https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bobcats&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:22:05ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_24&diff=1630512016 AMC 10A Problems/Problem 242021-10-02T16:15:37Z<p>Bobcats: /* Solution 12 */</p>
<hr />
<div>==Problem==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
<br />
==Solution 1 (Algebra)==<br />
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by <math>200</math> for now, then multiply it back at the end of our solution.<br />
<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(B,D,O,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$E$",E,WSW);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(B--D);<br />
draw(rightanglemark(C,E,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite.<br />
<br />
We set lengths <math>BE=ED</math> equal to <math>x</math> (Solution 1.1 begins from here). By the Pythagorean Theorem,<br />
<cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath><br />
<br />
We solve for <math>x</math>:<br />
<cmath>1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2</cmath><br />
<cmath>2\sqrt{(1-x^2)(2-x^2)}=2x^2-1</cmath><br />
<cmath>4(1-x^2)(2-x^2)=(2x^2-1)^2</cmath><br />
<cmath>8-12x^2+4x^4=4x^4-4x^2+1</cmath><br />
<cmath>8x^2=7</cmath><br />
<cmath>x=\frac{\sqrt{14}}{4}</cmath><br />
<br />
By Ptolemy's Theorem,<br />
<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath><br />
<br />
Substituting values,<br />
<cmath>1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2</cmath><br />
<cmath>1+AD=\frac{7}{2}</cmath><br />
<cmath>AD=\frac{5}{2}</cmath><br />
<br />
Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>.<br />
<br />
==Solution 2 (HARD Algebra)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
</asy><br />
<br />
Let quadrilateral <math>ABCD</math> be inscribed in circle <math>O</math>, where <math>AD</math> is the side of unknown length. Draw the radii from center <math>O</math> to all four vertices of the quadrilateral, and draw the altitude of <math>\triangle BOC</math> such that it passes through side <math>AD</math> at the point <math>G</math> and meets side <math>BC</math> at the point <math>H</math>.<br />
<br />
By the Pythagorean Theorem, the length of <math>OH</math> is<br />
<cmath>\begin{align*}<br />
\sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2}<br />
\\ &= \sqrt{80000 - 10000}<br />
\\ &= \sqrt{70000}<br />
\\ &= 100\sqrt{7}.<br />
\end{align*}</cmath><br />
<br />
Note that <math>[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].</math> Let the length of <math>OG</math> be <math>h</math> and the length of <math>AD</math> be <math>x</math>; then we have that <br />
<br />
<math> [AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].</math> <br />
<br />
Furthermore,<br />
<cmath>\begin{align*}<br />
h &= \sqrt{OD^2 - GD^2}<br />
\\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2}<br />
\\ &= \sqrt{80000 - \frac{x^2}{4}}<br />
\end{align*}</cmath><br />
<br />
Substituting this value of <math>h</math> into the previous equation and evaluating for <math>x</math>, we get:<br />
<cmath>\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}</cmath><br />
<cmath>\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}</cmath><br />
<cmath>60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)</cmath><br />
<cmath>40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)</cmath><br />
<cmath>7x^2 - 5600x + 1120000 = 320000 - x^2</cmath><br />
<cmath>8x^2 - 5600x + 800000 = 0</cmath><br />
<cmath>x^2 - 700x + 100000 = 0</cmath><br />
<br />
The roots of this quadratic are found by using the quadratic formula:<br />
<cmath>\begin{align*}<br />
x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1}<br />
\\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2}<br />
\\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2}<br />
\\ &= 350 \pm \frac{300}{2}<br />
\\ &= 200, 500<br />
\end{align*}</cmath><br />
<br />
If the length of <math>AD</math> is <math>200</math>, then quadrilateral <math>ABCD</math> would be a square and thus, the radius of the circle would be<br />
<cmath>\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}</cmath><br />
