math>C.</math> Since <math>\underline{ABC}\equiv0\pmod3,</math> we need <math>A+B+C\equiv0\pmod3</math> by the divisibility rule for <math>3.</math><br />
<br />
As <math>A\in\{1,2,4,5,6,7,8,9\}</math> and <math>C\in\{1,5,7,9\},</math> there are <math>8</math> possibilities for <math>A</math> and <math>4</math> possibilities for <math>C.</math> Note that each ordered pair <math>(A,C)</math> determines the value of <math>B</math> modulo <math>3,</math> so <math>B</math> can be any element in one of the sets <math>\{0,6,9\},\{1,4,7\},</math> or <math>\{2,5,8\}.</math> We conclude that there are always <math>3</math> possibilities for <math>B.</math><br />
<br />
By the Multiplication Principle, the answer is <math>8\cdot4\cdot3=\boxed{\textbf{(A) } 96}.</math><br />
<br />
~Plasma_Vortex ~MRENTHUSIASM<br />
<br />
== Solution 2 ==<br />
Let <math>\underline{ABC}</math> be one such odd positive <math>3</math>-digit integer with hundreds digit <math>A,</math> tens digit <math>B,</math> and ones digit <math>C.</math> Since <math>\underline{ABC}\equiv0\pmod3,</math> we need <math>A+B+C\equiv0\pmod3</math> by the divisibility rule for <math>3.</math><br />
<br />
As <math>A\in\{1,2,4,5,6,7,8,9\},B\in\{0,1,2,4,5,6,7,8,9\},</math> and <math>C\in\{1,5,7,9\},</math> note that:<br />
<ol style="margin-left: 1.5em;"><br />
<li>There are <math>2</math> possibilities for <math>A\equiv0\pmod3,</math> namely <math>A=6,9.</math> <p><br />
There are <math>3</math> possibilities for <math>A\equiv1\pmod3,</math> namely <math>A=1,4,7.</math> <p><br />
There are <math>3</math> possibilities for <math>A\equiv2\pmod3,</math> namely <math>A=2,5,8.</math> <p><br />
</li><br />
<li>There are <math>3</math> possibilities for <math>B\equiv0\pmod3,</math> namely <math>B=0,6,9.</math> <p><br />
There are <math>3</math> possibilities for <math>B\equiv1\pmod3,</math> namely <math>B=1,4,7.</math> <p><br />
There are <math>3</math> possibilities for <math>B\equiv2\pmod3,</math> namely <math>B=2,5,8.</math> <p><br />
</li><br />
<li>There are <math>1</math> possibility for <math>C\equiv0\pmod3,</math> namely <math>C=9.</math> <p><br />
There are <math>2</math> possibilities for <math>C\equiv1\pmod3,</math> namely <math>C=1,7.</math> <p><br />
There are <math>1</math> possibility for <math>C\equiv2\pmod3,</math> namely <math>C=5.</math> <p><br />
</li><br />
</ol><br />
We apply casework to <math>A+B+C\equiv0\pmod3:</math><br />
<cmath>\begin{array}{c|c|c||l}<br />
& & & \\ [-2.5ex]<br />
\boldsymbol{A\operatorname{mod}3} & \boldsymbol{B\operatorname{mod}3} & \boldsymbol{C\operatorname{mod}3} & \multicolumn{1}{c}{\textbf{Count}} \\ [0.5ex]<br />
\hline<br />
& & & \\ [-2ex]<br />
0 & 0 & 0 & 2\cdot3\cdot1=6 \\<br />
0 & 1 & 2 & 2\cdot3\cdot1=6 \\<br />
0 & 2 & 1 & 2\cdot3\cdot2=12 \\<br />
1 & 0 & 2 & 3\cdot3\cdot1=9 \\<br />
1 & 1 & 1 & 3\cdot3\cdot2=18 \\<br />
1 & 2 & 0 & 3\cdot3\cdot1=9 \\<br />
2 & 0 & 1 & 3\cdot3\cdot2=18 \\<br />
2 & 1 & 0 & 3\cdot3\cdot1=9 \\<br />
2 & 2 & 2 & 3\cdot3\cdot1=9<br />
\end{array}</cmath><br />
Together, the answer is <math>6+6+12+9+18+9+18+9+9=\boxed{\textbf{(A) } 96}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
== Solution 3 ==<br />
Analyze that the three-digit integers divisible by <math>3</math> start from <math>102.</math> In the <math>200</math>'s, it starts from <math>201.</math> In the <math>300</math>'s, it starts from <math>300.</math> We see that the units digits is <math>0, 1, </math> and <math>2.</math><br />
<br />
Write out the <math>1</math>- and <math>2</math>-digit multiples of <math>3</math> starting from <math>0, 1,</math> and <math>2.</math> Count up the ones that meet the conditions. Then, add up and multiply by <math>3,</math> since there are three sets of three from <math>1</math> to <math>9.</math> Then, subtract the amount that started from <math>0,</math> since the <math>300</math>'s ll contain the digit <math>3.</math><br />
<br />
Together, the answer is <math>3(12+12+12)-12=\boxed{\textbf{(A) } 96}.</math><br />
<br />
== Solution 4 ==<br />
<br />
Consider the number of <math>2</math>-digit numbers that do not contain the digit <math>3,</math> which is <math>90-18=72.</math> For any of these <math>2</math>-digit numbers, we can append <math>1,5,7,</math> or <math>9</math> to reach a desirable <math>3</math>-digit number. However, we have <math>7 \equiv 1\pmod{3},</math> and thus we need to count any <math>2</math>-digit number <math>\equiv 2\pmod{3}</math> twice. There are <math>(98-11)/3+1=30</math> total such numbers that have remainder <math>2,</math> but <math>6</math> of them <math>(23,32,35,38,53,83)</math> contain <math>3,</math> so the number we want is <math>30-6=24.</math> Therefore, the final answer is <math>72+24= \boxed{\textbf{(A) } 96}.</math><br />
<br />
== Solution 5 ==<br />
We need to take care of all restrictions. Ranging from <math>101</math> to <math>999,</math> there are <math>450</math> odd <math>3</math>-digit numbers. Exactly <math>\frac{1}{3}</math> of these numbers are divisible by <math>3,</math> which is <math>450\cdot\frac{1}{3}=150.</math> Of these <math>150</math> numbers, <math>\frac{4}{5}</math> <math>\textbf{do not}</math> have <math>3</math> in their ones (units) digit, <math>\frac{9}{10}</math> <math>\textbf{do not}</math> have <math>3</math> in their tens digit, and <math>\frac{8}{9}</math> <math>\textbf{do not}</math> have <math>3</math> in their hundreds digit. Thus, the total number of <math>3</math>-digit integers is <cmath>900\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{8}{9}=\boxed{\textbf{(A) } 96}.</cmath><br />
<br />
~mathpro12345<br />
<br />
==Solution 6==<br />
<br />
We will start with the numbers that could work. This numbers include _ _ <math>1</math>, _ _ <math>5</math>, _ _ <math>7</math>, _ _ <math>9</math>. Let's work case by case.<br />
<br />
Case <math>1</math>: _ _ <math>1</math>: The two blanks could be any number that is <math>2</math> mod <math>3</math> that does not include <math>3</math>. We have <math>24</math> cases for this case (we could count every case). <br />
<br />
Case <math>2</math>: _ _ <math>5</math>: The <math>2</math> blanks could be any number that is <math>1</math> mod <math>3</math> that does not include <math>3</math>. But we could see that this case has exactly the same solutions to case <math>1</math> because we have a <math>1-1</math> correspondence. We can do the exact same for case <math>3</math>. <br />
<br />
Cases <math>4</math>: _ _ <math>9</math>: We need the blanks to be a multiple of <math>3</math>, but does not contain 3. We have <math>(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)</math> which also contains <math>24</math> numbers. Therefore, we have <math>24 \cdot 4</math> which is equal to <math>\boxed{\textbf{(A) } 96}.</math><br />
<br />
~Arcticturn<br />
<br />
==Solution 7==<br />
<br />
This problem is solvable by inclusion exclusion principle. There are <math>\frac{999-105}{6} + 1 = 150</math> odd <math>3</math>-digit numbers divisible by <math>3</math>. We consider the number of <math>3</math>-digit numbers divisible by <math>3</math> that contain either <math>1, 2</math> or <math>3</math> digits of <math>3</math>.<br />
<br />
For <math>\underline{AB3}</math>, <math>AB</math> is any <math>2</math>-digit number divisible by <math>3</math>, which gives us <math>\frac{99-12}{3} + 1 = 30</math>. For <math>\underline{A3B}</math>, for each odd <math>B</math>, we have <math>3</math> values of <math>A</math> that give a valid case, thus we have <math>5(3) = 15</math> cases. For <math>\underline{3AB}</math>, we also have <math>15</math> cases, but when <math>B=3, 9</math>, <math>A</math> can equal <math>0</math>, so we have <math>17</math> cases. <br />
<br />
For <math>\underline{A33}</math>, we have <math>3</math> cases. For <math>\underline{3A3}</math>, we have <math>4</math> cases. For <math>\underline{33A}</math>, we have <math>2</math> cases. Finally, there is just one case for <math>\underline{333}</math>. <br />
<br />
By inclusion exclusion principle, we get <math>150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{\textbf{(A) } 96}</math> numbers.<br />
<br />
~Zeric<br />
<br />
== Video Solution ==<br />
https://youtu.be/mgEZOXgIZXs?t=448<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
[[Category:Introductory Number Theory Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_6&diff=1489951983 AIME Problems/Problem 62021-03-09T18:00:22Z<p>Zeric: </p>
<hr />
<div>== Problem ==<br />
Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Firstly, we try to find a relationship between the numbers we're provided with and <math>49</math>. We notice that <math>49=7^2</math>, and both <math>6</math> and <math>8</math> are greater or less than <math>7</math> by <math>1</math>. <br />
<br />
Thus, expressing the numbers in terms of <math>7</math>, we get <math>a_{83} = (7-1)^{83}+(7+1)^{83}</math>.<br />
<br />
Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisible by <math>49</math> except the final term.<br />
<br />
After some quick division, our answer is <math>\boxed{035}</math>.<br />
<br />
=== Solution 2 ===<br />
Since <math>\phi(49) = 42</math> (see [[Euler's totient function]]), [[Euler's Totient Theorem]] tells us that <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math> <br />
<math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math><br />
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.<br />
<br />
*Alternatively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\pmod {49}</math>, and can finish the same way.<br />
<br />
== Solution 3==<br />
<math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math><br />
<br />
Becuase <math>7|(6+8)</math>, we only consider <math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \pmod{7}</math><br />
<br />
<math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \equiv (-1)^{82} - (-1)^{81}+ \ldots - (-1)^1 + 1 = 83 \equiv 6 \pmod{49}</math><br />
<br />
<math>6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod{49}</math><br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_6&diff=1489941983 AIME Problems/Problem 62021-03-09T17:59:38Z<p>Zeric: </p>
<hr />
<div>== Problem ==<br />
Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Firstly, we try to find a relationship between the numbers we're provided with and <math>49</math>. We notice that <math>49=7^2</math>, and both <math>6</math> and <math>8</math> are greater or less than <math>7</math> by <math>1</math>. <br />
<br />
Thus, expressing the numbers in terms of <math>7</math>, we get <math>a_{83} = (7-1)^{83}+(7+1)^{83}</math>.<br />
<br />
Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisible by <math>49</math> except the final term.<br />
<br />
After some quick division, our answer is <math>\boxed{035}</math>.<br />
<br />
=== Solution 2 ===<br />
Since <math>\phi(49) = 42</math> (see [[Euler's totient function]]), [[Euler's Totient Theorem]] tells us that <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math> <br />
<math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math><br />
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.<br />
<br />
*Alternatively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\pmod {49}</math>, and can finish the same way.<br />
<br />
== Solution 3==<br />
<math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math><br />
<br />
<br />
Becuase <math>7|(6+8)</math>, we only consider <math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \pmod{7}</math><br />
<br />
<math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \equiv (-1)^{82} - (-1)^{81}+ \ldots - (-1)^1 + 1 = 83 \equiv 6 \pmod{49}</math><br />
<br />
<math>6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod{49}</math><br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_6&diff=1489931983 AIME Problems/Problem 62021-03-09T17:52:40Z<p>Zeric: </p>
<hr />
<div>== Problem ==<br />
Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
Firstly, we try to find a relationship between the numbers we're provided with and <math>49</math>. We notice that <math>49=7^2</math>, and both <math>6</math> and <math>8</math> are greater or less than <math>7</math> by <math>1</math>. <br />
<br />
Thus, expressing the numbers in terms of <math>7</math>, we get <math>a_{83} = (7-1)^{83}+(7+1)^{83}</math>.<br />
<br />
Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisible by <math>49</math> except the final term.<br />
<br />
After some quick division, our answer is <math>\boxed{035}</math>.<br />
<br />
=== Solution 2 ===<br />
Since <math>\phi(49) = 42</math> (see [[Euler's totient function]]), [[Euler's Totient Theorem]] tells us that <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math> <br />
<math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math><br />
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.<br />
<br />
*Alternatively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\pmod {49}</math>, and can finish the same way.<br />
<br />
== Solution 3==<br />
<math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math><br />
<br />
Becuase <math>7|(6+8)</math>, we only consider <math>(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82}) \text{mod} 7</math><br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_19&diff=1437092015 AMC 12B Problems/Problem 192021-01-29T03:53:28Z<p>Zeric: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
In <math>\triangle ABC</math>, <math>\angle C = 90^\circ</math> and <math>AB = 12</math>. Squares <math>ABXY</math> and <math>ACWZ</math> are constructed outside of the triangle. The points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle. What is the perimeter of the triangle?<br />
<br />
<math>\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32</math><br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
</asy><br />
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of <math>WZ</math> and <math>XY</math> and finding their intersection point. This point happens to be the midpoint of <math>AB</math>, the hypotenuse. Let this point be <math>M</math>. To find the radius, determine <math>MY</math>, where <math>MY^{2} = MA^2 + AY^2</math>, <math>MA = \frac{12}{2} = 6</math>, and <math>AY = AB = 12</math>. Thus, the radius <math>=r =MY = 6\sqrt5</math>.<br />
<br />
Next we let <math>AC = b</math> and <math>BC = a</math>. Consider the right triangle <math>ACB</math> first. Using the Pythagorean theorem, we find that <math>a^2 + b^2 = 12^2 = 144</math>. <br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
draw(E--Z--M--cycle,dashed);<br />
</asy><br />
Now, we let <math>E</math> be the midpoint of <math>WZ</math>, and we consider right triangle <math>ZEM</math>. By the Pythagorean theorem, we have that <math>\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180</math>. Expanding this equation, we get that<br />
<br />
<cmath>\frac{1}{4}(a^2+b^2) + a^2 + ab = 180</cmath><br />
<cmath>\frac{144}{4} + a^2 + ab = 180</cmath><br />
<cmath>a^2 + ab = 144 = a^2 + b^2</cmath><br />
<cmath>ab = b^2</cmath><br />
<cmath>b = a</cmath><br />
<br />
This means that <math>ABC</math> is a 45-45-90 triangle, so <math>a = b = \frac{12}{\sqrt2} = 6\sqrt2</math>. Thus the perimeter is <math>a + b + AB = 12\sqrt2 + 12</math> which is answer <math>\boxed{\textbf{(C)}\; 12 + 12\sqrt2}</math>.<br />
<br />
==Solution 2==<br />
The center of the circle on which <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie must be equidistant from each of these four points. Draw the perpendicular bisectors of <math>\overline{XY}</math> and of <math>\overline{WZ}</math>. Note that the perpendicular bisector of <math>\overline{WZ}</math> is parallel to <math>\overline{CW}</math> and passes through the midpoint of <math>\overline{AC}</math>. Therefore, the triangle that is formed by <math>A</math>, the midpoint of <math>\overline{AC}</math>, and the point at which this perpendicular bisector intersects <math>\overline{AB}</math> must be similar to <math>\triangle ABC</math>, and the ratio of a side of the smaller triangle to a side of <math>\triangle ABC</math> is 1:2. Consequently, the perpendicular bisector of <math>\overline{XY}</math> passes through the midpoint of <math>\overline{AB}</math>. The perpendicular bisector of <math>\overline{WZ}</math> must include the midpoint of <math>\overline{AB}</math> as well. Since all points on a perpendicular bisector of any two points <math>M</math> and <math>N</math> are equidistant from <math>M</math> and <math>N</math>, the center of the circle must be the midpoint of <math>\overline{AB}</math>.<br />
<br />
Now the distance between the midpoint of <math>\overline{AB}</math> and <math>Z</math>, which is equal to the radius of this circle, is <math>\sqrt{12^2 + 6^2} = \sqrt{180}</math>. Let <math>a=AC</math>. Then the distance between the midpoint of <math>\overline{AB}</math> and <math>Y</math>, also equal to the radius of the circle, is given by <math>\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}</math> (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have<br />
<br />
<cmath>\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180</cmath><br />
<cmath>144 - a^2 = a\sqrt{144-a^2}</cmath><br />
<cmath>(144-a^2)^2 = a^2(144-a^2)</cmath><br />
<br />
Since <math>a</math> cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by <math>(144-a^2)</math>, and arrive at <math>a = 6\sqrt{2}</math>. The length of other leg of the triangle must be <math>\sqrt{144-72} = 6\sqrt{2}</math>. Thus, the perimeter of the triangle is <math>12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
==Solution 3==<br />
In order to solve this problem, we can search for similar triangles. Begin by drawing triangle <math>ABC</math> and squares <math>ABXY</math> and <math>ACWZ</math>. Draw segments <math>\overline{YZ}</math> and <math>\overline{WX}</math>. Because we are given points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle, we can conclude that <math>WXYZ</math> forms a cyclic quadrilateral. Take <math>\overline{AC}</math> and extend it through a point <math>P</math> on <math>\overline{YZ}</math>. Now, we must do some angle chasing to prove that <math>\triangle WBX</math> is similar to <math>\triangle YAZ</math>. <br />
<br />
Let <math>\alpha</math> denote the measure of <math>\angle ABC</math>. Following this, <math>\angle BAC</math> measures <math>90 - \alpha</math>. By our construction, <math>\overline{CAP}</math> is a straight line, and we know <math>\angle YAB</math> is a right angle. Therefore, <math>\angle PAY</math> measures <math>\alpha</math>. Also, <math>\angle CAZ</math> is a right angle and thus, <math>\angle ZAP</math> is a right angle. Sum <math>\angle ZAP</math> and <math>\angle PAY</math> to find <math>\angle ZAY</math>, which measures <math>90 + \alpha</math>. We also know that <math>\angle WBY</math> measures <math>90 + \alpha</math>. Therefore, <math>\angle ZAY = \angle WBX</math>. <br />
<br />
Let <math>\beta</math> denote the measure of <math>\angle AZY</math>. It follows that <math>\angle WZY</math> measures <math>90 + \beta^\circ</math>. Because <math>WXYZ</math> is a cyclic quadrilateral, <math>\angle WZY + \angle YXW = 180^\circ</math>. Therefore, <math>\angle YXW</math> must measure <math>90 - \beta</math>, and <math>\angle BXW</math> must measure <math>\beta</math>. Therefore, <math>\angle AZY = \angle BXW</math>. <br />
<br />
<math>\angle ZAY = \angle WBX</math> and <math>\angle AZY = \angle BXW</math>, so <math>\triangle AZY \sim \triangle BXW</math>! Let <math>x = AC = WC</math>. By Pythagorean theorem, <math>BC = \sqrt{144-x^2}</math>. Now we have <math>WB = WC + BC = x + \sqrt{144-x^2}</math>, <math>BX = 12</math>, <math>YA = 12</math>, and <math>AZ = x</math>. <br />
We can set up an equation: <br />
<br />
<cmath>\frac{YA}{AZ} = \frac{WB}{BX}</cmath><br />
<cmath>\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}</cmath><br />
<cmath>144 = x^2 + x\sqrt{144-x^2}</cmath><br />
<cmath>12^2 - x^2 = x\sqrt{144-x^2}</cmath><br />
<cmath>12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4</cmath><br />
<cmath>2x^4 - 3(12^2)x^2 + 12^4 = 0</cmath><br />
<cmath>(2x^2 - 144)(x^2 - 144) = 0</cmath><br />
<br />
Solving for <math>x</math>, we find that <math>x = 6\sqrt{2}</math> or <math>x = 12</math>, which we omit. The perimeter of the triangle is <math>12 + x + \sqrt{144-x^2}</math>. Plugging in <math>x = 6\sqrt{2}</math>, we get <math>\boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
<br />
Alternatively, let <math>BC = S_1, AB = S_2</math> and <math>AC = x</math>. Because <math>\frac{WB}{BX} = \frac{AY}{AZ}</math>, we get that <math>S_2^2 = S_1^2 + S_1x</math>. <math>S_1 = x</math> satisfies the equation because of Pythagorean theorem, so <math>\triangle ABC</math> is right isosceles.<br />
<br />
==Solution 4==<br />
We claim that <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles right triangle.<br />
<br />
''Proof:'' If <math>\triangle ACB</math> is an isosceles right triangle, then <math>\angle WAY=180º</math>. Therefore, <math>W</math>, <math>A</math>, and <math>Y</math> are collinear. Since <math>WY</math> and <math>YX</math> form a right angle, <math>WX</math> is the diameter of the circumcircle of <math>\triangle WYX</math>. Similarly, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear, and <math>ZX</math> forms a right angle with <math>ZW</math>. Thus, <math>WX</math> is also the diameter of the circumcircle of <math>\triangle WZX</math>. Therefore, since <math>\triangle WYX</math> and <math>\triangle WZX</math> share a circumcircle, <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles triangle.<br />
<br />
If <math>\triangle ACB</math> is isosceles, then its legs have length <math>6\sqrt{2}</math>. The perimeter of <math>\triangle ACB</math> is <math>\boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
<br />
==Solution 5==<br />
Note that because <math>\angle WZA=\angle AYX=90^\circ</math>, these two angles inscribe the semicircle defined by diameter <math>WX</math>. Since <math>A</math> lies on line <math>ZA</math>, and <math>\angle XAY=45^\circ</math> (because <math>ABXY</math> is a square), we can find that <math>\angle ZAY= 180^\circ - 45^\circ = 135^\circ</math>.<br />
<br />
Now, we can see that <math>\angle BAC=45^\circ</math> to complete the full <math>360^\circ</math>. Therefore, <math>\triangle ABC</math> is and isosceles right triangle, and <math>AC=BC=6\sqrt2</math>. So Our answer is <math>6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_19&diff=1437072015 AMC 12B Problems/Problem 192021-01-29T03:53:07Z<p>Zeric: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
In <math>\triangle ABC</math>, <math>\angle C = 90^\circ</math> and <math>AB = 12</math>. Squares <math>ABXY</math> and <math>ACWZ</math> are constructed outside of the triangle. The points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle. What is the perimeter of the triangle?<br />
<br />
<math>\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32</math><br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
</asy><br />
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of <math>WZ</math> and <math>XY</math> and finding their intersection point. This point happens to be the midpoint of <math>AB</math>, the hypotenuse. Let this point be <math>M</math>. To find the radius, determine <math>MY</math>, where <math>MY^{2} = MA^2 + AY^2</math>, <math>MA = \frac{12}{2} = 6</math>, and <math>AY = AB = 12</math>. Thus, the radius <math>=r =MY = 6\sqrt5</math>.<br />
<br />
Next we let <math>AC = b</math> and <math>BC = a</math>. Consider the right triangle <math>ACB</math> first. Using the Pythagorean theorem, we find that <math>a^2 + b^2 = 12^2 = 144</math>. <br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
draw(E--Z--M--cycle,dashed);<br />
</asy><br />
Now, we let <math>E</math> be the midpoint of <math>WZ</math>, and we consider right triangle <math>ZEM</math>. By the Pythagorean theorem, we have that <math>\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180</math>. Expanding this equation, we get that<br />
<br />
<cmath>\frac{1}{4}(a^2+b^2) + a^2 + ab = 180</cmath><br />
<cmath>\frac{144}{4} + a^2 + ab = 180</cmath><br />
<cmath>a^2 + ab = 144 = a^2 + b^2</cmath><br />
<cmath>ab = b^2</cmath><br />
<cmath>b = a</cmath><br />
<br />
This means that <math>ABC</math> is a 45-45-90 triangle, so <math>a = b = \frac{12}{\sqrt2} = 6\sqrt2</math>. Thus the perimeter is <math>a + b + AB = 12\sqrt2 + 12</math> which is answer <math>\boxed{\textbf{(C)}\; 12 + 12\sqrt2}</math>.<br />
<br />
==Solution 2==<br />
The center of the circle on which <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie must be equidistant from each of these four points. Draw the perpendicular bisectors of <math>\overline{XY}</math> and of <math>\overline{WZ}</math>. Note that the perpendicular bisector of <math>\overline{WZ}</math> is parallel to <math>\overline{CW}</math> and passes through the midpoint of <math>\overline{AC}</math>. Therefore, the triangle that is formed by <math>A</math>, the midpoint of <math>\overline{AC}</math>, and the point at which this perpendicular bisector intersects <math>\overline{AB}</math> must be similar to <math>\triangle ABC</math>, and the ratio of a side of the smaller triangle to a side of <math>\triangle ABC</math> is 1:2. Consequently, the perpendicular bisector of <math>\overline{XY}</math> passes through the midpoint of <math>\overline{AB}</math>. The perpendicular bisector of <math>\overline{WZ}</math> must include the midpoint of <math>\overline{AB}</math> as well. Since all points on a perpendicular bisector of any two points <math>M</math> and <math>N</math> are equidistant from <math>M</math> and <math>N</math>, the center of the circle must be the midpoint of <math>\overline{AB}</math>.<br />
<br />
Now the distance between the midpoint of <math>\overline{AB}</math> and <math>Z</math>, which is equal to the radius of this circle, is <math>\sqrt{12^2 + 6^2} = \sqrt{180}</math>. Let <math>a=AC</math>. Then the distance between the midpoint of <math>\overline{AB}</math> and <math>Y</math>, also equal to the radius of the circle, is given by <math>\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}</math> (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have<br />
<br />
<cmath>\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180</cmath><br />
<cmath>144 - a^2 = a\sqrt{144-a^2}</cmath><br />
<cmath>(144-a^2)^2 = a^2(144-a^2)</cmath><br />
<br />
Since <math>a</math> cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by <math>(144-a^2)</math>, and arrive at <math>a = 6\sqrt{2}</math>. The length of other leg of the triangle must be <math>\sqrt{144-72} = 6\sqrt{2}</math>. Thus, the perimeter of the triangle is <math>12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
==Solution 3==<br />
In order to solve this problem, we can search for similar triangles. Begin by drawing triangle <math>ABC</math> and squares <math>ABXY</math> and <math>ACWZ</math>. Draw segments <math>\overline{YZ}</math> and <math>\overline{WX}</math>. Because we are given points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle, we can conclude that <math>WXYZ</math> forms a cyclic quadrilateral. Take <math>\overline{AC}</math> and extend it through a point <math>P</math> on <math>\overline{YZ}</math>. Now, we must do some angle chasing to prove that <math>\triangle WBX</math> is similar to <math>\triangle YAZ</math>. <br />
<br />
Let <math>\alpha</math> denote the measure of <math>\angle ABC</math>. Following this, <math>\angle BAC</math> measures <math>90 - \alpha</math>. By our construction, <math>\overline{CAP}</math> is a straight line, and we know <math>\angle YAB</math> is a right angle. Therefore, <math>\angle PAY</math> measures <math>\alpha</math>. Also, <math>\angle CAZ</math> is a right angle and thus, <math>\angle ZAP</math> is a right angle. Sum <math>\angle ZAP</math> and <math>\angle PAY</math> to find <math>\angle ZAY</math>, which measures <math>90 + \alpha</math>. We also know that <math>\angle WBY</math> measures <math>90 + \alpha</math>. Therefore, <math>\angle ZAY = \angle WBX</math>. <br />
<br />
Let <math>\beta</math> denote the measure of <math>\angle AZY</math>. It follows that <math>\angle WZY</math> measures <math>90 + \beta^\circ</math>. Because <math>WXYZ</math> is a cyclic quadrilateral, <math>\angle WZY + \angle YXW = 180^\circ</math>. Therefore, <math>\angle YXW</math> must measure <math>90 - \beta</math>, and <math>\angle BXW</math> must measure <math>\beta</math>. Therefore, <math>\angle AZY = \angle BXW</math>. <br />
<br />
<math>\angle ZAY = \angle WBX</math> and <math>\angle AZY = \angle BXW</math>, so <math>\triangle AZY \sim \triangle BXW</math>! Let <math>x = AC = WC</math>. By Pythagorean theorem, <math>BC = \sqrt{144-x^2}</math>. Now we have <math>WB = WC + BC = x + \sqrt{144-x^2}</math>, <math>BX = 12</math>, <math>YA = 12</math>, and <math>AZ = x</math>. <br />
We can set up an equation: <br />
<br />
<cmath>\frac{YA}{AZ} = \frac{WB}{BX}</cmath><br />
<cmath>\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}</cmath><br />
<cmath>144 = x^2 + x\sqrt{144-x^2}</cmath><br />
<cmath>12^2 - x^2 = x\sqrt{144-x^2}</cmath><br />
<cmath>12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4</cmath><br />
<cmath>2x^4 - 3(12^2)x^2 + 12^4 = 0</cmath><br />
<cmath>(2x^2 - 144)(x^2 - 144) = 0</cmath><br />
<br />
Solving for <math>x</math>, we find that <math>x = 6\sqrt{2}</math> or <math>x = 12</math>, which we omit. The perimeter of the triangle is <math>12 + x + \sqrt{144-x^2}</math>. Plugging in <math>x = 6\sqrt{2}</math>, we get <math>\boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
==Solution 4==<br />
We claim that <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles right triangle.<br />
<br />
''Proof:'' If <math>\triangle ACB</math> is an isosceles right triangle, then <math>\angle WAY=180º</math>. Therefore, <math>W</math>, <math>A</math>, and <math>Y</math> are collinear. Since <math>WY</math> and <math>YX</math> form a right angle, <math>WX</math> is the diameter of the circumcircle of <math>\triangle WYX</math>. Similarly, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear, and <math>ZX</math> forms a right angle with <math>ZW</math>. Thus, <math>WX</math> is also the diameter of the circumcircle of <math>\triangle WZX</math>. Therefore, since <math>\triangle WYX</math> and <math>\triangle WZX</math> share a circumcircle, <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles triangle.<br />
<br />
If <math>\triangle ACB</math> is isosceles, then its legs have length <math>6\sqrt{2}</math>. The perimeter of <math>\triangle ACB</math> is <math>\boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
<br />
==Solution 5==<br />
Note that because <math>\angle WZA=\angle AYX=90^\circ</math>, these two angles inscribe the semicircle defined by diameter <math>WX</math>. Since <math>A</math> lies on line <math>ZA</math>, and <math>\angle XAY=45^\circ</math> (because <math>ABXY</math> is a square), we can find that <math>\angle ZAY= 180^\circ - 45^\circ = 135^\circ</math>.<br />
<br />
Now, we can see that <math>\angle BAC=45^\circ</math> to complete the full <math>360^\circ</math>. Therefore, <math>\triangle ABC</math> is and isosceles right triangle, and <math>AC=BC=6\sqrt2</math>. So Our answer is <math>6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_19&diff=1437062015 AMC 12B Problems/Problem 192021-01-29T03:52:26Z<p>Zeric: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
In <math>\triangle ABC</math>, <math>\angle C = 90^\circ</math> and <math>AB = 12</math>. Squares <math>ABXY</math> and <math>ACWZ</math> are constructed outside of the triangle. The points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle. What is the perimeter of the triangle?<br />
<br />
<math>\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32</math><br />
<br />
==Solution 1==<br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
</asy><br />
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of <math>WZ</math> and <math>XY</math> and finding their intersection point. This point happens to be the midpoint of <math>AB</math>, the hypotenuse. Let this point be <math>M</math>. To find the radius, determine <math>MY</math>, where <math>MY^{2} = MA^2 + AY^2</math>, <math>MA = \frac{12}{2} = 6</math>, and <math>AY = AB = 12</math>. Thus, the radius <math>=r =MY = 6\sqrt5</math>.<br />
<br />
Next we let <math>AC = b</math> and <math>BC = a</math>. Consider the right triangle <math>ACB</math> first. Using the Pythagorean theorem, we find that <math>a^2 + b^2 = 12^2 = 144</math>. <br />
<asy><br />
pair A,B,C,M,E,W,Z,X,Y;<br />
A=(2,0);<br />
B=(0,2);<br />
C=(0,0);<br />
M=(A+B)/2;<br />
W=(-2,2);<br />
Z=(-2,-0);<br />
X=(2,4);<br />
Y=(4,2);<br />
E=(W+Z)/2;<br />
draw(A--B--C--cycle);<br />
draw(W--B--C--Z--cycle);<br />
draw(A--B--X--Y--cycle);<br />
dot(M);<br />
dot(E);<br />
label("W",W,NW);<br />
label("Z",Z,SW);<br />
label("C",C,S);<br />
label("A",A,S);<br />
label("B",B,N);<br />
label("X",X,NE);<br />
label("Y",Y,SE);<br />
label("E",E,1.5*plain.W);<br />
label("M",M,NE);<br />
draw(circle(M,sqrt(10)));<br />
draw(E--Z--M--cycle,dashed);<br />
</asy><br />
Now, we let <math>E</math> be the midpoint of <math>WZ</math>, and we consider right triangle <math>ZEM</math>. By the Pythagorean theorem, we have that <math>\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180</math>. Expanding this equation, we get that<br />
<br />
<cmath>\frac{1}{4}(a^2+b^2) + a^2 + ab = 180</cmath><br />
<cmath>\frac{144}{4} + a^2 + ab = 180</cmath><br />
<cmath>a^2 + ab = 144 = a^2 + b^2</cmath><br />
<cmath>ab = b^2</cmath><br />
<cmath>b = a</cmath><br />
<br />
This means that <math>ABC</math> is a 45-45-90 triangle, so <math>a = b = \frac{12}{\sqrt2} = 6\sqrt2</math>. Thus the perimeter is <math>a + b + AB = 12\sqrt2 + 12</math> which is answer <math>\boxed{\textbf{(C)}\; 12 + 12\sqrt2}</math>.<br />
<br />
==Solution 2==<br />
The center of the circle on which <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie must be equidistant from each of these four points. Draw the perpendicular bisectors of <math>\overline{XY}</math> and of <math>\overline{WZ}</math>. Note that the perpendicular bisector of <math>\overline{WZ}</math> is parallel to <math>\overline{CW}</math> and passes through the midpoint of <math>\overline{AC}</math>. Therefore, the triangle that is formed by <math>A</math>, the midpoint of <math>\overline{AC}</math>, and the point at which this perpendicular bisector intersects <math>\overline{AB}</math> must be similar to <math>\triangle ABC</math>, and the ratio of a side of the smaller triangle to a side of <math>\triangle ABC</math> is 1:2. Consequently, the perpendicular bisector of <math>\overline{XY}</math> passes through the midpoint of <math>\overline{AB}</math>. The perpendicular bisector of <math>\overline{WZ}</math> must include the midpoint of <math>\overline{AB}</math> as well. Since all points on a perpendicular bisector of any two points <math>M</math> and <math>N</math> are equidistant from <math>M</math> and <math>N</math>, the center of the circle must be the midpoint of <math>\overline{AB}</math>.<br />
<br />
Now the distance between the midpoint of <math>\overline{AB}</math> and <math>Z</math>, which is equal to the radius of this circle, is <math>\sqrt{12^2 + 6^2} = \sqrt{180}</math>. Let <math>a=AC</math>. Then the distance between the midpoint of <math>\overline{AB}</math> and <math>Y</math>, also equal to the radius of the circle, is given by <math>\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}</math> (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have<br />
<br />
<cmath>\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180</cmath><br />
<cmath>144 - a^2 = a\sqrt{144-a^2}</cmath><br />
<cmath>(144-a^2)^2 = a^2(144-a^2)</cmath><br />
<br />
Since <math>a</math> cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by <math>(144-a^2)</math>, and arrive at <math>a = 6\sqrt{2}</math>. The length of other leg of the triangle must be <math>\sqrt{144-72} = 6\sqrt{2}</math>. Thus, the perimeter of the triangle is <math>12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
<br />
Alternatively, let <math>BC = S_1, AB = S_2</math> and <math>AC = x</math>. Because <math>\frac{WB}{BX} = \frac{AY}{AZ}</math>, we get that <math>S_2^2 = S_1^2 + S_1x</math>. <math>S_1 = x</math> satisfies the equation because of Pythagorean theorem, so <math>\triangle ABC</math> is right isosceles.<br />
<br />
==Solution 3==<br />
In order to solve this problem, we can search for similar triangles. Begin by drawing triangle <math>ABC</math> and squares <math>ABXY</math> and <math>ACWZ</math>. Draw segments <math>\overline{YZ}</math> and <math>\overline{WX}</math>. Because we are given points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle, we can conclude that <math>WXYZ</math> forms a cyclic quadrilateral. Take <math>\overline{AC}</math> and extend it through a point <math>P</math> on <math>\overline{YZ}</math>. Now, we must do some angle chasing to prove that <math>\triangle WBX</math> is similar to <math>\triangle YAZ</math>. <br />
<br />
Let <math>\alpha</math> denote the measure of <math>\angle ABC</math>. Following this, <math>\angle BAC</math> measures <math>90 - \alpha</math>. By our construction, <math>\overline{CAP}</math> is a straight line, and we know <math>\angle YAB</math> is a right angle. Therefore, <math>\angle PAY</math> measures <math>\alpha</math>. Also, <math>\angle CAZ</math> is a right angle and thus, <math>\angle ZAP</math> is a right angle. Sum <math>\angle ZAP</math> and <math>\angle PAY</math> to find <math>\angle ZAY</math>, which measures <math>90 + \alpha</math>. We also know that <math>\angle WBY</math> measures <math>90 + \alpha</math>. Therefore, <math>\angle ZAY = \angle WBX</math>. <br />
<br />
Let <math>\beta</math> denote the measure of <math>\angle AZY</math>. It follows that <math>\angle WZY</math> measures <math>90 + \beta^\circ</math>. Because <math>WXYZ</math> is a cyclic quadrilateral, <math>\angle WZY + \angle YXW = 180^\circ</math>. Therefore, <math>\angle YXW</math> must measure <math>90 - \beta</math>, and <math>\angle BXW</math> must measure <math>\beta</math>. Therefore, <math>\angle AZY = \angle BXW</math>. <br />
<br />
<math>\angle ZAY = \angle WBX</math> and <math>\angle AZY = \angle BXW</math>, so <math>\triangle AZY \sim \triangle BXW</math>! Let <math>x = AC = WC</math>. By Pythagorean theorem, <math>BC = \sqrt{144-x^2}</math>. Now we have <math>WB = WC + BC = x + \sqrt{144-x^2}</math>, <math>BX = 12</math>, <math>YA = 12</math>, and <math>AZ = x</math>. <br />
We can set up an equation: <br />
<br />
<cmath>\frac{YA}{AZ} = \frac{WB}{BX}</cmath><br />
<cmath>\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}</cmath><br />
<cmath>144 = x^2 + x\sqrt{144-x^2}</cmath><br />
<cmath>12^2 - x^2 = x\sqrt{144-x^2}</cmath><br />
<cmath>12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4</cmath><br />
<cmath>2x^4 - 3(12^2)x^2 + 12^4 = 0</cmath><br />
<cmath>(2x^2 - 144)(x^2 - 144) = 0</cmath><br />
<br />
Solving for <math>x</math>, we find that <math>x = 6\sqrt{2}</math> or <math>x = 12</math>, which we omit. The perimeter of the triangle is <math>12 + x + \sqrt{144-x^2}</math>. Plugging in <math>x = 6\sqrt{2}</math>, we get <math>\boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>.<br />
<br />
==Solution 4==<br />
We claim that <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles right triangle.<br />
<br />
''Proof:'' If <math>\triangle ACB</math> is an isosceles right triangle, then <math>\angle WAY=180º</math>. Therefore, <math>W</math>, <math>A</math>, and <math>Y</math> are collinear. Since <math>WY</math> and <math>YX</math> form a right angle, <math>WX</math> is the diameter of the circumcircle of <math>\triangle WYX</math>. Similarly, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear, and <math>ZX</math> forms a right angle with <math>ZW</math>. Thus, <math>WX</math> is also the diameter of the circumcircle of <math>\triangle WZX</math>. Therefore, since <math>\triangle WYX</math> and <math>\triangle WZX</math> share a circumcircle, <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles triangle.<br />
<br />
If <math>\triangle ACB</math> is isosceles, then its legs have length <math>6\sqrt{2}</math>. The perimeter of <math>\triangle ACB</math> is <math>\boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
<br />
==Solution 5==<br />
Note that because <math>\angle WZA=\angle AYX=90^\circ</math>, these two angles inscribe the semicircle defined by diameter <math>WX</math>. Since <math>A</math> lies on line <math>ZA</math>, and <math>\angle XAY=45^\circ</math> (because <math>ABXY</math> is a square), we can find that <math>\angle ZAY= 180^\circ - 45^\circ = 135^\circ</math>.<br />
<br />
Now, we can see that <math>\angle BAC=45^\circ</math> to complete the full <math>360^\circ</math>. Therefore, <math>\triangle ABC</math> is and isosceles right triangle, and <math>AC=BC=6\sqrt2</math>. So Our answer is <math>6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}</math>.<br />
.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=20|num-b=18}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_24&diff=1436232015 AMC 12A Problems/Problem 242021-01-28T17:08:33Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
<br />
Rational numbers <math>a</math> and <math>b</math> are chosen at random among all rational numbers in the interval <math>[0,2)</math> that can be written as fractions <math>\frac{n}{d}</math> where <math>n</math> and <math>d</math> are integers with <math>1 \le d \le 5</math>. What is the probability that<br />
<cmath>(\text{cos}(a\pi)+i\text{sin}(b\pi))^4</cmath><br />
is a real number?<br />
<br />
<math> \textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}</math><br />
<br />
==Solution==<br />
<br />