Which is a contradiction. Therefore, our answer is <math>\boxed{500}.</math><br />
<br />
==Solution 3 (Trigonometry Bash)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Angle mark for BOC<br />
draw(anglemark(C,O,B));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on <math>\Delta BOC</math>; let <math> \theta = \angle BOC</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath><br />
Because <math>\angle AOB</math>, <math>\angle BOC</math>, and <math>\angle COD</math> are congruent, <math>\angle AOD = 3\theta</math>. To find the remaining side (<math>AD</math>), we simply have to apply the law of cosines to <math>\Delta AOD</math> . Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath><br />
<br />
==Solution 4 (Easier Trigonometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=foot(A,B,C);<br />
F=foot(D,B,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(A--E);<br />
draw(E--B);<br />
draw(C--F);<br />
draw(F--D);<br />
label("$E$",E,NW);<br />
label("$F$",F,NE);<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
draw(rightanglemark(A,E,B));<br />
draw(rightanglemark(C,F,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle <math>O</math> with <math>AD</math> being the desired side. Then, drop perpendiculars from <math>A</math> and <math>D</math> to the extended line of <math>\overline{BC}</math> and let these points be <math>E</math> and <math>F</math>, respectively. Also, let <math>\theta = \angle BOC</math>. From the [[Law of Cosines]] on <math>\triangle BOC</math>, we have <math>\cos \theta = \frac{3}{4}</math>.<br />
<br />
Now, since <math>\triangle BOC</math> is isosceles with <math>\overline{OB} \cong \overline{OC}</math>, we have that <math>\angle BCO = \angle CBO = 90 - \frac{\theta}{2}</math>. In addition, we know that <math>\overline{BC} \cong \overline{CD}</math> as they are both equal to <math>200</math> and <math>\overline{OB} \cong \overline{OC} \cong \overline{OD}</math> as they are both radii of the same circle. By SSS Congruence, we have that <math>\triangle OBC \cong \triangle OCD</math>, so we have that <math>\angle OCD = \angle BCO = 90 - \frac{\theta}{2}</math>, so <math>\angle DCF = \theta</math>.<br />
<br />
Thus, we have <math>\frac{FC}{DC} = \cos \theta = \frac{3}{4}</math>, so <math>FC = 150</math>. Similarly, <math>BE = 150</math>, and <math>AD = 150 + 200 + 150 = \boxed{500}</math>.<br />
<br />
==Solution 5 (Just Geometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
E=extension(B,O,A,D);<br />
<br />
label("$E$",E,NE);<br />
<br />
F=extension(C,O,A,D);<br />
<br />
label("$F$",F,NE);<br />
<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
<br />
</asy><br />
<br />
Let AD intersect OB at E and OC at F.<br />
<br />
<br />
<math>\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta</math><br />
<br />
<math>\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}</math><br />
<br />
<br />
From there, <math>\triangle{OAB} \sim \triangle{ABE}</math>, thus:<br />
<br />
<math>\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}</math><br />
<br />
<math>OA = OB</math> because they are both radii of <math>\odot{O}</math>. Since <math>\frac{OA}{AB} = \frac{OB}{AE}</math>, we have that <math>AB = AE</math>. Similarly, <math>CD = DF</math>.<br />
<br />
<math>OE = 100\sqrt{2} = \frac{OB}{2}</math> and <math>EF=\frac{BC}{2}=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}</math><br />
<br />
==Solution 6 (Ptolemy's Theorem)==<br />
<br />
<asy><br />
pathpen = black; pointpen = black;<br />
size(6cm);<br />
draw(unitcircle);<br />
pair A = D("A", dir(50), dir(50));<br />
pair B = D("B", dir(90), dir(90));<br />
pair C = D("C", dir(130), dir(130));<br />
pair D = D("D", dir(170), dir(170));<br />
pair O = D("O", (0,0), dir(-90));<br />
draw(A--C, red);<br />
draw(B--D, blue+dashed);<br />
draw(A--B--C--D--cycle);<br />
draw(A--O--C);<br />
draw(O--B);<br />
</asy><br />
<br />
Let <math>s = 200</math>. Let <math>O</math> be the center of the circle. Then <math>AC</math> is twice the altitude of <math>\triangle OBC</math> to <math>\overline{OB}</math>. Since <math>\triangle OBC</math> is isosceles we can compute its area to be <math>\frac{s^2 \sqrt{7}}{4}</math>, hence <math>CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}</math>.<br />