Let <math>\cos(a\pi) = x</math> and <math>\sin(b\pi) = y</math>. Consider the binomial expansion of the expression:<br />
<cmath>x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.</cmath><br />
<br />
We notice that the only terms with <math>i</math> are the second and the fourth terms. Thus for the expression to be a real number, either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> must be <math>0</math>, or the second term and the fourth term cancel each other out (because in the fourth term, you have <math>i^2 = -1</math>). <br />
<br />
<math>\text{Case~1:}</math> Either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> is <math>0</math>. <br />
<br />
The two <math>\text{a's}</math> satisfying this are <math>\tfrac{1}{2}</math> and <math>\tfrac{3}{2}</math>, and the two <math>\text{b's}</math> satisfying this are <math>0</math> and <math>1</math>. Because <math>a</math> and <math>b</math> can both be expressed as fractions with a denominator less than or equal to <math>5</math>, there are a total of <math>20</math> possible values for <math>a</math> and <math>b</math>:<br />
<br />
<cmath>0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},</cmath> <br />
<cmath>\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},</cmath> <br />
<cmath>\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},</cmath> <br />
<cmath>\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.</cmath><br />
<br />
Calculating the total number of sets of <math>(a,b)</math> results in <math>20 \cdot 20 = 400</math> sets.<br />
Calculating the total number of invalid sets (sets where <math>a</math> doesn't equal <math>\tfrac{1}{2}</math> or <math>\tfrac{3}{2}</math> and <math>b</math> doesn't equal <math>0</math> or <math>1</math>), resulting in <math>(20-2) \cdot (20-2) = 324</math>.<br />
<br />
Thus the number of valid sets is <math>76</math>.<br />
<br />
<math>\text{Case~2}</math>: The two terms cancel. <br />
<br />
We then have: <br />
<br />
<cmath>\cos^3(a\pi) \cdot \sin(b\pi) = \cos(a\pi) \cdot \sin^3(b\pi).</cmath><br />
<br />
So: <br />
<br />
<cmath>\cos^2(a\pi) = \sin^2(b\pi),</cmath><br />
<br />
which means for a given value of <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math>, there are <math>4</math> valid values(one in each quadrant).<br />
<br />
When either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> are equal to <math>1</math>, however, there are only two corresponding values. We don't count the sets where either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> equals <math>0</math>, for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if <math>a</math> is <math>\tfrac{1}{5}</math>, then <math>b</math> must be <math>\tfrac{3}{10}</math>, which we don't have). Thus the total number of sets for this case is <math>4 \cdot 4 + 2 \cdot 2 = 20</math>.<br />
<br />
Thus, our final answer is <math>\frac{(20 + 76)}{400} = \frac{6}{25}</math>, which is <math>\boxed{\text{(D)}}</math>.<br />
<br />
==Solution 2==<br />
<br />
Multiplying complex numbers is equivalent to multiplying their magnitudes and summing their angles. In order for <math>(\cos(a\pi)+i\sin(b\pi))^4</math> to be a real number, then the angle of <math>\cos(a\pi)+i\sin(b\pi)</math> must be a multiple of <math>45^{\circ}</math>, so <math>\cos(a\pi)+i\sin(b\pi)</math> satisfies <math>x=0</math>, <math>y=0</math>, <math>x=y</math>, or <math>x=-y</math>. <br />
<br />
There are <math>20</math> possible values of <math>a</math> and <math>b</math>. <math>\sin(x\pi) = 0</math> at <math>x = \{0,1\}</math> and <math>\cos(x\pi) = 0</math> at <math>x = \{\frac12, \frac32\}</math>. The probability of <math>\sin(b\pi) = 0</math> or <math>\cos(a\pi) = 0</math> is <math>\frac{1}{10} + \frac{1}{10} - \frac{1}{100}</math> (We overcounted the case where <math>\sin = \cos = 0</math>.)<br />
<br />
We also consider the case where <math>\sin(b\pi) = \pm \cos(a\pi)</math>. This only happens when <math>a = \{0,1\}, b = \{\frac12, \frac32\}</math> or <math>a,b = \{ \frac14, \frac34, \frac54, \frac74 \}</math>. The probability is <math>\frac{1}{100} + \frac{1}{25}</math> The total probability is <math>\frac15 + \frac{1}{25} = \frac{6}{25} \rightarrow\boxed{\text{(D)}}</math><br />
<br />
~zeric<br />
<br />
=== Video Solution by Richard Rusczyk ===<br />
<br />
https://artofproblemsolving.com/videos/amc/2015amc12a/401<br />
<br />
~ dolphin7<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2015|ab=A|num-b=23|num-a=25}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_16&diff=1432152002 AMC 12A Problems/Problem 162021-01-25T05:12:28Z<p>Zeric: New solution</p>
<hr />
<div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #16]] and [[2002 AMC 10A Problems|2002 AMC 10A #24]]}}<br />
<br />
<br />
==Problem==<br />
Tina randomly selects two distinct numbers from the set <math>\{ 1, 2, 3, 4, 5 \}</math>, and Sergio randomly selects a number from the set <math>\{ 1, 2, ..., 10 \}</math>. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina? <br />
<br />
<math>\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25</math><br />
<br />
== Video Solution ==<br />
https://youtu.be/8WrdYLw9_ns?t=381<br />
<br />
~ pi_is_3.14<br />
<br />
==Solution==<br />
<br />
=== Solution 1 ===<br />
<br />
This is not too bad using casework. <br />
<br />
Tina gets a sum of 3: This happens in only one way <math>(1,2)</math> and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.<br />
<br />
Tina gets a sum of 4: This once again happens in only one way <math>(1,3)</math>. Sergio can choose a number from 5 to 10, so 6 ways here.<br />
<br />
Tina gets a sum of 5: This can happen in two ways <math>(1,4)</math> and <math>(2,3)</math>. Sergio can choose a number from 6 to 10, so <math>2\cdot5=10</math> ways here.<br />
<br />
Tina gets a sum of 6: Two ways here <math>(1,5)</math> and <math>(2,4)</math>. Sergio can choose a number from 7 to 10, so <math>2\cdot4=8</math> here.<br />
<br />
Tina gets a sum of 7: Two ways here <math>(2,5)</math> and <math>(3,4)</math>. Sergio can choose from 8 to 10, so <math>2\cdot3=6</math> ways here.<br />
<br />
Tina gets a sum of 8: Only one way possible <math>(3,5</math>). Sergio chooses 9 or 10, so 2 ways here.<br />
<br />
Tina gets a sum of 9: Only one way <math>(4,5)</math>. Sergio must choose 10, so 1 way.<br />
<br />
In all, there are <math>7+6+10+8+6+2+1=40</math> ways. Tina chooses two distinct numbers in <math>\binom{5}{2}=10</math> ways while Sergio chooses a number in <math>10</math> ways, so there are <math>10\cdot 10=100</math> ways in all. Since <math>\frac{40}{100}=\frac{2}{5}</math>, our answer is <math>\boxed{\text{(A)}\frac{2}{5}}</math>.<br />
<br />
=== Solution 2 ===<br />
We want to find the average of the smallest possible chance of Sergio winning and the largest possible chance of Sergio winning. This is because the probability decreases linearly. The largest possibility of Sergio winning if Tina chooses a 1 and a 2. The chances of Sergio winning is then <math> \frac{7}{10}</math> . The smallest possibility of Sergio winning is if Tina chooses a 4 and a 5. The chances of Sergio winning then is <math> \frac{1}{10}</math>. The average of <math> \frac{7}{10} </math> and <math> \frac{1}{10} </math> is <math>\boxed{\text{(A)}\frac{2}{5}}</math>.<br />
<br />
=== Solution 3 ===<br />
We invoke some symmetry. Let <math>T</math> denote Tina's sum, and let <math>S</math> denote Sergio's number. Observe that, for <math>i = 2, 3, \ldots, 10</math>, <math>\text{Pr}(T=i) = \text{Pr}(T=12-i)</math>.<br />
<br />
If Tina's sum is <math>i</math>, then the probability that Sergio's number is larger than Tina's sum is <math>\frac{10-i}{10}</math>. Thus, the probability <math>P</math> is<br />
<br />
<cmath>P = \text{Pr}(S>T) = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{10-i}{10}</cmath><br />
<br />
Using the symmetry observation, we can also write the above sum as<br />
<cmath> P = \sum_{i=2}^{10} \text{Pr}(T=12-i) \times \frac{10-i}{10} = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{i-2}{10}</cmath><br />
where the last equality follows as we reversed the indices of the sum (by replacing <math>12-i</math> with <math>i</math>). Thus, adding the two equivalent expressions for <math>P</math>, we have<br />
<br />
<cmath><br />
\begin{align*}<br />
2P &= \sum_{i=2}^{10} \text{Pr}(T=i) \times \left(\frac{10-i}{10} + \frac{i-2}{10}\right) \\<br />
&= \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{4}{5} \\<br />
&= \frac{4}{5} \sum_{i=2}^{10} \text{Pr}(T=i) \\<br />
&= \frac{4}{5}<br />
\end{align*}<br />
</cmath><br />
<br />
Since this represents twice the desired probability, the answer is <math>P = \boxed{\textbf{(A)} \frac{2}{5}}</math>. -scrabbler94<br />
<br />
==Solution 4==<br />
We have 5 cases, if Tina choose <math>1, 2, 3, 4,</math> or <math>5.</math><br />
<br />
The number of ways of choosing 2 numbers from <math>5</math> are <math>\binom{5}{2}</math>.<br />
--------------------------<br />
Case 1: Tina chooses <math>1</math>.<br />
<br />
In this case, since the numbers are distinct, Tina can choose <math>(1, 2), (1, 3), (1, 4),</math> or <math>(1, 5).</math><br />
<br />
If Tina chooses <math>1</math> and <math>2</math> which some to <math>3</math>, Sergio only has <math>10-3=7</math> choices.<br />
<br />
Since the sum of the combined numbers increases by <math>1</math> every time for this specific case, Sergio has <math>1</math> less choice every time.<br />
<br />
Therefore, the probability of this is <math>\frac{7+6+5+4}{10 \cdot \binom{5}{2}}</math>.<br />
--------------------------<br />
If you do this over and over again you will see that you have <math>\frac{(7+6+5+4)+(7+5+4+3)+(6+5+3+2)+(5+4+3+1)+(4+3+2+1)}{10 \cdot \binom{5}{2}} = \frac{80}{100} = \frac{4}{5}</math> probability.<br />
<br />
But since we overcounted by 2 (e.g. <math>(1, 2)</math> and <math>(2, 1)</math>) we need to divide by <math>2.</math><br />
<br />
Thus our answer is <math>\frac{4}{5} \div 2 = \boxed{\textbf{(A)} \frac{2}{5}}.</math><br />
<br />
~mathboy282<br />
<br />
Note: I will add in all the cases soon, kind of busy today so yea.<br />
<br />
=== Solution 5 ===<br />
<br />
Assume Sergio chooses from <math>{2,3,\ldots,10}</math>. The probably of Tina getting a sum of <math>6+x</math> and <math>6-x</math> (<math>x \leq 4</math>) are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than <math>6+x</math> is equal to him choosing numbers lower/higher than <math>6-x</math>. Therefore over all of Tina's sums, the probability of Sergio choosing a number higher is equal to the probability of choosing a number lower. <br />
<br />
The probability that they get the same value is <math>1/9</math>, so the probability of Sergio getting a higher number is <math>\frac{(9-1)/2}{9} = \frac49</math>. <br />
<br />
Sergio never wins when choosing <math>1</math> so the probability is <math>\frac49 \frac{9}{10} + (0)\frac{1}{10} = \boxed{\textbf{(A)} \frac{2}{5}}.</math><br />
<br />
~zeric<br />
<br />
==See Also==<br />
{{AMC12 box|year=2002|ab=A|num-b=15|num-a=17}}<br />
{{AMC10 box|year=2002|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Probability Problems]]<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_17&diff=1428022000 AMC 12 Problems/Problem 172021-01-19T19:14:37Z<p>Zeric: New solution</p>
<hr />
<div>== Problem ==<br />
A circle centered at <math>O</math> has radius <math>1</math> and contains the point <math>A</math>. The segment <math>AB</math> is tangent to the circle at <math>A</math> and <math>\angle AOB = \theta</math>. If point <math>C</math> lies on <math>\overline{OA}</math> and <math>\overline{BC}</math> bisects <math>\angle ABO</math>, then <math>OC =</math><br />
<br />
<asy><br />
import olympiad;<br />
size(6cm);<br />
unitsize(1cm);<br />
defaultpen(fontsize(8pt)+linewidth(.8pt));<br />
labelmargin=0.2;<br />
dotfactor=3;<br />
pair O=(0,0);<br />
pair A=(1,0);<br />
pair B=(1,1.5);<br />
pair D=bisectorpoint(A,B,O);<br />
pair C=extension(B,D,O,A);<br />
draw(Circle(O,1));<br />
draw(O--A--B--cycle);<br />
draw(B--C);<br />
label("$O$",O,SW);<br />
dot(O);<br />
label("$\theta$",(0.1,0.05),ENE);<br />
dot(C);<br />
label("$C$",C,S);<br />
dot(A);<br />
label("$A$",A,E);<br />
dot(B);<br />
label("$B$",B,E);</asy><br />
<br />
<math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}</math><br />
<br />
== Solution 1 ==<br />
Since <math>\overline{AB}</math> is tangent to the circle, <math>\triangle OAB</math> is a right triangle. This means that <math>OA = 1</math>, <math>AB = \tan \theta</math> and <math>OB = \sec \theta</math>. By the [[Angle Bisector Theorem]], <cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by <math>\cos \theta</math> to simplify the trigonometric functions, <cmath> AC=OC \sin \theta </cmath> Since <math>AC + OC = 1</math>, <math>1 - OC = OC \sin \theta \Longrightarrow</math> <math>OC = \dfrac{1}{1+\sin \theta}</math>. Therefore, the answer is <math>\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}</math>.<br />
<br />
== Solution 2 ==<br />
Alternatively, one could notice that OC approaches the value 1/2 as theta gets close to 90 degrees. The only choice that is consistent with this is (D).<br />
<br />
== Solution 3 (with minimal trig) ==<br />
Let's assign a value to <math>\theta</math> so we don't have to use trig functions to solve. <math>60</math> is a good value for <math>\theta</math>, because then we have a <math>30-60-90 \triangle</math> -- <math>\angle BAC=90</math> because <math>AB</math> is tangent to Circle <math>O</math>.<br />
<br />
Using our special right triangle, since <math>AO=1</math>, <math>OB=2</math>, and <math>AB=\sqrt{3}</math>.<br />
<br />
Let <math>OC=x</math>. Then <math>CA=1-x</math>. since <math>BC</math> bisects <math>\angle ABO</math>, we can use the angle bisector theorem:<br />
<br />
<math>\frac{2}{x}=\frac{\sqrt{3}}{1-x}</math><br />
<br />
<math>2-2x=\sqrt{3}x</math><br />
<br />
<math>2=(\sqrt{3}+2)x</math><br />
<br />
<math>x=\frac{2}{\sqrt{3}+2}</math>.<br />
<br />
Now, we only have to use a bit of trig to guess and check: the only trig facts we need to know to finish the problem is:<br />
<br />
<math>\sin\theta =\frac{\text{Opposite}}{\text{Hypotenuse}}</math><br />
<br />
<math>\cos\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}</math><br />
<br />
<math>\tan\theta =\frac{\text{Opposite}}{\text{Adjacent}}</math>.<br />
<br />
With a bit of guess and check, we get that the answer is <math>\boxed{D}</math>.<br />
<br />
== Solution 4 ==<br />
Let <math>OC</math> = x, <math>OB</math> = h, and <math>AB</math> = y. <math>AC</math> = <math>OA</math> - <math>OC</math>.<br />
<br />
Because <math>OC</math> = x, and <math>OA</math> = 1 (given in the problem), <math>AC</math> = 1-x.<br />
<br />
Using the [[Angle Bisector Theorem]], <math>\frac{h}{y}</math> = <math>\frac{x}{1-x}</math> <math>\Longrightarrow</math> h(1-x) = xy. Solving for x gives us x = <math>\frac{h}{h+y}</math>.<br />
<br />
<math>\sin\theta = \frac{opposite}{hypotenuse} = \frac{y}{h}</math>. Solving for y gives us y = h <math>\sin\theta</math>.<br />
<br />
Substituting this for y in our initial equation yields x = <math>\dfrac{h}{h+\sin \theta}</math>.<br />
<br />
Using the distributive property, x = <math>\dfrac{h}{h(1+\sin \theta)}</math> and finally <math>\dfrac{1}{1+\sin \theta}</math> or <math>\boxed{\textbf{(D)}}</math><br />
<br />
== Solution 5 ==<br />
Since <math>\overline{AB}</math> is tangent to the circle, <math>\angle OAB=90^{\circ}</math> and thus we can use trig ratios directly. <br />
<br />
<cmath>\sin{\theta}=\frac{\overline{AB}}{\overline{BO}}, \cos{\theta}=\frac{1}{\overline{BO}}, \tan{\theta}=\overline{AB}</cmath><br />
<br />
By the angle bisector theorem, we have <br />
<br />
<cmath>\frac{\overline{OB}}{\overline{AB}}=\frac{\overline{OC}}{\overline{CA}}</cmath><br />
<br />
Seeing the resemblance of the ratio on the left-hand side to <math>\sin{\theta},</math> we turn the ratio around to allow us to plug in <math>\sin{\theta}.</math> Another source of motivation for this also lies in the idea of somehow adding 1 to the right-hand side so that we can substitute for a given value, i.e. <math>\overline{OA}=1</math>, and flipping the fraction will preserve the <math>\overline{OC}</math>, whilst adding one right now would make the equation remain in direct terms of <math>\overline{CA}.</math><br />
<br />
<cmath>\frac{\overline{AB}}{\overline{OB}}=\sin{\theta}=\frac{\overline{CA}}{\overline{OC}}\Rightarrow \sin{\theta}+1=\frac{\overline{CA}+\overline{OC}}{\overline{OC}}=\frac{1}{\overline{OC}}</cmath><br />
<br />
<cmath>\sin{\theta}+1=\frac{1}{\overline{OC}} \Rightarrow \boxed{\overline{OC}=\frac{1}{\sin{\theta}+1}}</cmath><br />
<br />
== Solution 6 (tangent half angle) ==<br />
<br />
<math>\angle CBO = 45^{\circ} - \frac{\theta}{2}, \angle ACB = 45^{\circ} + \frac{\theta}{2}, OB = \frac{1}{\cos(\theta)}</math>.<br />
By sine law, <math>\frac{OC}{\sin(\angle CBO)} = \frac{OB}{\sin(\angle OCB)} = \frac{OB}{\sin(\angle ACB)}</math><br />
<br />
<cmath>OC = \frac{\sin(45^{\circ} - \frac{\theta}{2})}{\sin(45^{\circ} + \frac{\theta}{2})}OB = \frac{\sin(45^{\circ} - \frac{\theta}{2})}{\cos(45^{\circ} - \frac{\theta}{2})}OB = \tan(45^{\circ} - \frac{\theta}{2})OB = \frac{1-\tan(\theta/2)}{1+\tan(\theta/2)}OB</cmath><br />
<br />
Let <math>t = \tan(\theta/2)</math>. <math>OC = \frac{1-t}{1+t}OB = \frac{1-t^2}{1+2t+t^2}</math>. Because <math>\sin(\theta) = \frac{2t}{1+t^2}</math> and <math>\cos(\theta) = \frac{1-t^2}{1+t^2}</math>, <cmath>OC = \frac{\cos(\theta)}{1+\sin(\theta)}OB = \boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}</cmath><br />
<br />
== See also ==<br />
{{AMC12 box|year=2000|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Introductory Trigonometry Problems]]<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_9&diff=1196172020 AIME I Problems/Problem 92020-03-17T05:44:30Z<p>Zeric: </p>
<hr />
<div><br />
== Problem ==<br />
Let <math>S</math> be the set of positive integer divisors of <math>20^9.</math> Three numbers are chosen independently and at random with replacement from the set <math>S</math> and labeled <math>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> divides <math>a_3</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m.</math><br />
<br />
== Solution ==<br />
<br />
<asy><br />
size(12cm);<br />
for (int x = 1; x < 18; ++x) {<br />
draw((x, 0) -- (x, 9), dotted);<br />
}<br />
for (int y = 1; y < 9; ++y) {<br />
draw((0, y) -- (18, y), dotted);<br />
}<br />
<br />
draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle);<br />
<br />
pair b1, b2, b3;<br />
pair c1, c2, c3;<br />
pair a1, a2, a3;<br />
b1 = (3, 0); b2 = (12, 0); b3 = (16, 0);<br />
c1 = (0, 2); c2 = (0, 4); c3 = (0, 8);<br />
a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3;<br />
<br />
draw(b1 -- a1 -- c1);<br />
draw(b2 -- a2 -- c2);<br />
draw(b3 -- a3 -- c3);<br />
<br />
dot(a1); dot(a2); dot(a3);<br />
label("$a_1$", a1, NE);<br />
label("$a_2$", a2, NE);<br />
label("$a_3$", a3, NE);<br />
label("$b_1$", b1, S);<br />
label("$b_2$", b2, S);<br />
label("$b_3$", b3, S);<br />
label("$c_1$", c1, W);<br />
label("$c_2$", c2, W);<br />
label("$c_3$", c3, W);<br />
<br />
</asy><br />
<br />
First, prime factorize <math>20^9</math> as <math>2^{18} \cdot 5^9</math>. Denote <math>a_1</math> as <math>2^{b_1} \cdot 5^{c_1}</math>, <math>a_2</math> as <math>2^{b_2} \cdot 5^{c_2}</math>, and <math>a_3</math> as <math>2^{b_3} \cdot 5^{c_3}</math>.<br />
<br />
In order for <math>a_1</math> to divide <math>a_2</math>, and for <math>a_2</math> to divide <math>a_3</math>, <math>b_1\le b_2\le b_3</math>, and <math>c_1\le c_2\le c_3</math>. We will consider each case separately. Note that the total amount of possibilities is <math>190^3</math>, as there are <math>(18+1)(9+1)=190</math> choices for each factor.<br />
<br />
We notice that if we add <math>1</math> to <math>b_2</math> and <math>2</math> to <math>b_3</math>, then we can reach the stronger inequality <math>0\le b_1<b_2+1<b_3+2\le 20</math>. Therefore, if we pick <math>3</math> integers from <math>0</math> to <math>20</math>, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers <math>b_1</math>, <math>b_2+1</math>, and <math>b_3+2</math>. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is <math>\dbinom{21}{3}</math>.<br />
<br />
The case for <math>c_1</math>,<math>c_2</math>, and <math>c_3</math> proceeds similarly for a result of <math>\dbinom{12}{3}</math>. Therefore, the probability of choosing three such factors is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>.<br />
<br />
-molocyxu<br />
<br />
== Solution 2==<br />
<br />
Same as before, say the factors have powers of <math>b</math> and <math>c</math>. <math>b_1, b_2, b_3</math> can either be all distinct, all equal, or two of the three are equal. As well, we must have <math>b_1 \leq b_2 \leq b_3</math>. If they are all distinct, the number of cases is simply <math>{19 \choose 3}</math>. If they are all equal, there are only <math>19</math> cases for the general value. If we have a pair equal, then we have <math>2 \cdot {19\choose 2}</math>. We need to multiply by <math>2</math> because if we have two values <math>b_i < b_j</math>, we can have either <math>(b_i, b_i, b_j)</math> or <math>(b_i, b_j, b_j)</math>. <br />