<br />
Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.</math> This gives us: <cmath>\boxed{\textbf{(E) } 500.}</cmath><br />
<br />
==Solution 7 (Trigonometry)==<br />
Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths <math>100,100\sqrt{7},200\sqrt{2}</math> by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is <math>\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}</math>. Similarly, the cosine is <math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}</math>.<br />
Since there are three sides, and since <math>\sin\theta=\sin\left(180-\theta\right)</math>,we seek to find <math>2r\sin 3\theta</math>.<br />
First, <math>\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}</math> and <math>\cos 2\theta=\frac{3}{4}</math> by Pythagorean.<br />
<cmath>\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}</cmath><br />
<cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}</cmath><br />
<br />
==Solution 8 (Area By Brahmagupta's Formula)==<br />
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=2</math> and <math>DA</math> is the missing side length. Let <math>DA=2x</math>. If <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>AD</math>, respectively, the height of the trapezoid is <math>OM-ON</math>. By the pythagorean theorem, <math>OM=\sqrt{OB^2-BM^2}=\sqrt7</math> and <math>ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}</math>. Thus the height of the trapezoid is <math>\sqrt7-\sqrt{8-x^2}</math>, so the area is <math>\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})</math>. By Brahmagupta's formula, the area is <math>\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Setting these two equal, we get <math>(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Dividing both sides by <math>x+1</math> and then squaring, we get <math>7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)</math>. Expanding the right hand side and canceling the <math>x^2</math> terms gives us <math>15-2(\sqrt7)(\sqrt{8-x^2})=2x+3</math>. Rearranging and dividing by two, we get <math>(\sqrt7)(\sqrt{8-x^2})=6-x</math>. Squaring both sides, we get <math>56-7x^2=x^2-12x+36</math>. Rearranging, we get <math>8x^2-12x-20=0</math>. Dividing by 4 we get <math>2x^2-3x-5=0</math>. Factoring we get, <math>(2x-5)(x+1)=0</math>, and since <math>x</math> cannot be negative, we get <math>x=2.5</math>. Since <math>DA=2x</math>, <math>DA=5</math>. Scaling up by 100, we get <math>\boxed{\textbf{(E)}\text{ 500}}</math>.<br />
<br />
==Solution 9 (Cheap Solution - For when you are running out of time.)==<br />
<br />
WLOG, let <math>AB=BC=CD=200</math>, and let ABCD be inscribed in a clrcle with radius <math>200\sqrt2</math>. We draw perpendiculars from <math>B</math> and <math>C</math> to <math>AD</math>, and label the intersections <math>E</math> and <math>F</math>, respectively. We can see that <math>EF=200</math> (because BCFE is a rectangle), and since <math>AD</math> is clearly greater than 200, and and since <math>EF</math>, which is part of segment <math>AD</math>, is an integer, than we conclude that <math>AD</math> is also an integer or of the form <math>200+2*AE</math>. There is no reason for <math>AE</math> to be of the form <math>a\sqrt{b} - 100</math> because it seems too arbitrary. The only other integer choice is <math>\boxed{\textbf{(E)}\text{ 500}}</math>.<br />
<br />
==Solution 10 (Similar Triangles)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations, L is used to write alpha= statement<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O, L;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(A,D,O,B);<br />
F=extension(A,D,O,C);<br />
L=midpoint(C--D);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad = A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,NW);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,NE);<br />
label("$E$",E,SW);<br />
label("$F$",F,SE);<br />
label("$O$",O,SE);<br />
dot(O,linewidth(5));<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0));<br />
draw(anglemark(C,O,B));<br />