<br />
<cmath>{19 \choose 3} + 2 \cdot {19 \choose 2} + 19 = 1330</cmath><br />
<br />
Likewise for <math>c</math>, we get<br />
<br />
<cmath>{10 \choose 3} + 2 \cdot {10 \choose 2} + 10 = 220</cmath><br />
<br />
The final probability is simply <math>\frac{1330 \cdot 220}{190^3}</math>. Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>.<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_8&diff=1171342020 AMC 12B Problems/Problem 82020-02-08T01:22:44Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
<br />
How many ordered pairs of integers <math>(x,y)</math> satisfy the equation <cmath>x^{2020}+y^2=2y</cmath><br />
<br />
==Solution==<br />
Set it up as a quadratic in terms of y:<br />
<cmath>y^2-2y+x^{2020}=0</cmath><br />
Then the discriminant is<br />
<cmath>\Delta = 4-4x^{2020}</cmath><br />
This will clearly only yield real solutions when <math>x^{2020} \leq 1</math>, because it is always positive.<br />
Then <math>x=-1,0,1</math>. Checking each one:<br />
<math>-1</math> and <math>1</math> are the same when raised to the 2020th power:<br />
<cmath>y^2-2y+1=(y-1)^2=0</cmath><br />
This has only has solutions <math>1</math>, so <math>(\pm 1,1)</math> are solutions.<br />
Next, if <math>x=0</math>:<br />
<cmath>y^2-2y=0</cmath><br />
Which has 2 solutions, so <math>(0,2)</math> and <math>(0,0)</math><br />
<br />
These are the only 4 solutions, so <math>\boxed{D}</math><br />
<br />
==Solution 2==<br />
<br />
Move the <math>y^2</math> term to the other side to get <math>x^{2020}=2y-y^2 = y(2-y)</math>. Because <math>x^{2020} \geq 0</math> for all <math>x</math>, then <math>y(2-y) \geq 0 \Rightarrow y = 0,1,2</math>. If <math>y=0</math> or <math>y=2</math>, the right side is <math>0</math> and therefore <math>x=0</math>. When <math>y=1</math>, the right side become <math>1</math>, therefore <math>x=1,-1</math>. Our solutions are <math>(0,2)</math>, <math>(0,0)</math>, <math>(1,1)</math>, <math>(-1,1)</math>. There are <math>4</math> solutions, so the answer is <math>\boxed{D}</math><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_4&diff=1171302020 AMC 10B Problems/Problem 42020-02-08T01:17:40Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
<br />
The acute angles of a right triangle are <math>a^{\circ}</math> and <math>b^{\circ}</math>, where <math>a>b</math> and both <math>a</math> and <math>b</math> are prime numbers. What is the least possible value of <math>b</math>?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11</math><br />
<br />
==Solution==<br />
<br />
Since the three angles of a triangle add up to <math>180^{\circ}</math> and one of the angles is <math>90^{\circ}</math> because it's a right triangle, then <math>a^{\circ} + b^{\circ} = 90^{\circ}</math>.<br />
<br />
The greatest prime number less than <math>90</math> is <math>89</math>. If <math>a=89^{\circ}</math>, then <math>b=90^{\circ}-89^{\circ}=1^{\circ}</math>, which is not prime.<br />
<br />
The next greatest prime number less than <math>90</math> is <math>83</math>. If <math>a=83^{\circ}</math>, then <math>b=7^{\circ}</math>, which IS prime, so we have our answer <math>\boxed{\textbf{(D)}\ 7}</math> ~quacker88<br />
<br />
<br />
==Solution 2==<br />
<br />
Looking at the answer choices, only <math>7</math> and <math>11</math> are coprime to <math>90</math>. Testing <math>7</math> makes the other angle <math>83</math> which is prime, therefore our answer is <math>\boxed{\textbf{(D)}\ 7}</math><br />
==Video Solution==<br />
https://youtu.be/Gkm5rU5MlOU<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_17&diff=1160382020 AMC 10A Problems/Problem 172020-02-01T03:07:22Z<p>Zeric: </p>
<hr />
<div>Define<cmath>P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).</cmath>How many integers <math>n</math> are there such that <math>P(n)\leq 0</math>?<br />
<br />
<math>\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100</math><br />
<br />
==Solution 1==<br />
<br />
Notice that <math>P(x)</math> is a product of many integers. We either need one factor to be 0 or an odd number of negative factors.<br />
Case 1: There are 100 integers <math>n</math> for which <math>P(x)=0</math><br />
Case 2: For there to be an odd number of negative factors, <math>n</math> must be between an odd number squared and an even number squared. This means that there are <math>2+6+\dots+10</math> total possible values of <math>n</math>. Simplifying, there are <math>5000</math> possible numbers.<br />
<br />
Summing, there are <math>\boxed{\textbf{(E) } 5100}</math> total possible values of <math>n</math>.<br />
<br />
==Solution 2==<br />
<br />
Notice that <math>P(x)</math> is nonpositive when <math>x</math> is between <math>100^2</math> and <math>99^2</math>, <math>98^2</math> and <math>97^2 \ldots</math> , <math>2^2</math> and <math>1^2</math> (inclusive), which means that the amount of values equals <math>((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \ldots + ((2+1)(2-1)+1)</math>.<br />
<br />
This reduces to <math>200 + 196 + 192 + \ldots + 4 = 4(1+2+\ldots + 50) = 4 \frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}</math><br />
<br />
~Zeric<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_5&diff=1160212020 AMC 10A Problems/Problem 52020-02-01T02:57:01Z<p>Zeric: </p>
<hr />
<div>==Problem 5==<br />
What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math><br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25</math><br />
<br />
== Solution 1== <br />
<br />
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.<br />
<br />
The first case yields <math>x^2-12x+34=2</math>, which is equal to <math>(x-4)(x-8)=0</math>. Therefore, the two values for the positive case is <math>4</math> and <math>8</math>.<br />
<br />
Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>-x^2+12x-34=2</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>.<br />
<br />
Summing all the values results in <math>4+8+6=\boxed{\text{(C) }18}</math>.<br />
<br />
== Solution 2== <br />
We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36=0</math>.<br />
<br />
Notice that the second is a perfect square with a double root at <math>x=6</math>, and the first has real roots. By Vieta's, the sum of the roots of the first equation is <math>12</math>. <math>12+6=\boxed{\text{(C) }18}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_5&diff=1160202020 AMC 10A Problems/Problem 52020-02-01T02:56:31Z<p>Zeric: </p>
<hr />
<div>==Problem 5==<br />
What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math><br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25</math><br />
<br />
== Solution 1== <br />
<br />
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.<br />
<br />
The first case yields <math>x^2-12x+34=2</math>, which is equal to <math>(x-4)(x-8)=0</math>. Therefore, the two values for the positive case is <math>4</math> and <math>8</math>.<br />
<br />
Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>-x^2+12x-34=2</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>.<br />
<br />
Summing all the values results in <math>4+8+6=\boxed{\text{(C) }18}</math>.<br />
<br />
== Solution 2== <br />
We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36=2</math>.<br />
<br />
Notice that the second is a perfect square with a double root at <math>x=6</math>, and the first has real roots. By Vieta's, the sum of the roots of the first equation is <math>12</math>. <math>12+6=\boxed{\text{(C) }18}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_5&diff=1160182020 AMC 10A Problems/Problem 52020-02-01T02:56:06Z<p>Zeric: </p>
<hr />
<div>==Problem 5==<br />
What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math><br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25</math><br />
<br />
== Solution 1== <br />
<br />
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.<br />
<br />
The first case yields <math>x^2-12x+34=2</math>, which is equal to <math>(x-4)(x-8)=0</math>. Therefore, the two values for the positive case is <math>4</math> and <math>8</math>.<br />
<br />
Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>-x^2+12x-34=2</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>.<br />
<br />
Summing all the values results in <math>4+8+6=\boxed{\text{(C) }18}</math>.<br />
<br />
== Solution 2== <br />
We have the equations <math>x^2-12x+32=0</math> and x^2-12x+36=2<math>.<br />
<br />
Notice that the second is a perfect square with a double root at </math>x=6<math>, and the first has real roots. By Vieta's, the sum of the roots of the first equation is </math>12<math>. </math>12+6=\boxed{\text{(C) }18}$.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_21&diff=1155382018 AMC 12A Problems/Problem 212020-01-25T21:02:56Z<p>Zeric: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Which of the following polynomials has the greatest real root?<br />
<math>\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018 </math><br />
<br />
==Solution 1==<br />
<br />
We can see that our real solution has to lie in the open interval <math>(-1,0)</math>. From there, note that <math>x^a < x^b</math> if <math>a</math>, <math>b</math> are odd positive integers if <math>a<b</math>, so hence it can only either be B or E(as all of the other polynomials will be larger than the polynomial B). E gives the solution <math>x=-\frac{2018}{2019}</math>. We can approximate the root for B by using <math>x=-\frac 12</math>. <cmath> (- \frac {1}{2}) ^{17} - \frac{2018}{2048} + 1 \approx 0</cmath> therefore the root for B is approximately <math>-\frac 12</math>. The answer is <math>\fbox{B}</math>. (cpma213)<br />
<br />
==Solution 2 (Calculus version of solution 1)==<br />
<br />
Note that <math>a(-1)=b(-1)=c(-1)=d(-1) < 0</math> and <math>a(0)=b(0)=c(0)=d(0) > 0</math>. Calculating the definite integral for each function on the interval <math>[-1,0]</math>, we see that <math>B(x)\rvert^{0}_{-1}</math> gives the most negative value. To maximize our real root, we want to maximize the area of the curve under the x-axis, which means we want our integral to be as negative as possible and thus the answer is <math>\fbox{B}</math>.<br />
<br />
==Solution 3 (Alternate Calculus Version)==<br />
Newton's Method is used to approximate the zero <math>x_{1}</math> of any real valued function given an estimation for the root <math>x_{0}</math>: <math>x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}\,.</math> After looking at all the options, <math>x_{0}=-1</math> gives a reasonable estimate. For options A to D, <math>f(-1) = -1</math> and the estimation becomes <math>x_{1}=-1+{\frac {1}{f'(-1)}}\,.</math> Thus we need to minimize the derivative, giving us B. Now after comparing B and E through Newton's method, we see that B has the higher root, so the answer is <math>\fbox{B}</math>. (Qcumber)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_6&diff=1155342018 AMC 12A Problems/Problem 62020-01-25T18:15:56Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
For positive integers <math>m</math> and <math>n</math> such that <math>m+10<n+1</math>, both the mean and the median of the set <math>\{m, m+4, m+10, n+1, n+2, 2n\}</math> are equal to <math>n</math>. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24</math><br />
<br />
==Solution 1==<br />
The mean and median are <cmath>\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,</cmath>so <math>3m+17=2n</math> and <math>m+11=n</math>. Solving this gives <math>\left(m,n\right)=\left(5,16\right)</math> for <math>m+n=\boxed{21}</math>. (trumpeter)<br />
<br />
==Solution 2==<br />
This is an alternate solution if you don't want to solve using algebra. First, notice that the median <math>n</math> is the average of <math>m+10</math> and <math>n+1</math>. Therefore, <math>n=m+11</math>, so the answer is <math>m+n=2m+11</math>, which must be odd. This leaves two remaining options: <math>{(B) 21}</math> and <math>{(D) 23}</math>. Notice that if the answer is <math>(B)</math>, then <math>m</math> is odd, while <math>m</math> is even if the answer is <math>(D)</math>. Since the average of the set is an integer <math>n</math>, the sum of the terms must be even. <math>4+10+1+2+2n</math> is odd by definition, so we know that <math>3m+2n</math> must also be odd, thus with a few simple calculations <math>m</math> is odd. Because all other answers have been eliminated, <math>(B)</math> is the only possibility left. Therefore, <math>m+n=\boxed{21}</math>. ∎ --anna0kear<br />
<br />
==Solution 3==<br />
Since the median is <math>n</math>, then <math>\frac{m+10+n+1}{2} = n \Rightarrow m+11 = n</math>, or <math>m= n-11</math>. Plug this in for <math>m</math> values to get <math>\frac{7n-16}{6} = n \Rightarrow 7n-16 = 6n \Rightarrow n= 16</math>. Plug it back in to get <math>m = 5</math>, thus <math>16 + 5 = \boxed{21}</math>.<br />
<br />
~ iron<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_25&diff=1152782019 AMC 12B Problems/Problem 252020-01-23T21:06:17Z<p>Zeric: /* Solution 1 (vectors) */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCD</math> be a convex quadrilateral with <math>BC=2</math> and <math>CD=6.</math> Suppose that the centroids of <math>\triangle ABC,\triangle BCD,</math> and <math>\triangle ACD</math> form the vertices of an equilateral triangle. What is the maximum possible value of <math>ABCD</math>?<br />
<br />
<math>\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30</math><br />
<br />
==Solution 1 (vectors)==<br />
Place an origin at <math>A</math>, and assign position vectors of <math>B = \vec{p}</math> and <math>D = \vec{q}</math>. Since <math>AB</math> is not parallel to <math>AD</math>, vectors <math>\vec{p}</math> and <math>\vec{q}</math> are linearly independent, so we can write <math>C = m\vec{p} + n\vec{q}</math> for some constants <math>m</math> and <math>n</math>. Now, recall that the centroid of a triangle <math>\triangle XYZ</math> has position vector <math>\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)</math>. <br />
<br />
Thus the centroid of <math>\triangle ABC</math> is <math>g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}</math>; the centroid of <math>\triangle BCD</math> is <math>g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>; and the centroid of <math>\triangle ACD</math> is <math>g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>. <br />
<br />
Hence <math>\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}</math>, <math>\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}</math>, and <math>\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}</math>. For <math>\triangle G_{1}G_{2}G_{3}</math> to be equilateral, we need <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD</math>. Further, <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD</math>. Hence we have <math>AB = AD = BD</math>, so <math>\triangle ABD</math> is equilateral.<br />
<br />
Now let the side length of <math>\triangle ABD</math> be <math>k</math>, and let <math>\angle BCD = \theta</math>. By the Law of Cosines in <math>\triangle BCD</math>, we have <math>k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}</math>. Since <math>\triangle ABD</math> is equilateral, its area is <math>\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}</math>, while the area of <math>\triangle BCD</math> is <math>\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}</math>. Thus the total area of <math>ABCD</math> is <math>10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\sin{\theta} \frac{1}{2} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}</math>, where in the last step we used the subtraction formula for <math>\sin</math>. Alternatively, we can use calculus to find the local maximum. Observe that <math>\sin{\left(\theta-60^{\circ}\right)}</math> has maximum value <math>1</math> when e.g. <math>\theta = 150^{\circ}</math>, which is a valid configuration, so the maximum area is <math>10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}</math>.<br />
<br />
==Solution 2==<br />
<br />
Let <math>G_1</math>, <math>G_2</math>, <math>G_3</math> be the centroids of <math>ABC</math>, <math>BCD</math>, and <math>CDA</math> respectively, and let <math>M</math> be the midpoint of <math>BC</math>. <math>A</math>, <math>G_1</math>, and <math>M</math> are collinear due to well-known properties of the centroid. Likewise, <math>D</math>, <math>G_2</math>, and <math>M</math> are collinear as well. Because (as is also well-known) <math>AG_1 = 3AM</math> and <math>DG_2 = 3DM</math>, we have <math>\triangle MG_1G_2\sim\triangle MAD</math>. This implies that <math>AD</math> is parallel to <math>G_1G_2</math>, and in terms of lengths, <math>AD = 3G_1G_2</math>.<br />
<br />
We can apply the same argument to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>, concluding that <math>AB</math> is parallel to <math>G_2G_3</math> and <math>AB = 3G_2G_3</math>. Because <math>3G_1G_2 = 3G_2G_3</math> (due to the triangle being equilateral), <math>AB = AD</math>, and the pair of parallel lines preserve the <math>60^{\circ}</math> angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore <math>\triangle BAD</math> is equilateral.<br />
<br />
At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:<br />
<br />
Let <math>BD = 2x</math>, where <math>2 < x < 4</math> due to the Triangle Inequality in <math>\triangle BCD</math>. By breaking the quadrilateral into <math>\triangle ABD</math> and <math>\triangle BCD</math>, we can create an expression for the area of <math>ABCD</math>. We use the formula for the area of an equilateral triangle given its side length to find the area of <math>\triangle ABD</math> and Heron's formula to find the area of <math>\triangle BCD</math>.<br />
<br />
After simplifying,<br />
<br />
<cmath>[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}</cmath><br />
<br />
Substituting <math>k = x^2 - 10</math>, the expression becomes<br />
<br />
<cmath>[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}</cmath><br />
<br />
We can ignore the <math>10\sqrt{3}</math> for now and focus on <math>k\sqrt{3} + \sqrt{36 - k^2}</math>.<br />
<br />
By the Cauchy-Schwarz inequality,<br />
<br />
<cmath>\left(k\sqrt 3 + \sqrt{36-k^2}\right)^2 \leq \left(\left(\sqrt{3}\right)^2+1^2\right)\left(\left(k\right)^2 + \left(\sqrt{36-k^2}\right)^2\right)</cmath><br />
<br />
The RHS simplifies to <math>12^2</math>, meaning the maximum value of <math>k\sqrt{3} + \sqrt{36 - k^2}</math> is <math>12</math>.<br />
<br />
Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_19&diff=1151732018 AMC 10B Problems/Problem 192020-01-23T03:47:15Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is <math>1</math> year older than Chloe, and Zoe is exactly <math>1</math> year old today. Today is the first of the <math>9</math> birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?<br />
<br />
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11 </math><br />
<br />
==Solution 1==<br />
Let Joey's age be <math>j</math>, Chloe's age be <math>c</math>, and we know that Zoe's age is <math>1</math>. <br />
<br />
We know that there must be <math>9</math> values <math>k\in\mathbb{Z}</math> such that <math>c+k=a(1+k)</math> where <math>a</math> is an integer.<br />
<br />
Therefore, <math>c-1+(1+k)=a(1+k)</math> and <math>c-1=(1+k)(a-1)</math>. Therefore, we know that, as there are <math>9</math> solutions for <math>k</math>, there must be <math>9</math> solutions for <math>c-1</math>. We know that this must be a perfect square. Testing perfect squares, we see that <math>c-1=36</math>, so <math>c=37</math>. Therefore, <math>j=38</math>. Now, since <math>j-1=37</math>, by similar logic, <math>37=(1+k)(a-1)</math>, so <math>k=36</math> and Joey will be <math>38+36=74</math> and the sum of the digits is <math>\boxed{\text{(E) }11}</math><br />
<br />
== Solution 2 ==<br />
<br />
Here's a different way of saying the above solution:<br />
<br />
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has <math>9</math> factors. Therefore, the difference between Chloe and Zoe's age is <math>36</math>, so Chloe is <math>37</math>, and Joey is <math>38</math>. The common factor that will divide both of their ages is <math>37</math>, so Joey will be <math>74</math>. <math>7 + 4 = \boxed{\text{(E) }11}</math><br />
<br />
==Solution 3==<br />
Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y)<br />
<br />