label("$\theta$",O,3N);<br />
draw(anglemark(E,F,O));<br />
label("$\alpha$",F,3SW);<br />
draw(anglemark(D,F,C));<br />
label("$\alpha$",F,3NE);<br />
draw(anglemark(F,C,D));<br />
label("$\alpha$",C,3SSE);<br />
draw(anglemark(C,D,F));<br />
label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW);<br />
</asy><br />
Label the points as shown, and let <math>\angle{EOF} = \theta</math>. Since <math>\overline{OB} = \overline{OC}</math>, and <math>\triangle{OFE} \sim \triangle{OCB}</math>, we get that <math>\angle{EFO} = 90-\frac{\theta}{2}</math>. We assign <math>\alpha</math> to <math>90-\frac{\theta}{2}</math> for simplicity. <br />
From here, by vertical angles <math>\angle{CFD} = \alpha</math>. Also, since <math>\triangle{OCB} \cong \triangle{ODC}</math>, <math>\angle{OCD} = \alpha</math>. This means that <math>\angle{CDF} = 180-2\alpha = \theta</math>, which leads to <math>\triangle{OCB} \sim \triangle{DCF}</math>. <br />
Since we know that <math>\overline{CD} = 200</math>, <math>\overline{DF} = 200</math>, and by similar reasoning <math>\overline{AE} = 200</math>. <br />
Finally, again using similar triangles, we get that <math>\overline{CF} = 100\sqrt{2}</math>, which means that <math>\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}</math>. We can again apply similar triangles (or use Power of a Point) to get <math>\overline{EF} = 100</math>, and finally <math>\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}</math> - ColtsFan10<br />
<br />
==Solution 11 (Parameshwara’s Formula for Circumradius) ==<br />
<br />
Scale down by <math>100</math>. We know that the semiperimeter of the quadrilateral is <math>\frac{(2 + 2 + 2 + x)}{2}</math> where <math>x = \overline {AD}</math>. Simplifying we get <math>\frac{6 + x}{2}</math>. Now, the radius is <math>2\sqrt {2}</math>, so <br />
<math>2\sqrt {2} = \frac{1}{4} \sqrt \frac {(4 + 2x)^{3}}{(\frac{2 + x}{2})^{3} (\frac {6 - x}{2})^{3}}</math>.<br />
<br />
Simplifying we get <math>x = 5</math>. So the answer is <math>500</math>.<br />
<br />
==Solution 12 (Cheaper Solution - For when you have no time.)==<br />
<br />
Note that once we have drawn out the quadrilateral, the fourth side length looks similar to the diameter of the circle. Therefore, this leaves us with only D and E as viable options. With a bit of luck, choose E which is our final answer.<br />
<br />
~yk2007<br />
<br />
==Solution 13 (Complex Numbers)==<br />
<br />
We first scale down by a factor of <math>200\sqrt{2}</math>. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, so that <math>AD</math> is the length of the fourth side. We draw this in the complex plane so that <math>D</math> corresponds to the complex number <math>1</math>, and we let <math>C</math> correspond to the complex number <math>z</math>. Then, <math>A</math> corresponds to <math>z^3</math> and <math>B</math> corresponds to <math>z^2</math>. We are given that <math>\lvert z \rvert = 1</math> and <math>\lvert z-1 \rvert = 1/\sqrt{2}</math>, and we wish to find <math>\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert</math>. Let <math>z=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers. Then, <math>a^2+b^2=1</math> and <math>a^2-2a+1+b^2=1/2</math>; solving for <math>a</math> and <math>b</math> yields <math>a=3/4</math> and <math>b=\sqrt{7}/4</math>. Thus, <math>AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}</math>. Scaling back up gives us a final answer of <math>\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{\textbf{(E)} 500}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
==Video Solution by AoPS (Deven Ware)==<br />
https://www.youtube.com/watch?v=hpSyHZwsteM<br />
<br />
==Video Solution by Walt S.==<br />
https://www.youtube.com/watch?v=3iDqR9YNNkU<br />
== Video Solution (Ptolemy’s Theorem) ==<br />
https://youtu.be/NsQbhYfGh1Q?t=5094<br />
<br />
~ pi_is_3.14<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/gCmQlaiEG5A<br />
<br />
~IceMatrix<br />
==See Also==<br />
<br />
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_24&diff=1630492016 AMC 10A Problems/Problem 242021-10-02T16:14:18Z<p>Bobcats: /* Solution 10 */</p>
<hr />
<div>==Problem==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