Let <math>C+n</math> denote Chloe's age, <math>J+n</math> denote Joey's age, and <math>Z+n</math> denote Zoe's age, where <math>n</math> is the number of years from now. We are told that <math>C+n</math> is a multiple of <math>Z+n</math> exactly nine times. Because <math>Z+n</math> is <math>1</math> at <math>n=0</math> and will increase until greater than <math>C-Z</math>, it will hit every natural number less than <math>C-Z</math>, including every factor of <math>C-Z</math>. For <math>C+n</math> to be an integral multiple of <math>Z+n</math>, the difference <math>C-Z</math> must also be a multiple of <math>Z</math>, which happens if <math>Z</math> is a factor of <math>C-Z</math>. Therefore, <math>C-Z</math> has nine factors. The smallest number that has nine positive factors is <math>2^23^2=36</math> . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know <math>Z=1</math> and <math>J=C+1</math>. Thus, <cmath>C-Z=36</cmath> <cmath>J-Z=37</cmath> By our above logic, the next time <math>J-Z</math> is a multiple of <math>Z+n</math> will occur when <math>Z+n</math> is a factor of <math>J-Z</math>. Because <math>37</math> is prime, the next time this happens is at <math>Z+n=37</math>, when <math>J+n=74</math>. <math>7+4=\boxed{(\textbf{E}) 11}</math><br />
<br />
==Solution 4==<br />
Denote Zoe's age with <math>n</math>, then Chloe's age is <math>C-1+n</math> where <math>C</math> represents Chloe's age when Zoe is one. We must have <math>n | C-1+n</math>. Obviously <math>n|n</math>, therefore, <math>n | C-1</math> for 9 values of <math>n</math>, and therefore, <math>C-1</math> has <math>9</math> factors. <math>C-1</math> either takes the form of <math>a^8</math> (which is too large) or <math>a^2 b^2</math>. <math>C</math> must be less than <math>100</math> and <math>a</math> and <math>b</math> must be prime, therefore the only answer is <math>C-1 = 36</math>. Joey's age is <math>37+n</math>, which is divisible by <math>n</math> when <math>n|37</math>, therefore the answer occurs when <math>n=37</math> and Joey is <math>74</math>. <math>7+4=\boxed{(\textbf{E}) 11}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2018|ab=B|num-b=13|num-a=15}}<br />
<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&diff=1056932008 AMC 12A Problems/Problem 222019-05-06T01:34:26Z<p>Zeric: </p>
<hr />
<div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #22]] and [[2008 AMC 10A Problems/Problem 25|2008 AMC 10A #25]]}}<br />
==Problem==<br />
A round table has radius <math>4</math>. Six rectangular place mats are placed on the table. Each place mat has width <math>1</math> and length <math>x</math> as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length <math>x</math>. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is <math>x</math>?<br />
<br />
<asy>unitsize(4mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(1\)",(-0.5,3.8),S);</asy><br />
<br />
<math>\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}</math><br />
<br />
==Solution==<br />
=== Solution 1 (trigonometry) ===<br />
Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown: <br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(x\)",(0.03,1.5),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(B\)",(-3.15,1.35),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(D\)",(-0.75,4.15),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-0.1,3.8),S);<br />
label("\(4\)",(-0.4,2.2),S);<br />
draw((0,0)--(0,3.103));<br />
draw((0,0)--(-2.687,1.5513));<br />
draw((0,0)--(-0.5,3.9686));</asy><br />
<br />
Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]]. So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>. <br />
<br />
By the [[Law of Cosines]]: <math>4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}</math>. <br />
<br />
Since <math>x</math> must be positive, <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C</math>.<br />
<br />
=== Solution 2 (without trigonometry) ===<br />
Draw <math>OD</math> and <math>OC</math> as in the diagram. Draw the altitude from <math>O</math> to <math>DC</math> and call the intersection <math>E</math><br />
<br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(x\)",(0.03,1.5),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(B\)",(-3.15,1.35),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(D\)",(-0.75,4.15),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-0.1,3.8),S);<br />
label("\(4\)",(-0.4,2.2),S);<br />
draw((0,0)--(0,3.103));<br />
draw((0,0)--(-2.687,1.5513));<br />
draw((0,0)--(-0.5,3.9686));</asy><br />
<br />
As proved in the first solution, <math> \angle OCD = 150^\circ</math>. <br />
That makes <math>\Delta OCE</math> a <math>30-60-90</math> triangle, so <math>OE = \frac{x}{2}</math> and <math>CE= \frac{x\sqrt 3}{2}</math><br />
<br />
Since <math> \Delta OEC</math> is a right triangle, <br />
<math>\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0</math><br />
<br />
Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C </math><br />
<br />
==Solution 3==<br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.95,3),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(E\)",(0.40,-3.60),E);<br />
label("\(B\)",(-0.75,4.15),E);<br />
label("\(D\)",(-2.62,1.5),E);<br />
label("\(F\)",(-2.64,-1.43),E);<br />
label("\(G\)",(-0.2,-2.8),E);<br />
label("\( \sqrt{3}x\)",(-1.5,-0.5),E);<br />
label("\(M\)",(-2,-0.9),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-2.7,2.3),S);<br />
label("\(1\)",(0.1,-3.4),S);<br />
label("\(8\)",(-0.3,0),S);<br />
draw((0,-3.103)--(-2.687,1.5513));<br />
draw((0.5,-3.9686)--(-0.5,3.9686));</asy><br />
<br />
Looking at the diagram above, we know that <math>BE</math> is a diameter of circle <math>O</math> due to symmetry. Due to Thales' theorem, triangle <math>ABE</math> is a right triangle with <math>A = 90 ^\circ</math>. <math>AE</math> lies on <math>AD</math> and <math>GE</math> because <math>BAD</math> is also a right angle. To find the length of <math>DG</math>, notice that if we draw a line from <math>F</math> to <math>M</math>, the midpoint of line <math>DG</math>, it creates two <math>30</math> - <math>60</math> - <math>90</math> triangles. Therefore, <math>MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x</math>. <math>AE = 2 + \sqrt{3}x</math><br />
<br />
Use the Pythagorean theorem on triangle <math>ABE</math>, we get <cmath>(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0</cmath> Using the pythagorean theorem to solve, we get <cmath>x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}</cmath> <math>x</math> must be positive, therefore <cmath>x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow C</cmath><br />
<br />
~Zeric Hang<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}}<br />
{{AMC10 box|year=2008|ab=A|num-b=24|after=Last Question}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Introductory Trigonometry Problems]]<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1049812019 AIME I Problems/Problem 82019-03-27T18:57:25Z<p>Zeric: </p>
<hr />
<div>==Problem 8==<br />
Let <math>x</math> be a real number such that <math>\sin^{10}x+\cos^{10} x = \tfrac{11}{36}</math>. Then <math>\sin^{12}x+\cos^{12} x = \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 />
We can substitute <math>y = \sin^2{x}</math>. Since we know that <math>\cos^2{x}=1-\sin^2{x}</math>, we can do some simplification. <br />
<br />
This yields <math>y^5+(1-y)^5=\frac{11}{36}</math>. From this, we can substitute again to get some cancellation through binomials. If we let <math>z=1/2-y</math>, we can simplify the equation to <math>(1/2+z)^5+(1/2-z)^5=\frac{11}{36}</math>. After using binomial theorem, this simplifies to <math>\frac{1}{16}(80z^4+40z^2+1)=11/36</math>. If we use the quadratic formula, we obtain the that <math>z^2=\frac{1}{12}</math>, so <math>z=\pm\frac{1}{2\sqrt{3}}</math>. By plugging z into <math>(1/2-z)^6+(1/2+z)^6</math> (which is equal to <math>\sin^{12}{x}+\cos^{12}{x}</math>, we can either use binomial theorem or sum of cubes to simplify, and we end up with <math>\frac{13}{54}</math>. Therefore, the answer is <math>\boxed{067}</math>.<br />
<br />
eric2020, inspired by Tommy2002<br />
<br />
==Solution 2==<br />
<br />
First, for simplicity, let <math>a=\sin{x}</math> and <math>b=\cos{x}</math>. Note that <math>a^2+b^2=1</math>. We then bash the rest of the problem out. Take the tenth power of this expression and get <math>a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>. Note that we also have <math>\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)</math>. So, it suffices to compute <math>a^2b^2(a^8+b^8)</math>. Let <math>y=a^2b^2</math>. We have from cubing <math>a^2+b^2=1</math> that <math>a^6+b^6+3a^2b^2(a^2+b^2)=1</math> or <math>a^6+b^6=1-3y</math>. Next, using <math>\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>, we get <math>a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}</math> or <math>y(1-3y)+2y^2=y-y^2=\frac{5}{36}</math>. Solving gives <math>y=\frac{5}{6}</math> or <math>y=\frac{1}{6}</math>. Clearly <math>y=\frac{5}{6}</math> is extraneous, so <math>y=\frac{1}{6}</math>. Now note that <math>a^4+b^4=(a^2+b^2)-2a^2b^2=\frac{2}{3}</math>, and <math>a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}</math>. Thus we finally get <math>a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}*\frac{1}{6}=\frac{13}{54}</math>, giving <math>\boxed{067}</math>.<br />
<br />
-Emathmaster<br />
<br />
==Solution 3 (Newton Sums)==<br />
Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in solution 2. Let <math>\sin^2x</math> and <math>\cos^2x</math> be the roots of some polynomial <math>F(a)</math>. Then, <math>F(a)=a^2-a+b</math> for some <math>b=\sin^2x\cdot\cos^2x</math>.<br />
<br />
Let <math>S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k</math>. We want to find <math>S_6</math>. Clearly <math>S_1=1</math> and <math>S_2=1-2b</math>. Newton sums tells us that <math>S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}</math> where <math>k\ge 3</math> for our polynomial <math>F(a)</math>. <br />
<br />
Bashing, we have<br />
<cmath>S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b</cmath><br />
<cmath>S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1</cmath><br />
<cmath>S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}</cmath><br />
<br />
Thus <cmath>5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0</cmath><br />
<math>b=\frac{1}{6} \text{ or } \frac{5}{6}</math>. Clearly, <math>\sin^2x\cdot\cos^2x\not=\frac{5}{6}</math> so <math>\sin^2x\cdot\cos^2x=b=\frac{1}{6}</math>.<br />
<br />
Note <math>S_4=\frac{7}{18}</math>. Solving for <math>S_6</math>, we get <math>S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}</math>. Finally, <math>13+54=\boxed{067}</math>.<br />
<br />
==Solution 4 (Not Fun)==<br />
We let <math>a = \sin^2(x)</math> and <math>b = \cos^2(x)</math>, so we have <math>a+b=1</math> and <math>a^5 + b^5 = \frac{11}{36}</math>. Noticing that <math>ab</math> might be a useful value to find, we let <math>c = ab</math>. Then we can work our way up to find <math>c</math>.<br />
<cmath>a + b = 1</cmath><br />
<cmath>(a+b)(a+b) = 1(a+b)</cmath><br />
<cmath>a^2 + 2ab + b^2 = 1</cmath><br />
<cmath>a^2 + b^2 = -2c + 1</cmath><br />
<cmath>(a+b)(a^2 + b^2) = -2c + 1</cmath><br />
<cmath>a^3 + ab^2 + a^2b + b^3 = -2c + 1</cmath><br />
<cmath>a^3 + b^3 + ab(a + b) = -2c + 1</cmath><br />
<cmath>a^3 + b^3 + c = -2c + 1</cmath><br />
<cmath>a^3 + b^3 = -3c + 1</cmath><br />
<cmath>(a + b)(a^3 + b^3) = -3c + 1</cmath><br />
<cmath>a^4 + ab^3 + a^3b + b^4 = -3c + 1</cmath><br />
<cmath>a^4 + b^4 + ab(a^2 + b^2) = -3c + 1</cmath><br />
<cmath>a^4 + b^4 + c(-2c + 1) = -3c + 1</cmath><br />
<cmath>a^4 + b^4 - 2c^2 + c = -3c + 1</cmath><br />
<cmath>a^4 + b^4 = 2c^2 - 4c + 1</cmath><br />
<cmath>a^5 + ab^4 + a^4b + b^5 = 2c^2 - 4c + 1</cmath><br />
<cmath>a^5 + b^5 + ab(a^3 + b^3) = 2c^2 - 4c + 1</cmath><br />
<cmath>\frac{11}{36} + c(-3c + 1) = 2c^2 - 4c + 1</cmath><br />
<cmath>\frac{11}{36} = 5c^2 - 5c + 1</cmath><br />
<cmath>5c^2 - 5c + \frac{25}{36} = 0</cmath><br />
using quadform you get <math>c = \frac{1}{6}</math> or <math>c = \frac{5}{6}</math>. Since <math>c = \sin^2(x)\cos^2(x) = (\sin(x)\cos(x))^2 = (\frac{\sin(2x)}{2})^2</math>, and since <math>\sin(2x)</math> can't exceed 1, <math>c</math> can't exceed <math>(\frac{1}{2})^2 = \frac{1}{4}</math>. Clearly, <math>c = \frac{1}{6}</math>. And finally,<br />
<cmath>a^5 + b^5 = (a + b)(a^5 + b^5)</cmath><br />
<cmath>a^5 + b^5 = a^6 + ab^5 + a^5b + b^6</cmath><br />
<cmath>\frac{11}{36} = a^6 + b^6 + c(a^4 + b^4)</cmath><br />
looking back to previous results, we see that <math>a^4 + b^4 = 2c^2 - 4c + 1 = \frac{14}{36}</math> (it's easier not to simplify the fraction).<br />
<cmath>\frac{11}{36} = a^6 + b^6 + c(\frac{14}{36}) = a^6 + b^6 + \frac{14}{216}</cmath><br />
<cmath>a^6 + b^6 = \frac{11}{36} - \frac{14}{216} = \frac{13}{54}</cmath><br />
which yields the answer <math>\boxed{067}</math>.<br />
<br />
~PCampbell<br />
<br />
==Solution 5==<br />
Factor the first equation. <cmath>\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)</cmath><br />
First of all, <math>\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x</math> because <math>\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x</math><br />
We group the 1st, 3rd and 5th term and 2nd and 4th term. The 1st group: <cmath> \sin^8+\sin^4x\cos^4x+\cos^8x = (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x)<br />
=1+4\sin^4x\cos^4x-4\sin^2x\cos^2x</cmath> The 2nd group: <cmath>-\sin^6x\cos^2x-\sin^2x\cos^6x = -\sin^2x\cos^2x(\sin^4x+\cos^4x)=-\sin^2x\cos^2x(1-2\sin^2x\cos^2x) = -\sin^2x\cos^2x+2\sin^4x\cos^4x</cmath> Add the two together to make <cmath>1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x</cmath> Because this equals <math>\frac{11}{36}</math>, we have <cmath>5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0</cmath> Let <math>\sin^2x\cos^2x = a</math> so we get <cmath>5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}</cmath> Solving the quadratic gives us <cmath>a = \frac{1 \pm \frac{2}{3}}{2}</cmath> Because <math>\sin^2x\cos^2x \le \frac{1}{4}</math>, we finally get <math>a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}</math>. <br />
<br />
Now from the second equation, <cmath>\sin^{12}x + \cos^{12}x = (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)=(1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)=(1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x)</cmath> Plug in <math>\sin^2x\cos^2x = \frac{1}{6}</math> to get <cmath>(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}</cmath><br />
which yields the answer <math>\boxed{067}</math><br />
<br />
~ZericHang<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_1&diff=1046932018 AIME I Problems/Problem 12019-03-19T13:44:11Z<p>Zeric: /* Solution 5 */</p>
<hr />
<div>==Problem 1==<br />
Let <math>S</math> be the number of ordered pairs of integers <math>(a,b)</math> with <math>1 \leq a \leq 100</math> and <math>b \geq 0</math> such that the polynomial <math>x^2+ax+b</math> can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when <math>S</math> is divided by <math>1000</math>.<br />
<br />
==Solution==<br />
<br />
You let the linear factors be as <math>(x+c)(x+d)</math>.<br />
<br />
Then, obviously <math>a=c+d</math> and <math>b=cd</math>.<br />
<br />
We know that <math>1\le a\le 100</math> and <math>b\ge 0</math>, so <math>c</math> and <math>d</math> both have to be non-negative<br />
<br />
However, <math>a</math> cannot be <math>0</math>, so at least one of <math>c</math> and <math>d</math> must be greater than <math>0</math>, ie positive.<br />
<br />
Also, <math>a</math> cannot be greater than <math>100</math>, so <math>c+d</math> must be less than or equal to <math>100</math>.<br />
<br />
Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices <math>(0,0), (0, 100),</math> and <math>(100,0)</math>. Remember that <math>(0,0)</math> does not work, so there is a square with top right corner <math>(1,1)</math>.<br />
<br />
Note that <math>c</math> and <math>d</math> are interchangeable, since they end up as <math>a</math> and <math>b</math> in the end anyways. Thus, we simply draw a line from <math>(1,1)</math> to <math>(50,50)</math>, designating one of the halves as our solution (since the other side is simply the coordinates flipped).<br />
<br />
We note that the pattern from <math>(1,1)</math> to <math>(50,50)</math> is <math>2+3+4+\dots+51</math> solutions and from <math>(51, 49)</math> to <math>(100,0)</math> is <math>50+49+48+\dots+1</math> solutions, since we can decrease the <math>y</math>-value by <math>1</math> until <math>0</math> for each coordinate.<br />
<br />
Adding up gives <cmath>\dfrac{2+51}{2}\cdot 50+\dfrac{50+1}{2}\cdot 50.</cmath><br />
This gives us <math>2600</math>, and <math>2600\equiv 600 \bmod{1000}.</math><br />
<br />
Thus, the answer is: <cmath>\boxed{600}.</cmath><br />
<br />
==Solution 2==<br />
<br />
Similar to the previous problem, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is 100+51+51-3, or just 200. Using Pick's theorem, we know that the area of the half-triangle, which is 2500, is just I+100-1. That means that there are 2401 interior points, plus 200 boundary points, which is 2601. However, (0,0) does not work, so the answer is <cmath>\boxed{600}.</cmath><br />
<br />
==Solution 3 (less complicated)==<br />
Notice that for <math>x^2+ax+b</math> to be true, for every <math>a</math>, <math>b</math> will always be the product of the possibilities of how to add two integers to <math>a</math>. For example, if <math>a=3</math>, <math>b</math> will be the product of <math>(3,0)</math> and <math>(2,1)</math>, as those two sets are the only possibilities of adding two integers to <math>a</math>. Note that order does not matter. If we just do some simple casework, we find out that: <br />
<br />
if <math>a</math> is odd, there will always be <math>\left\lceil\frac{a}{2}\right\rceil</math> <math>\left(\text{which is also }\frac{a+1}{2}\right)</math> possibilities of adding two integers to <math>a</math>.<br />
<br />
if <math>a</math> is even, there will always be <math>\frac{a}{2}+1</math> possibilities of adding two integers to <math>a</math>. <br />
<br />
Using the casework, we have <math>1+2+2+3+3+...50+50+51</math> possibilities. This will mean that the answer is <cmath>\frac{(1+51)\cdot100}{2}\Rightarrow52\cdot50=2600</cmath> possibilities.<br />
<br />
Thus, our solution is <math>2600\bmod {1000}\equiv\boxed{600}</math>.<br />
<br />
Solution by IronicNinja~<br />
<br />
==Solution 4==<br />
<br />
Let's write the linear factors as <math>(x+n)(x+m)</math>.<br />
<br />
Then we can write them as: <math>a=n+m, b=nm</math>.<br />
<br />
<math>n</math> or <math>m</math> has to be a positive integer as a cannot be 0.<br />
<br />
<math>n+m</math> has to be between <math>1</math> and <math>100</math>, as a cannot be over <math>100</math>.<br />
<br />
Excluding <math>a=1</math>, we can see there is always a pair of <math>2</math> a-values for a certain amount of b-values. <br />
<br />
For instance, <math>a=2</math> and <math>a=3</math> both have <math>2</math> b-values. <math>a=4</math> and <math>a=5</math> both have <math>3</math> b-values.<br />
<br />
We notice the pattern of the number of b-values in relation to the a-values:<br />
<math>1, 2, 2, 3, 3, 4, 4…</math><br />
<br />
The following link is the URL to the graph I drew showing the relationship between a-values and b-values<br />
http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file<br />
<br />
The pattern continues until <math>a=100</math>, and in total, there are <math>49</math> pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the (<math>a=1</math>, amount of b-values=1) in the beginning, and (<math>a=100</math>, amount of b-values=51) in the end. <br />
<br />
Then, we add numbers from the opposite ends of the spectrum, and quickly notice that there are <math>50</math> pairs each with a sum of <math>52</math>. <math>52\cdot50</math> gives <math>2600</math> ordered pairs:<br />
<br />
<math>1+51, 2+50, 2+50, 3+49, 3+49, 4+48, 4+48…</math><br />
<br />
When divided by <math>1000</math>, it gives the remainder <math>\boxed{600}</math>, the answer.<br />
<br />
Solution provided by- Yonglao<br />
<br />
==Solution 5==<br />
<br />
Let's say that the quadratic <math>x^2 + ax + b</math> can be factored into <math>(x+c)(x+d)</math> where <math>c</math> and <math>d</math> are non-negative numbers. We can't have both of them zero because <math>a</math> would not be within bounds. Also, <math>c+d \leq 100</math>. Assume that <math>c < d</math>. <math>d</math> can be written as <math>c + x</math> where <math>x \geq 0</math>. Therefore, <math>c + d = 2c + x \leq 100</math>. To find the amount of ordered pairs, we must consider how many values of <math>x</math> are possible for each value of <math>c</math>. The amount of possible values of <math>x</math> is given by <math>100 - 2c + 1</math>. The <math>+1</math> is the case where <math>c = d</math>. We don't include the case where <math>c = d = 0</math>, so we must subtract a case from our total. The amount of ordered pairs of <math>(a,b)</math> is:<br />
<cmath>\left(\sum_{c=0}^{50} (100 - 2c + 1)\right) - 1</cmath><br />
This is an arithmetic progression. <cmath>\frac{(101 + 1)(51)}{2} - 1 = 2601 - 1 = 2600</cmath> <br />
When divided by <math>1000</math>, it gives the remainder <math>\boxed{600}</math><br />
<br />
~Zeric Hang<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_1&diff=1046922018 AIME I Problems/Problem 12019-03-19T13:43:54Z<p>Zeric: </p>
<hr />
<div>==Problem 1==<br />
Let <math>S</math> be the number of ordered pairs of integers <math>(a,b)</math> with <math>1 \leq a \leq 100</math> and <math>b \geq 0</math> such that the polynomial <math>x^2+ax+b</math> can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when <math>S</math> is divided by <math>1000</math>.<br />
<br />
==Solution==<br />
<br />
You let the linear factors be as <math>(x+c)(x+d)</math>.<br />
<br />
Then, obviously <math>a=c+d</math> and <math>b=cd</math>.<br />
<br />
We know that <math>1\le a\le 100</math> and <math>b\ge 0</math>, so <math>c</math> and <math>d</math> both have to be non-negative<br />
<br />
However, <math>a</math> cannot be <math>0</math>, so at least one of <math>c</math> and <math>d</math> must be greater than <math>0</math>, ie positive.<br />
<br />
Also, <math>a</math> cannot be greater than <math>100</math>, so <math>c+d</math> must be less than or equal to <math>100</math>.<br />
<br />
Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices <math>(0,0), (0, 100),</math> and <math>(100,0)</math>. Remember that <math>(0,0)</math> does not work, so there is a square with top right corner <math>(1,1)</math>.<br />
<br />
Note that <math>c</math> and <math>d</math> are interchangeable, since they end up as <math>a</math> and <math>b</math> in the end anyways. Thus, we simply draw a line from <math>(1,1)</math> to <math>(50,50)</math>, designating one of the halves as our solution (since the other side is simply the coordinates flipped).<br />
<br />
We note that the pattern from <math>(1,1)</math> to <math>(50,50)</math> is <math>2+3+4+\dots+51</math> solutions and from <math>(51, 49)</math> to <math>(100,0)</math> is <math>50+49+48+\dots+1</math> solutions, since we can decrease the <math>y</math>-value by <math>1</math> until <math>0</math> for each coordinate.<br />
<br />
Adding up gives <cmath>\dfrac{2+51}{2}\cdot 50+\dfrac{50+1}{2}\cdot 50.</cmath><br />
This gives us <math>2600</math>, and <math>2600\equiv 600 \bmod{1000}.</math><br />
<br />
Thus, the answer is: <cmath>\boxed{600}.</cmath><br />
<br />
==Solution 2==<br />
<br />
Similar to the previous problem, we plot the triangle and cut it in half. Then, we find the number of boundary points, which is 100+51+51-3, or just 200. Using Pick's theorem, we know that the area of the half-triangle, which is 2500, is just I+100-1. That means that there are 2401 interior points, plus 200 boundary points, which is 2601. However, (0,0) does not work, so the answer is <cmath>\boxed{600}.</cmath><br />
<br />
==Solution 3 (less complicated)==<br />
Notice that for <math>x^2+ax+b</math> to be true, for every <math>a</math>, <math>b</math> will always be the product of the possibilities of how to add two integers to <math>a</math>. For example, if <math>a=3</math>, <math>b</math> will be the product of <math>(3,0)</math> and <math>(2,1)</math>, as those two sets are the only possibilities of adding two integers to <math>a</math>. Note that order does not matter. If we just do some simple casework, we find out that: <br />
<br />
if <math>a</math> is odd, there will always be <math>\left\lceil\frac{a}{2}\right\rceil</math> <math>\left(\text{which is also }\frac{a+1}{2}\right)</math> possibilities of adding two integers to <math>a</math>.<br />
<br />
if <math>a</math> is even, there will always be <math>\frac{a}{2}+1</math> possibilities of adding two integers to <math>a</math>. <br />
<br />
Using the casework, we have <math>1+2+2+3+3+...50+50+51</math> possibilities. This will mean that the answer is <cmath>\frac{(1+51)\cdot100}{2}\Rightarrow52\cdot50=2600</cmath> possibilities.<br />
<br />
Thus, our solution is <math>2600\bmod {1000}\equiv\boxed{600}</math>.<br />
<br />
Solution by IronicNinja~<br />
<br />
==Solution 4==<br />
<br />
Let's write the linear factors as <math>(x+n)(x+m)</math>.<br />
<br />
Then we can write them as: <math>a=n+m, b=nm</math>.<br />
<br />
<math>n</math> or <math>m</math> has to be a positive integer as a cannot be 0.<br />
<br />
<math>n+m</math> has to be between <math>1</math> and <math>100</math>, as a cannot be over <math>100</math>.<br />
<br />
Excluding <math>a=1</math>, we can see there is always a pair of <math>2</math> a-values for a certain amount of b-values. <br />
<br />
For instance, <math>a=2</math> and <math>a=3</math> both have <math>2</math> b-values. <math>a=4</math> and <math>a=5</math> both have <math>3</math> b-values.<br />
<br />
We notice the pattern of the number of b-values in relation to the a-values:<br />
<math>1, 2, 2, 3, 3, 4, 4…</math><br />
<br />
The following link is the URL to the graph I drew showing the relationship between a-values and b-values<br />
http://artofproblemsolving.com/wiki/index.php?title=File:Screen_Shot_2018-04-30_at_8.15.00_PM.png#file<br />
<br />
The pattern continues until <math>a=100</math>, and in total, there are <math>49</math> pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the (<math>a=1</math>, amount of b-values=1) in the beginning, and (<math>a=100</math>, amount of b-values=51) in the end. <br />
<br />
Then, we add numbers from the opposite ends of the spectrum, and quickly notice that there are <math>50</math> pairs each with a sum of <math>52</math>. <math>52\cdot50</math> gives <math>2600</math> ordered pairs:<br />
<br />
<math>1+51, 2+50, 2+50, 3+49, 3+49, 4+48, 4+48…</math><br />
<br />
When divided by <math>1000</math>, it gives the remainder <math>\boxed{600}</math>, the answer.<br />
<br />
Solution provided by- Yonglao<br />
<br />
==Solution 5==<br />
<br />
Let's say that the quadratic <math>x^2 + ax + b</math> can be factored into <math>(x+c)(x+d)</math> where <math>c</math> and <math>d</math> are non-negative numbers. We can't have both of them zero because <math>a</math> would not be within bounds. Also, <math>c+d \leq 100</math>. Assume that <math>c < d</math>. <math>d</math> can be written as <math>c + x</math> where <math>x \geq 0</math>. Therefore, <math>c + d = 2c + x \leq 100</math>. To find the amount of ordered pairs, we must consider how many values of <math>x</math> are possible for each value of <math>c</math>. The amount of possible values of <math>x</math> is given by <math>100 - 2c + 1</math>. The <math>+1</math> is the case where <math>c = d</math>. We don't include the case where <math>c = d = 0</math>, so we must subtract a case from our total. The amount of ordered pairs of <math>(a,b)</math> is:<br />
<cmath>\left(\sum_{c=0}^{50} (100 - 2c + 1)\right) - 1</cmath><br />
This is an arithmetic progression. <cmath>\frac{(101 + 1)(51)}{2} - 1 = 2601 - 1 = 2600</cmath> <br />
When divided by <math>1000</math>, it gives the remainder <math>\boxed{600}</math><br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_21&diff=1019042016 AMC 10A Problems/Problem 212019-02-12T15:27:36Z<p>Zeric: </p>
<hr />
<div>Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
<br />
==Solution 1== <br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
pair P,Q,R,Pp,Qp,Rp;<br />
pair A,B;<br />
<br />
//Variable Definitions<br />
A=(-5, 0);<br />
B=(8, 0);<br />
P=(-2.828,1);<br />
Q=(0,2);<br />
R=(4.899,3);<br />
Pp=foot(P,A,B);<br />
Qp=foot(Q,A,B);<br />
Rp=foot(R,A,B);<br />
path PQR = P--Q--R--cycle;<br />
//Initial Diagram<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(Pp);<br />
dot(Qp);<br />
dot(Rp);<br />
draw(Circle(P, 1), linewidth(0.8));<br />
draw(Circle(Q, 2), linewidth(0.8));<br />
draw(Circle(R, 3), linewidth(0.8));<br />
draw(A--B,Arrows);<br />
label("$P$",P,N);<br />
label("$Q$",Q,N);<br />
label("$R$",R,N);<br />
label("$P'$",Pp,S);<br />
label("$Q'$",Qp,S);<br />
label("$R'$",Rp,S);<br />
label("$l$",B,E);<br />
<br />
//Added lines<br />
draw(PQR);<br />
draw(P--Pp);<br />
draw(Q--Qp);<br />
draw(R--Rp);<br />
<br />
//Angle marks<br />
draw(rightanglemark(P,Pp,B));<br />
draw(rightanglemark(Q,Qp,B));<br />
draw(rightanglemark(R,Rp,B));<br />
</asy><br />
Notice that we can find <math>[P'PQRR']</math> in two different ways: <math>[P'PQQ']+[Q'QRR']</math> and <math>[PQR]+[P'PRR']</math>, so <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']</math> <br />
<math>\break</math><br />
<br />
<math>P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}</math>. Additionally, <math>Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}</math>. Therefore, <math>[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}</math>. Similarly, <math>[Q'QRR']=5\sqrt6</math>. We can calculate <math>[P'PRR']</math> easily because <math>P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}</math>. <math>[P'PRR']=4\sqrt{2}+4\sqrt{6}</math>. <math>\newline</math><br />
<br />
Plugging into first equation, the two sums of areas, <math>3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]</math>. <math>\newline</math><br />
<br />
<math>[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}</math>.<br />
<br />
==Solution 2== <br />
<br />
Use the [[Shoelace Theorem]].<br />
<br />
Let the center of the first circle of radius 1 be at <math>(0, 1)</math>. <br />
<br />
Draw the trapezoid <math>PQQ'P'</math> and using the Pythagorean Theorem, we get that <math>P'Q' = 2\sqrt{2}</math> so the center of the second circle of radius 2 is at <math>(2\sqrt{2}, 2)</math>.<br />
<br />
Draw the trapezoid <math>QRR'Q'</math> and using the Pythagorean Theorem, we get that <math>Q'R' = 2\sqrt{6}</math> so the center of the third circle of radius 3 is at <math>(2\sqrt{2}+2\sqrt{6}, 3)</math>.<br />
<br />
Now, we may use the Shoelace Theorem!<br />
<br />
<math>(0,1)</math><br />
<br />
<math>(2\sqrt{2}, 2)</math><br />
<br />
<math>(2\sqrt{2}+2\sqrt{6}, 3)</math><br />
<br />
<math>\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|</math><br />
<br />
<math>= \sqrt{6}-\sqrt{2}</math> <math>\fbox{D}</math>.<br />
<br />
==Solution 3==<br />
<math>PQ = 3</math> and <math>QR = 5</math> because they are the sum of two radii. <math>QQ' - PP' = 1</math> and <math>RR' - QQ' = 1</math>, the difference of the radii. Using pythagorean theorem, we find that <math>P'Q'</math> and <math>Q'R'</math> are <math>\sqrt{8}</math> and <math>\sqrt{24}</math>, <math>P'R' = \sqrt{8} + \sqrt{24}</math>. <br />
<br />
Draw a perpendicular from <math>P</math> to line <math>RR'</math>, then we can use the Pythagorean theorem to find <math>PR</math>. <math>RR' - PP' = 2</math>. We get <cmath>PR^2 = (\sqrt{8} + \sqrt{24})^2 + 4 = 36 + 16\sqrt{3} \Rightarrow PR = \sqrt{36 + 16\sqrt{3}} = 2\sqrt{9 + 4\sqrt{3}}</cmath><br />
<br />
To make our calculations easier, let <math>\sqrt{9 + 4\sqrt{3}} = a</math>. The semi-perimeter of our triangle is <math>\frac{3 + 5 + 2a}{2} = 4 + a</math>. Symbolize the area of the triangle with <math>A</math>. Using Heron's formula, we have <cmath>A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = (4 + a)(4 - a)(a + 1)(a - 1) = (16 - a^2)(a^2 - 1)</cmath> We can remove the outer root of a. <cmath>A^2 = (16 - 9 - 4\sqrt{3})(9 + 4\sqrt{3} - 1) = (7 - 4\sqrt{3})(8 + 4\sqrt{3}) = 8 - 4\sqrt{3} \rightarrow A = \sqrt{8 - 4\sqrt{3}}</cmath><br />
<br />
We solve the nested root. We want to turn <math>8 - 4\sqrt{3}</math> into the square of something. If we have <math>(a - b) ^ 2 = 8 - 4\sqrt{3}</math>, then we get <cmath>\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}</cmath> Solving the system of equations, we get <math>a = \sqrt{6}</math> and <math>b = \sqrt{2}</math>. Alternatively, you can square all the possible solutions until you find one that is equal to <math>8 - 4\sqrt{3}</math>. <cmath>A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2} \rightarrow \fbox{D}</cmath><br />
~ZericH<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}<br />
{{AMC12 box|year=2016|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_6&diff=1005862015 AMC 12B Problems/Problem 62019-01-19T00:51:57Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
Back in 1930, Tillie had to memorize her multiplication facts from <math>0 \times 0</math> to <math>12 \times 12</math>. The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?<br />
<br />
<math>\textbf{(A)}\; 0.21 \qquad\textbf{(B)}\; 0.25 \qquad\textbf{(C)}\; 0.46 \qquad\textbf{(D)}\; 0.50 \qquad\textbf{(E)}\; 0.75</math><br />
<br />
==Solution 1==<br />
There are a total of <math>(12+1) \times (12+1) = 169</math> products, and a product is odd if and only if both its factors are odd. There are <math>6</math> odd numbers between <math>0</math> and <math>12</math>, namely <math>1, 3, 5, 7, 9, 11,</math> hence the number of odd products is <math>6 \times 6 = 36</math>. Therefore the answer is <math>36/169 \doteq \boxed{\textbf{(A)} \, 0.21}</math>.<br />
<br />
==Solution 2==<br />
Note that if we had an <math>11</math> by <math>11</math> multiplication table, the fraction of odd products becomes <math>0.25</math>. If we add the <math>12</math> by <math>12</math>, the fraction of odd products decreases. Because <math>0.21</math> is the only option less than <math>0.25</math>, our answer is <math>\boxed{\textbf{(A)} \, 0.21}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=7|num-b=5}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_12&diff=1005852015 AMC 12B Problems/Problem 122019-01-19T00:29:13Z<p>Zeric: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Let <math>a</math>, <math>b</math>, and <math>c</math> be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation <math>(x-a)(x-b)+(x-b)(x-c)=0</math> ?<br />
<br />
<math>\textbf{(A)}\; 15 \qquad\textbf{(B)}\; 15.5 \qquad\textbf{(C)}\; 16 \qquad\textbf{(D)} 16.5 \qquad\textbf{(E)}\; 17</math><br />
<br />
==Solution 1==<br />
The left-hand side of the equation can be factored as <math>(x-b)(x-a+x-c) = (x-b)(2x-(a+c))</math>, from which it follows that the roots of the equation are <math>x=b</math>, and <math>x=\tfrac{a+c}{2}</math>. The sum of the roots is therefore <math>b + \tfrac{a+c}{2}</math>, and the maximum is achieved by choosing <math>b=9</math>, and <math>\{a,c\}=\{7,8\}</math>. Therefore the answer is <math>9 + \tfrac{7+8}{2} = 9 + 7.5 = \boxed{\textbf{(D)}\; 16.5}.</math><br />
<br />
==Solution 2==<br />
Expand the polynomial. We get <math>(x-a)(x-b)+(x-b)(x-c)=x^2-(a+b)x+ab+x^2-(b+c)x+bc=2x^2-(a+2b+c)x+(ab+bc).</math> <br />
<br />
Now, consider a general quadratic equation <math>ax^2+bx+c=0.</math> The two solutions to this are <cmath>\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}, \dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}.</cmath> The sum of these roots is <cmath>\dfrac{-b}{a}.</cmath> <br />
<br />
Therefore, reconsidering the polynomial of the problem, the sum of the roots is <cmath>\dfrac{a+2b+c}{2}.</cmath> Now, to maximize this, it is clear that <math>b=9.</math> Also, we must have <math>a=8, b=7</math> (or vice versa). The reason <math>a,b</math> have to equal these values instead of larger values is because each of <math>a,b,c</math> is distinct.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=13|num-b=11}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_13&diff=992282000 AMC 12 Problems/Problem 132018-12-02T17:43:44Z<p>Zeric: </p>
<hr />
<div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #13]] and [[2000 AMC 10 Problems|2000 AMC 10 #22]]}}<br />
<br />
== Problem ==<br />
One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?<br />
<br />
<math>\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
<br />
Let <math>c</math> be the total amount of coffee, <math>m</math> of milk, and <math>p</math> the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so <br />
<cmath>\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m</cmath><br />
Regrouping, we get <math>2c(6-p)=3m(p-4)</math>. Since both <math>c,m</math> are positive, it follows that <math>6-p</math> and <math>p-4</math> are also positive, which is only possible when <math>p = 5\ \mathrm{(C)}</math>.<br />
<br />
<br />
=== Solution 2 (less rigorous) ===<br />
<br />
One could notice that (since there are only two components to the mixture) Angela must have more than her "fair share" of milk and less then her "fair share" of coffee in order to ensure that everyone has <math>8</math> ounces. The "fair share" is <math>1/p.</math> So,<br />
<br />
<cmath>\frac{1}{6} < \frac{1}{p}<\frac{1}{4}</cmath><br />
<br />
Which requires that <math>p</math> be <math>p = 5\ \mathrm{(C)},</math> since <math>p</math> is a whole number.<br />
<br />
<br />
=== Solution 3 ===<br />
<br />
Again, let <math>c,</math> <math>m,</math> and <math>p</math> be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is <math>8p,</math> and also <math>c+m.</math> Thus, <math>c+m = 8p,</math> so <math>m = 8p-c</math> and <math>c = 8p-m.</math><br />
<br />
We also know that the amount Angela drank, which is <math>\frac{c}{6} + \frac{m}{4},</math> is equal to <math>8</math> ounces, thus <math>\frac{c}{6} + \frac{m}{4} = 8.</math> Rearranging gives <cmath>24p - c = 96.</cmath> Now notice that <math>c > 0</math> (by the problem statement). In addition, <math>m > 0,</math> so <math>c = 8p-m < 8p.</math> Therefore, <math>0 < c < 8p,</math> and so <math>24p > 24p-c > 16p.</math> We know that <math>24p-c = 96,</math> so <cmath>24p > 96 > 16p.</cmath> From the leftmost inequality, we get <math>p > 4,</math> and from the rightmost inequality, we get <math>p < 6.</math> The only possible value of <math>p</math> is <math>p = 5\ \mathrm{(C)}</math>.<br />
<br />
=== Solution 4 ===<br />
<br />
Let <math>c,</math> <math>m,</math> and <math>p</math> be the total amount of coffee, total amount of milk, and number of people in the family, respectively. <math>c</math> and <math>m</math> obviously can't be <math>0</math>. <br />
We know <math>\frac{c}{6} + \frac{m}{4} = 8</math> or <math>3c + 2m = 96</math> and <math>c + m = 8p</math> or <math>2c + 2m = 16p</math>. <br />
Then, <cmath>(2c + 2m) + c = 16p + c = 96</cmath>Because <math>16p</math> and <math>96</math> are both divisible by <math>16</math>, <math>c</math> must also be divisible by <math>16</math>. Let <math>c = 16k</math>. Now, <cmath>3(16k) + 2m = 48k + 2m = 96</cmath> <math>k</math> can't be <math>0</math>, otherwise <math>c</math> is <math>0</math>, and <math>k</math> can't be <math>2</math>, otherwise <math>m</math> is <math>0</math>. Therefore <math>k</math> must be <math>1</math>, <math>c = 16</math> and <math>m = 24</math>. <math>c + m = 16 + 24 = 40 = 8p</math>. Therefore, <math>p = 5\ \mathrm{(C)}</math>.<br />
<br />
~grtuit<br />
<br />
==Sidenote==<br />
If we now solve for <math>c</math> and <math>m</math>, we find that <math>m=16</math> and <math>c=24</math>. Thus in total the family drank <math>16</math> ounces of milk and <math>24</math> ounces of coffee. Angela drank exactly <math>4</math> ounces of milk and <math>4</math> ounces of coffee.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2000|num-b=12|num-a=14}}<br />
{{AMC10 box|year=2000|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_23&diff=973122006 AMC 10B Problems/Problem 232018-08-19T19:13:13Z<p>Zeric: </p>
<hr />
<div>== Problem ==<br />
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br />
pair F = intersectionpoint( A--D, B--Ep );<br />
<br />
draw( A -- B -- C -- cycle );<br />
draw( A -- D );<br />
draw( B -- Ep );<br />
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br />
<br />
label("$7$",(1.25,0.2));<br />
label("$7$",(2.2,0.45));<br />
label("$3$",(0.45,0.35));<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math><br />
<br />
== Solution 1==<br />
<br />
Label the points in the figure as shown below, and draw the segment <math>CF</math>. This segment divides the quadrilateral into two triangles, let their areas be <math>x</math> and <math>y</math>.<br />
<br />
<asy><br />
unitsize(2cm);<br />
defaultpen(.8);<br />
<br />
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br />
pair F = intersectionpoint( A--D, B--Ep );<br />
<br />
draw( A -- B -- C -- cycle );<br />
draw( A -- D );<br />
draw( B -- Ep );<br />
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br />
<br />
label("$7$",(1.45,0.15));<br />
label("$7$",(2.2,0.45));<br />
label("$3$",(0.45,0.35));<br />
<br />
draw( C -- F, dashed );<br />
<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$D$",D,NE);<br />
label("$E$",Ep,NW);<br />
label("$F$",F,S);<br />
<br />
label("$x$",(1,1));<br />
label("$y$",(1.6,1));<br />
</asy><br />
<br />
Since [[triangle]]s <math>AFB</math> and <math>DFB</math> share an [[altitude]] from <math>B</math> and have equal area, their bases must be equal, hence <math>AF=DF</math>.<br />
<br />
Since triangles <math>AFC</math> and <math>DFC</math> share an altitude from <math>C</math> and their respective bases are equal, their areas must be equal, hence <math>x+3=y</math>. <br />
<br />