<br />
==Solution 1 (Algebra)==<br />
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by <math>200</math> for now, then multiply it back at the end of our solution.<br />
<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(B,D,O,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$E$",E,WSW);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(B--D);<br />
draw(rightanglemark(C,E,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Let the intersection of <math>BD</math> and <math>OC</math> be point <math>E</math>. Notice that <math>BD</math> and <math>OC</math> are perpendicular because <math>BCDO</math> is a kite.<br />
<br />
We set lengths <math>BE=ED</math> equal to <math>x</math> (Solution 1.1 begins from here). By the Pythagorean Theorem,<br />
<cmath>\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}</cmath><br />
<br />
We solve for <math>x</math>:<br />
<cmath>1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2</cmath><br />
<cmath>2\sqrt{(1-x^2)(2-x^2)}=2x^2-1</cmath><br />
<cmath>4(1-x^2)(2-x^2)=(2x^2-1)^2</cmath><br />
<cmath>8-12x^2+4x^4=4x^4-4x^2+1</cmath><br />
<cmath>8x^2=7</cmath><br />
<cmath>x=\frac{\sqrt{14}}{4}</cmath><br />
<br />
By Ptolemy's Theorem,<br />
<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2</cmath><br />
<br />
Substituting values,<br />
<cmath>1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2</cmath><br />
<cmath>1+AD=\frac{7}{2}</cmath><br />
<cmath>AD=\frac{5}{2}</cmath><br />
<br />
Finally, we multiply back the <math>200</math> that we divided by at the beginning of the problem to get <math>AD=\boxed{500 (E)}</math>.<br />
<br />
==Solution 2 (HARD Algebra)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
</asy><br />
<br />
Let quadrilateral <math>ABCD</math> be inscribed in circle <math>O</math>, where <math>AD</math> is the side of unknown length. Draw the radii from center <math>O</math> to all four vertices of the quadrilateral, and draw the altitude of <math>\triangle BOC</math> such that it passes through side <math>AD</math> at the point <math>G</math> and meets side <math>BC</math> at the point <math>H</math>.<br />
<br />
By the Pythagorean Theorem, the length of <math>OH</math> is<br />
<cmath>\begin{align*}<br />
\sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2}<br />
\\ &= \sqrt{80000 - 10000}<br />
\\ &= \sqrt{70000}<br />
\\ &= 100\sqrt{7}.<br />
\end{align*}</cmath><br />
<br />
Note that <math>[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].</math> Let the length of <math>OG</math> be <math>h</math> and the length of <math>AD</math> be <math>x</math>; then we have that <br />
<br />
<math> [AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].</math> <br />
<br />
Furthermore,<br />
<cmath>\begin{align*}<br />
h &= \sqrt{OD^2 - GD^2}<br />
\\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2}<br />
\\ &= \sqrt{80000 - \frac{x^2}{4}}<br />
\end{align*}</cmath><br />
<br />
Substituting this value of <math>h</math> into the previous equation and evaluating for <math>x</math>, we get:<br />
<cmath>\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}</cmath><br />
<cmath>\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}</cmath><br />
<cmath>60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)</cmath><br />
<cmath>40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}</cmath><br />
<cmath>7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)</cmath><br />
<cmath>7x^2 - 5600x + 1120000 = 320000 - x^2</cmath><br />
<cmath>8x^2 - 5600x + 800000 = 0</cmath><br />
<cmath>x^2 - 700x + 100000 = 0</cmath><br />
<br />
The roots of this quadratic are found by using the quadratic formula:<br />
<cmath>\begin{align*}<br />
x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1}<br />
\\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2}<br />
\\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2}<br />
\\ &= 350 \pm \frac{300}{2}<br />
\\ &= 200, 500<br />
\end{align*}</cmath><br />
<br />
If the length of <math>AD</math> is <math>200</math>, then quadrilateral <math>ABCD</math> would be a square and thus, the radius of the circle would be<br />