Since triangles <math>EFA</math> and <math>BFA</math> share an altitude from <math>A</math> and their respective areas are in the ratio <math>3:7</math>, their bases must be in the same ratio, hence <math>EF:FB = 3:7</math>.<br />
<br />
Since triangles <math>EFC</math> and <math>BFC</math> share an altitude from <math>C</math> and their respective bases are in the ratio <math>3:7</math>, their areas must be in the same ratio, hence <math>x:(y+7) = 3:7</math>, which gives us <math>7x = 3(y+7)</math>.<br />
<br />
Substituting <math>y=x+3</math> into the second equation we get <math>7x = 3(x+10)</math>, which solves to <math>x=\frac{15}{2}</math>. Then <math>y=x+3 = \frac{15}{2}+3 = \frac{21}{2}</math>, and the total area of the quadrilateral is <math>x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}</math>.<br />
<br />
==Solution 2==<br />
<br />
Connect points <math>E</math> and <math>D</math>. Triangles <math>EFA</math> and <math>FAB</math> share an altitude and their areas are in the ration <math>3:7</math>. Their bases, <math>EF</math> and <math>FB</math>, must be in the same <math>3:7</math> ratio. <br />
<br />
Triangles <math>EFD</math> and <math>FBD</math> share an altitude and their bases are in a <math>3:7</math> ratio. Therefore, their areas are in a <math>3:7</math> ratio and the area of triangle <math>EFD</math> is <math>3</math>.<br />
<br />
Triangle <math>CED</math> and <math>DEA</math> share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases <math>CE</math> and <math>EA</math>. The ratio is <math>A:(3+3) \Rightarrow A:6</math> where <math>A</math> is the area of triangle <math>CED</math><br />
<br />
Triangles <math>CEB</math> and <math>EAB</math> also share an altitude. The ratio of their areas is also equal to the ratio of bases <math>CE</math> and <math>EA</math>. The ratio is <math>(A+3+7):(3+7) \Rightarrow (A+10):10</math><br />
<br />
Because the two ratios are equal, we get the equation <math>\frac{A}{6} = \frac {A+10}{10} \Rightarrow 10A = 6A+60 \Rightarrow A = 15</math>. We add the area of triangle <math>EDF</math> to get that the total area of the quadrilateral is <math>\boxed{\textbf{(D) }18}</math>.<br />
<br />
~Zeric Hang<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_16&diff=962602013 AMC 10A Problems/Problem 162018-07-15T19:52:41Z<p>Zeric: </p>
<hr />
<div>==Problem==<br />
<br />
A triangle with vertices <math>(6, 5)</math>, <math>(8, -3)</math>, and <math>(9, 1)</math> is reflected about the line <math>x=8</math> to create a second triangle. What is the area of the union of the two triangles?<br />
<br />
<math> \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
Let <math>A</math> be at <math>(6, 5)</math>, B be at <math>(8, -3)</math>, and <math>C</math> be at <math>(9, 1)</math>. Reflecting over the line <math>x=8</math>, we see that <math>A' = D = (10,5)</math>, <math>B' = B</math> (as the x-coordinate of B is 8), and <math>C' = E = (7, 1)</math>. Line <math>AB</math> can be represented as <math>y=-4x+29</math>, so we see that <math>E</math> is on line <math>AB</math>. <br />
<br />
<asy><br />
pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1);<br />
draw(A--B--C--cycle^^D--E--B--cycle);<br />
dot(A^^B^^C^^D^^E);<br />
label("$A$",A,NW);<br />
label("$B$",B,S);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",E,W);<br />
<br />
<br />
</asy><br />
<br />
<br />
We see that if we connect <math>A</math> to <math>D</math>, we get a line of length <math>4</math> (between <math>(6, 5)</math> and <math>(10,5)</math>). The area of <math>\triangle ABD</math> is equal to <math>\frac{bh}{2} = \frac{4(8)}{2} = 16</math>.<br />
<br />
Now, let the point of intersection between <math>AC</math> and <math>DE</math> be <math>F</math>. If we can just find the area of <math>\triangle ADF</math> and subtract it from 16, we are done.<br />
<br />
We realize that because the diagram is symmetric over <math>x = 8</math>, the intersection of lines <math>AC</math> and <math>DE</math> should intersect at an x-coordinate of <math>8</math>. We know that the slope of <math>DE</math> is <math>\frac{5-1}{10-7} = \frac{4}{3}</math>. Thus, we can represent the line going through <math>E</math> and <math>D</math> as <math>y - 1=\frac{4}{3}(x - 7)</math>. Plugging in <math>x = 8</math>, we find that the y-coordinate of F is <math>\frac{7}{3}</math>. Thus, the height of <math>\triangle ADF</math> is <math>5 - \frac{7}{3} = \frac{8}{3}</math>. Using the formula for the area of a triangle, the area of <math>\triangle ADF</math> is <math>\frac{16}{3}</math>. <br />
<br />
To get our final answer, we must subtract this from <math>16</math>. <math>[ABD] - [ADF] = 16 - \frac{16}{3} = \boxed{\textbf{(E) }\frac{32}{3}}</math><br />
<br />
==Solution 2==<br />
<br />
First, realize that <math>E</math> is the midpoint of <math>AB</math> and <math>C</math> is the midpoint of <math>BD</math>. Connect <math>A</math> to <math>D</math> to form <math>\triangle ABD</math>. Let the midpoint of <math>AD</math> be <math>G</math>. Connect <math>B</math> to <math>G</math>. <math>BG</math> is a median of <math>\triangle ABD</math>. <br />
<br />
<br />
Because <math>\triangle ABD</math> is isosceles, <math>BG</math> is also an altitude of <math>\triangle ABD</math>. We know the length of <math>AD</math> and <math>BG</math> from the given coordinates. The area of <math>\triangle ABD</math> is <math>\frac{bh}{2} = \frac{4(8)}{2} = 16</math>. <br />
<br />
<br />
Let the intesection of <math>AC</math>, <math>DE</math> and <math>BG</math> be <math>F</math>. <math>F</math> is the centroid of <math>\triangle ABD</math>. Therefore, it splits <math>BG</math> into <math>BF={2 \over 3}(BG)</math> and <math>FG={1\over 3}(BG)</math>. The area of quadrilateral <math>ABDF = 16\cdot {2 \over 3} = \boxed{\textbf{(E) }\frac{32}{3}}</math><br />
<br />
~Zeric Hang<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2013|ab=A|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_18&diff=961842018 AMC 10B Problems/Problem 182018-07-14T00:49:35Z<p>Zeric: </p>
<hr />
<div>Three young brother-sister pairs from different families need tot take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?<br />
<br />
<math>\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}</math><br />
<br />
==Solution 1 (Casework)==<br />
<br />
We can begin to put this into cases. Let's call the pairs <math>a</math>, <math>b</math> and <math>c</math>, and assume that a member of pair <math>a</math> is sitting in the leftmost seat of the second row. We can have the following cases then.<br />
<br />
Case 1: <br />
Second Row: a b c<br />
Third Row: b c a<br />
<br />
Case 2:<br />
Second Row: a c b<br />
Third Row: c b a<br />
<br />
Case 3: <br />
Second Row: a b c<br />
Third Row: c a b<br />
<br />
Case 4: <br />
Second Row: a c b<br />
Third Row: b a c<br />
<br />
For each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has <math>2 \cdot 2 \cdot 2 = 8</math> possibilities. Since there are four cases, when pair <math>a</math> has someone in the leftmost seat of the second row, there are 32 ways to rearrange it. However, someone from either pair <math>a</math>, <math>b</math>, or <math>c</math> could be sitting in the leftmost seat of the second row. So, we have to multiply it by 3 to get our answer of <math>32 \cdot 3 = 96</math>. So, the correct answer is <math>\boxed{\textbf{(D)} \text{ 96}}</math>.<br />
<br />
Written By: Archimedes15<br />
<br />
==Solution 2==<br />
<br />
Call the siblings <math>A_1</math>, <math>A_2</math>, <math>B_1</math>, <math>B_2</math>, <math>C_1</math>, and <math>C_2</math>.<br />
<br />
There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose [[Without loss of generality]] that <math>A_1</math> takes it (a <math>\circ</math> is an empty seat):<br />
<br />
<math>A_1 \circ \circ \\ \circ \ \circ \ \circ </math><br />
<br />
Then there are 4 choices for the second seat (<math>B_1</math>, <math>B_2</math>, <math>C_1</math>, or <math>C_2</math>). Again, it doesn't matter who takes the seat, so WLOG suppose it is <math>B_1</math>:<br />
<br />
<math>A_1 B_1 \circ \\ \circ \ \circ \ \circ</math><br />
<br />
The last seat in the first row cannot be <math>A_2</math> because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be <math>C_1</math> or <math>C_2</math>. Suppose WLOG that it is <math>C_1</math>. There are two ways to create a second row:<br />
<br />
<math>A_1 B_1 C_1 \\ B_2 C_2 A_2</math><br />
<br />
<math>A_1 B_1 C_1 \\ C_2 A_2 B_2</math><br />
<br />
Therefore, there are <math>6 \cdot 4 \cdot 2 \cdot 2= \boxed{\textbf{(D)} \text{ 96}}</math> possible seating arrangements.<br />
<br />
Written by: R1ceming<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_9&diff=961362018 AMC 10B Problems/Problem 92018-07-12T15:48:38Z<p>Zeric: /* Note */</p>
<hr />
<div>The faces of each of <math>7</math> standard dice are labeled with the integers from <math>1</math> to <math>6</math>. Let <math>p</math> be the probabilities that when all <math>7</math> dice are rolled, the sum of the numbers on the top faces is <math>10</math>. What other sum occurs with the same probability as <math>p</math>?<br />
<br />
<math>\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}</math><br />
<br />
==Solution 1==<br />
It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.<br />
<br />
So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows:<br />
<br />
(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...<br />
<br />
However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math>, and we are done.<br />
<br />
Written By: Archimedes15<br />
<br />
==Solution 2==<br />
<br />
Let's call the unknown value <math>x</math>. By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and <math>x</math>. So, <br />
<br />
<math>10 - 7 = 42- x </math><br />
<br />
<math> x = 39 </math> and our answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
By: Soccer_JAMS<br />
<br />
==Solution 3 (Simple Logic)==<br />
<br />
For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) x 7 = 49. Since having 10 is similar to not having 10, you just subtract 10 from the expected total sum. 49 - 10 = 39 so the answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
<br />
By: epicmonster<br />
<br />
==Solution 4==<br />
The expected value of the sums of the die rolls is <math>3.5*7=24.5</math>, and since the probabilities should be distributed symmetrically on both sides of <math>24.5</math>, the answer is <math>24.5+24.5-10=39</math>, which is <math>\boxed{\textbf{(D)} \text{ 39}}</math>.<br />
<br />
By: dajeff<br />
<br />
<br />
<br />
=== Note ===<br />
<br />
Calculating the probability of getting a sum of <math>10</math> is also easy. There are <math>3</math> cases:<br />
<br />
<br />
Case <math>1</math>: <math>\{1,1,1,1,1,1,4\}</math><br />
<br />
<br />
<math>\frac{7!}{6!}=7</math> cases<br />
<br />
<br />
Case <math>2</math>: <math>\{1,1,1,1,1,2,3\}</math><br />
<br />
<br />
<math>\frac{7!}{5!}=6*7=42</math> cases<br />
<br />
<br />
Case <math>3</math>: <math>\{1,1,1,1,2,2,2\}</math><br />
<br />
<br />
<math>\frac{7!}{4!3!}=5*7=35</math> cases<br />
<br />
<br />
The probability is <math>{84 \over 6^7} = \frac{14}{6^6}</math>. <br />
<br />
Calculating <math>6^6</math>: <br />
<br />
<math>6^6=(6^3)^2=216^2=46656</math> <br />
<br />
Therefore, the probability is <math>{14 \over 46656} = \boxed{{7 \over 23328}}</math><br />
<br />
~Zeric Hang<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_9&diff=961352018 AMC 10B Problems/Problem 92018-07-12T15:43:52Z<p>Zeric: </p>
<hr />
<div>The faces of each of <math>7</math> standard dice are labeled with the integers from <math>1</math> to <math>6</math>. Let <math>p</math> be the probabilities that when all <math>7</math> dice are rolled, the sum of the numbers on the top faces is <math>10</math>. What other sum occurs with the same probability as <math>p</math>?<br />
<br />
<math>\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}</math><br />
<br />
==Solution 1==<br />
It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.<br />
<br />
So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows:<br />
<br />
(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...<br />
<br />
However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math>, and we are done.<br />
<br />
Written By: Archimedes15<br />
<br />
==Solution 2==<br />
<br />
Let's call the unknown value <math>x</math>. By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and <math>x</math>. So, <br />
<br />
<math>10 - 7 = 42- x </math><br />
<br />
<math> x = 39 </math> and our answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
By: Soccer_JAMS<br />
<br />
==Solution 3 (Simple Logic)==<br />
<br />
For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) x 7 = 49. Since having 10 is similar to not having 10, you just subtract 10 from the expected total sum. 49 - 10 = 39 so the answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
<br />
By: epicmonster<br />
<br />
==Solution 4==<br />
The expected value of the sums of the die rolls is <math>3.5*7=24.5</math>, and since the probabilities should be distributed symmetrically on both sides of <math>24.5</math>, the answer is <math>24.5+24.5-10=39</math>, which is <math>\boxed{\textbf{(D)} \text{ 39}}</math>.<br />
<br />
By: dajeff<br />
<br />
<br />
<br />
=== Note ===<br />
<br />
Calculating the probability of getting a sum of <math>10</math> is also easy. There are <math>3</math> cases:<br />
<br />
<br />
Case <math>1</math>: <math>\{1,1,1,1,1,1,4\}</math><br />
<br />
<br />
<math>\frac{7!}{6!}=7</math> cases<br />
<br />
<br />
Case <math>2</math>: <math>\{1,1,1,1,1,2,3\}</math><br />
<br />
<br />
<math>\frac{7!}{5!}=6*7=42</math> cases<br />
<br />
<br />
Case <math>3</math>: <math>\{1,1,1,1,2,2,2\}</math><br />
<br />
<br />
<math>\frac{7!}{4!3!}=5*7=35</math> cases<br />
<br />
<br />
The probability is <math>{84 \over 6^7} = \frac{14}{6^6}</math>. <br />
<br />
Calculating <math>6^6</math>: <br />
<br />
<math>6^6=(6^3)^2=216^2=46656</math> <br />
<br />
Therefore, the probability is <math>{14 \over 46656} = \boxed{{7 \over 23328}}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_9&diff=961342018 AMC 10B Problems/Problem 92018-07-12T15:18:12Z<p>Zeric: </p>
<hr />
<div>The faces of each of <math>7</math> standard dice are labeled with the integers from <math>1</math> to <math>6</math>. Let <math>p</math> be the probabilities that when all <math>7</math> dice are rolled, the sum of the numbers on the top faces is <math>10</math>. What other sum occurs with the same probability as <math>p</math>?<br />
<br />
<math>\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}</math><br />
<br />
==Solution 1==<br />
It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.<br />
<br />
So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows:<br />
<br />
(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...<br />
<br />
However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math>, and we are done.<br />
<br />
Written By: Archimedes15<br />
<br />
==Solution 2==<br />
<br />
Let's call the unknown value <math>x</math>. By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and <math>x</math>. So, <br />
<br />
<math>10 - 7 = 42- x </math><br />
<br />
<math> x = 39 </math> and our answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
By: Soccer_JAMS<br />
<br />
==Solution 3 (Simple Logic)==<br />
<br />
For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) x 7 = 49. Since having 10 is similar to not having 10, you just subtract 10 from the expected total sum. 49 - 10 = 39 so the answer is <math>\boxed{\textbf{(D)} \text{ 39}}</math><br />
<br />
By: epicmonster<br />
<br />
==Solution 4==<br />
The expected value of the sums of the die rolls is <math>3.5*7=24.5</math>, and since the probabilities should be distributed symmetrically on both sides of <math>24.5</math>, the answer is <math>24.5+24.5-10=39</math>, which is <math>\boxed{\textbf{(D)} \text{ 39}}</math>.<br />
<br />
By: dajeff<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_8&diff=961332018 AMC 10B Problems/Problem 82018-07-12T15:12:48Z<p>Zeric: </p>
<hr />
<div>Sara makes a staircase out of toothpicks as shown:<asy><br />
size(150);<br />
defaultpen(linewidth(0.8));<br />
path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45);<br />
for(int i=0;i<=2;i=i+1)<br />
{<br />
for(int j=0;j<=3-i;j=j+1)<br />
{<br />
filldraw(shift((i,j))*h,black);<br />
filldraw(shift((j,i))*v,black);<br />
}<br />
}</asy><br />
This is a 3-step staircase and uses 18 toothpicks. How many steps would be in a staircase that used 180 toothpicks?<br />
<br />
<math>\textbf{(A)}\ 10\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30</math><br />
<br />
== Solution ==<br />
<br />
A staircase with <math>n</math> steps contains <math>4 + 6 + 8 + ... + 2n + 2</math> toothpicks. This can be rewritten as <math>(n+1)(n+2) -2</math>. <br />
<br />
So, <math>(n+1)(n+2) - 2 = 180</math><br />
<br />
So, <math>(n+1)(n+2) = 182.</math> <br />
<br />
Inspection could tell us that <math>13 * 14 = 182</math>, so the answer is <math>13 - 1 = \boxed {(C) 12}</math><br />
<br />
== Solution 2 ==<br />
<br />
Layer <math>1</math>: <math>4</math> steps<br />
<br />
Layer <math>1,2</math>: <math>10</math> steps<br />
<br />
Layer <math>1,2,3</math>: <math>18</math> steps<br />
<br />
Layer <math>1,2,3,4</math>: <math>28</math> steps<br />
<br />
From inspection, we can see that with each increase in layer the difference in toothpicks between the current layer and the previous increases by <math>2</math>. Using this pattern:<br />
<br />
<math> 4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180 </math><br />
<br />
From this we see that the solution is indeed <math>\boxed {(C) 12}</math><br />
<br />
By: Soccer_JAMS<br />
<br />
== Solution 3 ==<br />
<br />
We can find a function that gives us the number of toothpicks for every layer. Using finite difference, we know that the degree must be <math>2</math> and the leading coefficient is <math>1</math>. The function is <math>f(n)=n^2+3n</math> where <math>n</math> is the layer and <math>f(n)</math> is the number of toothpicks.<br />
<br />
<br />
We have to solve for <math>n</math> when <math>n^2+3n=180\Rightarrow n^2+3n-180=0</math>. Factor to get <math>(n-12)(n+15)</math>. The roots are <math>12</math> and <math>-15</math>. Clearly <math>-15</math> is impossible so the answer is <math>\boxed {(C) 12}</math>.<br />
<br />
~Zeric Hang<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_23&diff=961142012 AMC 10A Problems/Problem 232018-07-11T15:56:17Z<p>Zeric: /* Solution */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #19]] and [[2012 AMC 10A Problems|2012 AMC 10A #23]]}}<br />
<br />
== Problem ==<br />
<br />
Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?<br />
<br />
<br />
<math> \text{(A)}\ 60\qquad\text{(B)}\ 170\qquad\text{(C)}\ 290\qquad\text{(D)}\ 320\qquad\text{(E)}\ 660 </math><br />
<br />
== Solution ==<br />
<br />
Note that if <math>n</math> is the number of friends each person has, then <math>n</math> can be any integer from <math>1</math> to <math>4</math>, inclusive.<br />
<br />
Also note that one person can only have at most 5 friends. That's right, 5 friends. <br />
<br />
Also note that the cases of <math>n=1</math> and <math>n=4</math> are the same, since a map showing a solution for <math>n=1</math> can correspond one-to-one with a map of a solution for <math>n=4</math> by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with <math>n=2</math> when compared to configurations of <math>n=3</math>. Thus, we have two cases to examine, <math>n=1</math> and <math>n=2</math>, and we count each of these combinations twice.<br />
<br />
For <math>n=1</math>, if everyone has exactly one friend, that means there must be <math>3</math> pairs of friends, with no other interconnections. The first person has <math>5</math> choices for a friend. There are <math>4</math> people left. The next person has <math>3</math> choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are <math>15</math> configurations with <math>n=1</math>.<br />
<br />
For <math>n=2</math>, there are two possibilities. The group of <math>6</math> can be split into two groups of <math>3</math>, with each group creating a friendship triangle. The first person has <math>\binom{5}{2} = 10</math> ways to pick two friends from the other five, while the other three are forced together. Thus, there are <math>10</math> triangular configurations. <br />
<br />
However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon [[without loss of generality]]. This person has <math>\binom{5}{2} = 10</math> choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are <math>10 \cdot 3! = 60</math> hexagonal configurations, and in total <math>70</math> configurations for <math>n=2</math>.<br />
<br />
As stated before, <math>n=3</math> has <math>70</math> configurations, and <math>n=4</math> has <math>15</math> configurations. This gives a total of <math>(70 + 15)\cdot 2 = 170</math> configurations, which is option <math>\boxed{\textbf{(B)}\ 170}</math>.<br />
<br />
=== Note ===<br />
<br />
We can also calculate the triangular configurations by applying <math>\frac{\binom{6}{3}}{2} = \frac{20}{2}=10</math> (Because choosing <math>A</math>,<math>B</math> and <math>C</math> is the same as choosing <math>D</math>,<math>E</math> and <math>F</math>.<br />
<br />
<br />
For the hexagonal configurations, we know that the total amount of combinations is <math>6!=720</math>. However, we must subtract the number of cases for rotation and reflection. <math>{720 \over 6 \cdot 2} = 60</math><br />
<br />
~Zeric Hang<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=22|num-a=24}}<br />
<br />
{{AMC12 box|year=2012|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_22&diff=961132012 AMC 10A Problems/Problem 222018-07-11T15:11:52Z<p>Zeric: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
<br />
The sum of the first <math>m</math> positive odd integers is 212 more than the sum of the first <math>n</math> positive even integers. What is the sum of all possible values of <math>n</math>?<br />
<br />
<math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math><br />
<br />
== Solution 1==<br />
<br />
The sum of the first <math>m</math> odd integers is given by <math>m^2</math>. The sum of the first <math>n</math> even integers is given by <math>n(n+1)</math>.<br />
<br />
Thus, <math>m^2 = n^2 + n + 212</math>. Since we want to solve for n, rearrange as a quadratic equation: <math>n^2 + n + (212 - m^2) = 0</math>.<br />
<br />
Use the quadratic formula: <math>n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}</math>. <math>n</math> is clearly an integer, so <math>1 - 4(212 - m^2) = 4m^2 - 847</math> must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2), <math>4m^2 - 847</math> must be odd.<br />
<br />
Let <math>x</math> = <math>\sqrt{4m^2 - 847}</math>. (Note that this means that <math>n = \frac{-1 + x}{2}</math>.) This can be rewritten as <math>x^2 = 4m^2 - 847</math>, which can then be rewritten to <math>4m^2 - x^2 = 847</math>. Factor the left side by using the difference of squares. <math>(2m + x)(2m - x) = 847 = 7\cdot11^2</math>.<br />
<br />
Our goal is to find possible values for <math>x</math>, then use the equation above to find <math>n</math>. The difference between the factors is <math>(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.</math> We have three pairs of factors, <math>847\cdot1, 7\cdot121,</math> and <math>11\cdot77</math>. The differences between these factors are <math>846</math>, <math>114</math>, and <math>66</math> - those are all possible values for <math>2x</math>. Thus the possibilities for <math>x</math> are <math>423</math>, <math>57</math>, and <math>33</math>. <br />
<br />
Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. The answer is <math>\boxed{\textbf{(A)}\ 255}</math>.<br />
<br />
==Solution 2==<br />
<br />
As above, start off by noting that the sum of the first <math>m</math> odd integers <math>= m^2</math> and the sum of the first <math>n</math> even integers <math>= n(n+1)</math>. Clearly <math>m > n</math>, so let <math>m = n + a</math>, where <math>a</math> is some positive integer. We have:<br />
<br />
<math>(n+a)^2 = n(n+1) + 212</math>.<br />
Expanding, grouping like terms and factoring, we get:<br />
<math>n = \frac{(212 - a^2)}{(2a - 1)}</math>.<br />
<br />
We know that <math>n</math> and <math>a</math> are both positive integers, so we need only check values of <math>a</math> from <math>1</math> to <math>14</math> (<math>14^2 = 196 < 212 < 15^2 = 225</math>). Plugging in, the only values of <math>a</math> that give integral solutions are <math>1, 4,</math> and <math>6</math>. These gives <math>n</math> values of <math>211, 28,</math> and <math>16</math>, respectively. <math>211 + 28 + 16 = 255</math>. Hence, the answer is <math>\boxed{\textbf{(A)}\ 255}</math>.<br />
<br />
==Solution 3==<br />
Using the closed forms for the sums, we get <math>m^2=n(n+1)+212</math>, or <math>m^2=n^2+n+212</math>. We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now <math>4m^2=4n^2+4n+848</math>. Complete the square on the right hand side: <math>4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847</math>. Move over the <math>(2n+1)^2</math> and factor to get <math>(2m-2n-1)(2m+2n+1)=847=7\cdot11\cdot11</math>. The second factor is clearly greater than the first, and the only possible factor pairs are <math>1</math> and <math>847</math>, <math>7</math> and <math>121</math>, <math>11</math> and <math>77</math>. In each of these cases, solve for <math>m</math> and <math>n</math> and we find the solutions <math>(m,n)=(212,211), (32,28), (22,16)</math>. The sum of all possible values of <math>n</math> is <math>211+28+16=\boxed{\textbf{(A)}\ 255}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_8&diff=960962012 AMC 10A Problems/Problem 82018-07-10T15:15:45Z<p>Zeric: /* Solution */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #6]] and [[2012 AMC 10A Problems|2012 AMC 10A #8]]}}<br />
<br />
== Problem ==<br />
The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math><br />
<br />
==Solution==<br />
Let the three numbers be equal to <math>a</math>, <math>b</math>, and <math>c</math>. We can now write three equations:<br />
<br />
<math>a+b=12</math><br />
<br />
<math>b+c=17</math><br />
<br />
<math>a+c=19</math><br />
<br />
Adding these equations together, we get that<br />
<br />
<math>2(a+b+c)=48</math> and<br />
<br />
<math>a+b+c=24</math><br />
<br />
Substituting the original equations into this one, we find<br />
<br />
<math>c+12=24</math><br />
<br />
<math>a+17=24</math><br />
<br />
<math>b+19=24</math><br />
<br />
Therefore, our numbers are 12, 7, and 5. The middle number is <math>\boxed{\textbf{(D)}\ 7}</math><br />
<br />
==Solution 2 (Faster)==<br />
<br />
Let the three numbers be <math>a</math>, <math>b</math> and <math>c</math> and <math>a<b<c</math>. We get the three equations: <br />
<br />
<math>a+b=12</math><br />
<br />
<math>a+c=17</math><br />
<br />
<math>b+c=19</math><br />
<br />
We add the first and last equations and then subtract the second one.<br />
<br />
<math>(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7</math><br />
<br />
Because <math>b</math> is the middle number, the middle number is <math>\boxed{\textbf{(D)}\ 7}</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=7|num-a=9}}<br />
{{AMC12 box|year=2012|ab=A|num-b=5|num-a=7}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_24&diff=960292017 AMC 10B Problems/Problem 242018-07-08T13:56:50Z<p>Zeric: /* Solution */</p>
<hr />
<div>==Problem 24==<br />
The vertices of an equilateral triangle lie on the hyperbola <math>xy=1</math>, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169</math><br />
<br />
==Solution==<br />
WLOG, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, <math>A = (1,1)</math>, so <math>AI = BI = CI = 2\sqrt{2}</math>, so since <math>\triangle AIB</math> is isosceles and <math>\angle AIB = 120^{\circ}</math>, then by Law of Cosines, <math>AB = 2\sqrt{6}</math>. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to <math>\frac {s}{\sqrt{3}}</math>. Therefore, the area of the triangle is <math>\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}</math>, so the square of the area of the triangle is <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 2==<br />
WLOG, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. WLOG, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = (a,\frac{1}{a})</math> and <math>C=(\frac{1}{a},a)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_22&diff=960282017 AMC 10B Problems/Problem 222018-07-08T13:48:59Z<p>Zeric: /* Solutions */</p>
<hr />
<div>==Problem==<br />
The diameter <math>\overline{AB}</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>\overline{AE}</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
==Solutions==<br />
<center><asy><br />
size(10cm);<br />
pair A, B, C, D, E, O;<br />
A = (-2,0);<br />
B = (2,0);<br />
C = (2*cos(1.24),2*sin(1.24));<br />
D = (5,0);<br />
E = (5,5);<br />
O = (A+B)/2;<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(E);<br />
dot(O);<br />
draw(Circle((A+B)/2,2));<br />
draw(A--D--E--C--A);<br />
draw(C--B);<br />
draw(rightanglemark(A,C,B,5));<br />
draw(rightanglemark(A,D,E,5));<br />
label("$A$",A,W);<br />
label("$B$",B,SE);<br />
label("$D$",D,SE);<br />
label("$E$",E,NE);<br />
label("$C$",C,N);<br />
label("$2$",(O+B)/2,S);<br />
label("$3$",(B+D)/2,S);<br />
label("$5$",(D+E)/2,NE);<br />
</asy></center><br />
===Solution 1===<br />
Notice that <math>ADE</math> and <math>ABC</math> are right triangles. Then <math>AE = \sqrt{7^2+5^2} = \sqrt{74}</math>. <math>\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}</math>, so <math>BC = \frac{20}{\sqrt{74}}</math>. We also find that <math>AC = \frac{28}{\sqrt{74}}</math>, and thus the area of <math>ABC</math> is <math>\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
<br />
===Solution 2===<br />
We note that <math>\triangle ACB ~ \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2</math>, so the area of <math>\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
<br />
===Solution 3===<br />
As stated before, note that <math>\triangle ACB ~ \triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2 = 4^2</math>. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Mulitplying <math>x</math> by <math>7</math> and <math>5</math> eventually gives us <math>\frac {14\sqrt{74}}{37}</math> <math>\frac {10\sqrt{74}}{37}</math>. We divide this by 2, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
<br />
===Solution 4===<br />
Let's call the center of the circle that segment <math>AB</math> is the diameter of, <math>O</math>. Note that <math>\triangle ODE</math> is an isosceles right triangle. Solving for side <math>OE</math>, using the Pythagorean theorem, we find it to be <math>5\sqrt{2}</math>. Calling the point where segment <math>OE</math> intersects circle <math>O</math>, the point <math>I</math>, segment <math>IE</math> would be <math>5\sqrt{2}-2</math>. Also, noting that <math>\triangle ADE</math> is a right triangle, we solve for side <math>AE</math>, using the Pythagorean Theorem, and get <math>\sqrt{74}</math>. Using Power of Point on point <math>E</math>, we can solve for <math>CE</math>. We can subtract <math>CE</math> from <math>AE</math> to find <math>AC</math> and then solve for <math>CB</math> using Pythagorean theorem once more.<br />
<br />
<math>(AE)(CE)</math> = (Diameter of circle <math>O</math> + <math>IE</math>)<math>(IE)</math> <math>\rightarrow</math> <math>{\sqrt{74}}(CE)</math> = <math>(5\sqrt{2}+2)(5\sqrt{2}-2)</math> <math>\Rightarrow</math> <math>CE</math> = <math>\frac{23\sqrt{74}}{37}</math><br />
<br />
<math>AC = AE - CE</math> <math>\rightarrow</math> <math>AC</math> = <math>{\sqrt74}</math> - <math>\frac{23\sqrt{74}}{37}</math> <math>\Rightarrow</math> <math>AC</math> = <math>\frac{14\sqrt{74}}{37}</math><br />
<br />
Now to solve for <math>CB</math>:<br />
<br />
<math>AB^2</math> - <math>AC^2</math> = <math>CB^2</math> <math>\rightarrow</math> <math>4^2</math> + <math>\frac{14\sqrt{74}}{37}^2</math> = <math>CB^2</math> <math>\Rightarrow</math> <math>CB</math> = <math>\frac{10\sqrt{74}}{37}</math><br />
<br />
Note that <math>\triangle ABC</math> is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases <math>AC</math> and <math>BC</math>, we get the area of triangle <math>ABC</math> to be <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_25&diff=957442010 AMC 10A Problems/Problem 252018-07-02T17:53:15Z<p>Zeric: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Jim starts with a positive integer <math>n</math> and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with <math>n = 55</math>, then his sequence contains <math>5</math> numbers:<br />
<br />
<br />
<cmath>\begin{array}{ccccc}<br />
{}&{}&{}&{}&55\\<br />
55&-&7^2&=&6\\<br />
6&-&2^2&=&2\\<br />
2&-&1^2&=&1\\<br />
1&-&1^2&=&0\\<br />
\end{array}</cmath><br />
<br />
Let <math>N</math> be the smallest number for which Jim’s sequence has <math>8</math> numbers. What is the units digit of <math>N</math>?<br />
<br />
<math><br />
\mathrm{(A)}\ 1<br />
\qquad<br />
\mathrm{(B)}\ 3<br />
\qquad<br />
\mathrm{(C)}\ 5<br />
\qquad<br />
\mathrm{(D)}\ 7<br />
\qquad<br />
\mathrm{(E)}\ 9<br />
</math><br />
<br />
==Solution==<br />
<br />
We can find the answer by working backwards. We begin with <math>1-1^2=0</math> on the bottom row, then the <math>1</math> goes to the right of the equal's sign in the row above. We find the smallest value <math>x</math> for which <math>x-1^2=1</math> and <math>x>1^2</math>, which is <math>x=2</math>.<br />
<br />
We repeat the same procedure except with <math>x-1^2=1</math> for the next row and <math>x-1^2=2</math> for the row after that. However, at the fourth row, we see that solving <math>x-1^2=3</math> yields <math>x=4</math>, in which case it would be incorrect since <math>1^2=1</math> is not the greatest perfect square less than or equal to <math>x</math> . So we make it a <math>2^2</math> and solve <math>x-2^2=3</math>. We continue on using this same method where we increase the perfect square until <math>x</math> can be made bigger than it. When we repeat this until we have <math>8</math> rows, we get:<br />
<br />
<cmath> \begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array} </cmath><br />
<br />
Hence the solution is the last digit of <math>7223</math>, which is <math>\boxed{\textbf{(B)}\ 3}</math>.<br />
<br />
Note: We can go up to <math>167</math>, and then notice the pattern of units digits alternating between <math>3</math> and <math>7</math>, so we do not need to calculate <math>7223</math>.<br />
<br />
==Solution 2==<br />
Notice that to get the previous term, we must add the smallest square number, (let's call it <math>n^2</math>) such that the sum is less than <math>(n+1)^2</math>. Otherwise, instead of subtracting <math>n^2</math> from the previous term, we're subtracting a greater square number. <br />
<br />
<br />
Remember that <math>(x+1)^2 = x^2 + x + (x+1)</math>. Recall that to find the previous term, we must add a square number such that it is less than the next square number. <math>a + n^2 < (n+1)^2</math>. For this to be true, <math>a < n + (n+1)</math>. What that means is that given a term <math>a</math>, we can find the previous term by adding <math>n^2</math> where <math>n > \frac {a-1}{2}</math>. <br />
<br />
<br />
For example, to find the term that precedes <math>167</math>, we know that <math>n>166/2 = 83</math>. Therefore, <math>n=84</math> and the previous term is <math>167 + 84^2 = 7223</math>. The last digit of <math>7223</math> is <math>3 \Rightarrow \boxed{\textbf{(B)}\ 3}</math><br />
<br />
~Zeric Hang<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|num-b=24|after=Last Question|ab=A}}<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_19&diff=957372010 AMC 10A Problems/Problem 192018-07-02T16:21:24Z<p>Zeric: /* Solution 3 (no trig) */</p>
<hr />
<div>== Problem ==<br />
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on triangle ABC, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>.<br />
<br />
If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore<br />
<br />
<math>\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)</math>.<br />
<br />
<br />
Based on the initial conditions,<br />
<br />
<cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath><br />
<br />
Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] (or Girard identities, or Newton-Girard identities) we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
===Solution 2===<br />
As above, we find that the area of <math>\triangle ACE</math> is <math>\frac{\sqrt3}4(r^2+r+1)</math>.<br />
<br />
We also find by the sine [[triangle]] area formula that <math>ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4</math>, and thus<br />
<cmath>\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4\right)}=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}</cmath><br />
This simplifies to <math>r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
===Solution 3 (no trig)===<br />
<br />
Extend <math>AB</math> so that it creates right triangle <math>\triangle AEC</math> where <math>\angle E = 90^\circ</math>. It is given that the hexagon is equiangular, therefore <math>\angle ABC = 120^\circ</math>. <math>\angle ABC</math> and <math>\angle EBC</math> are supplementary so <math>\angle EBC = 60^\circ</math>. <br />
<br />
<br />
We can use either Pythagorean theorem or the properties of a <math>30-60-90</math> triangle to find the length of <math>BE={r \over 2}</math> and <math>CE = {\sqrt 3 \over 2 }r</math>. The legs of <math>\triangle AEC</math> are <math>1 + {r \over 2}</math> and <math>{\sqrt 3 \over 2 }r</math>. <br />
<br />
<br />
Using Pythagorean theorem, we get <math>AC = (r^2+r+1)</math>. We can then follow <math>\textbf {Solution 1}</math> to solve for <math>r</math>. <math>\boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
<br />
Alternatively, we can find the area of <math>\triangle ABC</math>. We know that the three smaller triangles: <math>\triangle ABC</math>, <math>\triangle CDE</math>, and <math>\triangle EFA</math> are congruent because of <math>S-A-S</math>. Therefore one of the smaller triangles accounts for <math>10\%</math> of the total area. The height of the smaller triangle <math>\triangle ABC</math> is just <math>CE</math> so the area is <math>{1 \cdot {\sqrt 3 \over 2 }r \over 2}</math>. We can then find the area of the hexagon using <math>\textbf {Solution 1}</math>.<br />
<br />
<br />
We can even find the area of <math>\triangle ACE</math> and <math>\triangle ABC</math> and solve for <math>r</math> because the ratio of the areas is <math>7</math> to <math>1</math>.<br />
<br />
~Zeric Hang<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|num-b=18|num-a=20|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Zerichttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_19&diff=957342010 AMC 10A Problems/Problem 192018-07-02T14:09:46Z<p>Zeric: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on triangle ABC, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>.<br />
<br />
If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore<br />
<br />
<math>\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)</math>.<br />
<br />
<br />
Based on the initial conditions,<br />
<br />
<cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath><br />
<br />
Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] (or Girard identities, or Newton-Girard identities) we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
===Solution 2===<br />
As above, we find that the area of <math>\triangle ACE</math> is <math>\frac{\sqrt3}4(r^2+r+1)</math>.<br />
<br />
We also find by the sine [[triangle]] area formula that <math>ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4</math>, and thus<br />
<cmath>\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4\right)}=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}</cmath><br />
This simplifies to <math>r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
===Solution 3 (no trig)===<br />
<br />
Extend <math>AB</math> so that it creates right triangle <math>\triangle AEC</math> where <math>\angle E = 90^\circ</math>. It is given that the hexagon is equiangular, therefore <math>\angle ABC = 120^\circ</math>. <math>\angle ABC</math> and <math>\angle EBC</math> are supplementary so <math>\angle EBC = 60^\circ</math>. <br />
<br />
<br />
We can use either Pythagorean theorem or the properties of a <math>30-60-90</math> triangle to find the length of <math>BE={r \over 2}</math> and <math>CE = {\sqrt 3 \over 2 }r</math>. The legs of <math>\triangle AEC</math> are <math>1 + {r \over 2}</math> and <math>{\sqrt 3 \over 2 }r</math>. <br />
<br />
<br />
Using Pythagorean theorem, we get <math>AC = (r^2+r+1)</math>. We can then follow <math>\textbf {Solution 1}</math> to solve for <math>r</math>. <math>\boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
<br />
Alternatively, we can find the area of <math>\triangle ABC</math>. We know that the three smaller triangles: <math>\triangle ABC</math>, <math>\triangle CDE</math>, and <math>\triangle EFA</math> are congruent because of <math>S-A-S</math>. Therefore one of the smaller triangles accounts for <math>10\%</math> of the total area. The height of the smaller triangle <math>\triangle ABC</math> is just <math>CE</math> so the area is <math>{1 \cdot {\sqrt 3 \over 2 }r \over 2}</math>. We can then find the area of the hexagon using <math>\textbf {Solution 1}</math>.<br />
<br />
<br />
We can even find the area of <math>\triangle ACE</math> and <math>\triangle ABC</math> and solve for <math>r</math> because the ratio of the areas is <math>7</math> to <math>1</math>.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|num-b=18|num-a=20|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Zeric