<cmath>\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}</cmath><br />
Which is a contradiction. Therefore, our answer is <math>\boxed{500}.</math><br />
<br />
==Solution 3 (Trigonometry Bash)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,E);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Angle mark for BOC<br />
draw(anglemark(C,O,B));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle with <math>AD</math> being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center <math>O</math> to <math>A,B,C,</math> and <math>D</math>. Apply law of cosines on <math>\Delta BOC</math>; let <math> \theta = \angle BOC</math>. We get the following equation: <cmath>(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta</cmath> Substituting the values in, we get <cmath>(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta</cmath> Canceling out, we get <cmath>\cos\theta=\frac{3}{4}</cmath><br />
Because <math>\angle AOB</math>, <math>\angle BOC</math>, and <math>\angle COD</math> are congruent, <math>\angle AOD = 3\theta</math>. To find the remaining side (<math>AD</math>), we simply have to apply the law of cosines to <math>\Delta AOD</math> . Now, to find <math>\cos 3\theta</math>, we can derive a formula that only uses <math>\cos\theta</math>: <cmath>\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta</cmath> <cmath>\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)</cmath> <cmath>\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta</cmath> Plugging in <math>\cos\theta=\frac{3}{4}</math>, we get <math>\cos 3\theta= -\frac{9}{16}</math>. Now, applying law of cosines on triangle <math>OAD</math>, we get <cmath>(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}</cmath> <cmath>\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}</cmath> <cmath>AD=200 \cdot \frac{5}{2}=\boxed{500}</cmath><br />
<br />
==Solution 4 (Easier Trigonometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=foot(A,B,C);<br />
F=foot(D,B,C);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
draw(A--E);<br />
draw(E--B);<br />
draw(C--F);<br />
draw(F--D);<br />
label("$E$",E,NW);<br />
label("$F$",F,NE);<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
draw(rightanglemark(A,E,B));<br />
draw(rightanglemark(C,F,D));<br />
</asy><br />
<br />
Construct quadrilateral <math>ABCD</math> on the circle <math>O</math> with <math>AD</math> being the desired side. Then, drop perpendiculars from <math>A</math> and <math>D</math> to the extended line of <math>\overline{BC}</math> and let these points be <math>E</math> and <math>F</math>, respectively. Also, let <math>\theta = \angle BOC</math>. From the [[Law of Cosines]] on <math>\triangle BOC</math>, we have <math>\cos \theta = \frac{3}{4}</math>.<br />
<br />
Now, since <math>\triangle BOC</math> is isosceles with <math>\overline{OB} \cong \overline{OC}</math>, we have that <math>\angle BCO = \angle CBO = 90 - \frac{\theta}{2}</math>. In addition, we know that <math>\overline{BC} \cong \overline{CD}</math> as they are both equal to <math>200</math> and <math>\overline{OB} \cong \overline{OC} \cong \overline{OD}</math> as they are both radii of the same circle. By SSS Congruence, we have that <math>\triangle OBC \cong \triangle OCD</math>, so we have that <math>\angle OCD = \angle BCO = 90 - \frac{\theta}{2}</math>, so <math>\angle DCF = \theta</math>.<br />
<br />
Thus, we have <math>\frac{FC}{DC} = \cos \theta = \frac{3}{4}</math>, so <math>FC = 150</math>. Similarly, <math>BE = 150</math>, and <math>AD = 150 + 200 + 150 = \boxed{500}</math>.<br />
<br />
==Solution 5 (Just Geometry)==<br />
<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad= A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(Circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,W);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,ENE);<br />
label("$O$",O,S);<br />
label("$\theta$",O,3N);<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
E=extension(B,O,A,D);<br />
<br />
label("$E$",E,NE);<br />
<br />
F=extension(C,O,A,D);<br />
<br />
label("$F$",F,NE);<br />
<br />
<br />
//Angle marks<br />
draw(anglemark(C,O,B));<br />
<br />
</asy><br />
<br />
Let AD intersect OB at E and OC at F.<br />
<br />
<br />
<math>\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta</math><br />
<br />
<math>\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}</math><br />
<br />
<br />
From there, <math>\triangle{OAB} \sim \triangle{ABE}</math>, thus:<br />
<br />
<math>\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}</math><br />
<br />
<math>OA = OB</math> because they are both radii of <math>\odot{O}</math>. Since <math>\frac{OA}{AB} = \frac{OB}{AE}</math>, we have that <math>AB = AE</math>. Similarly, <math>CD = DF</math>.<br />
<br />
<math>OE = 100\sqrt{2} = \frac{OB}{2}</math> and <math>EF=\frac{BC}{2}=100</math> , so <math>AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}</math><br />
<br />
==Solution 6 (Ptolemy's Theorem)==<br />
<br />
<asy><br />
pathpen = black; pointpen = black;<br />
size(6cm);<br />
draw(unitcircle);<br />
pair A = D("A", dir(50), dir(50));<br />
pair B = D("B", dir(90), dir(90));<br />
pair C = D("C", dir(130), dir(130));<br />
pair D = D("D", dir(170), dir(170));<br />
pair O = D("O", (0,0), dir(-90));<br />
draw(A--C, red);<br />
draw(B--D, blue+dashed);<br />
draw(A--B--C--D--cycle);<br />
draw(A--O--C);<br />
draw(O--B);<br />
</asy><br />
<br />
Let <math>s = 200</math>. Let <math>O</math> be the center of the circle. Then <math>AC</math> is twice the altitude of <math>\triangle OBC</math> to <math>\overline{OB}</math>. Since <math>\triangle OBC</math> is isosceles we can compute its area to be <math>\frac{s^2 \sqrt{7}}{4}</math>, hence <math>CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}</math>.<br />
<br />
Now by Ptolemy's Theorem we have <math>CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.</math> This gives us: <cmath>\boxed{\textbf{(E) } 500.}</cmath><br />
<br />
==Solution 7 (Trigonometry)==<br />
Since all three sides equal <math>200</math>, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths <math>100,100\sqrt{7},200\sqrt{2}</math> by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is <math>\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}</math>. Similarly, the cosine is <math>\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}</math>.<br />
Since there are three sides, and since <math>\sin\theta=\sin\left(180-\theta\right)</math>,we seek to find <math>2r\sin 3\theta</math>.<br />
First, <math>\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}</math> and <math>\cos 2\theta=\frac{3}{4}</math> by Pythagorean.<br />
<cmath>\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}</cmath><br />
<cmath>2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}</cmath><br />
<br />
==Solution 8 (Area By Brahmagupta's Formula)==<br />
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be <math>ABCD</math>, where <math>AB=BC=CD=2</math> and <math>DA</math> is the missing side length. Let <math>DA=2x</math>. If <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>AD</math>, respectively, the height of the trapezoid is <math>OM-ON</math>. By the pythagorean theorem, <math>OM=\sqrt{OB^2-BM^2}=\sqrt7</math> and <math>ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}</math>. Thus the height of the trapezoid is <math>\sqrt7-\sqrt{8-x^2}</math>, so the area is <math>\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})</math>. By Brahmagupta's formula, the area is <math>\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Setting these two equal, we get <math>(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}</math>. Dividing both sides by <math>x+1</math> and then squaring, we get <math>7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)</math>. Expanding the right hand side and canceling the <math>x^2</math> terms gives us <math>15-2(\sqrt7)(\sqrt{8-x^2})=2x+3</math>. Rearranging and dividing by two, we get <math>(\sqrt7)(\sqrt{8-x^2})=6-x</math>. Squaring both sides, we get <math>56-7x^2=x^2-12x+36</math>. Rearranging, we get <math>8x^2-12x-20=0</math>. Dividing by 4 we get <math>2x^2-3x-5=0</math>. Factoring we get, <math>(2x-5)(x+1)=0</math>, and since <math>x</math> cannot be negative, we get <math>x=2.5</math>. Since <math>DA=2x</math>, <math>DA=5</math>. Scaling up by 100, we get <math>\boxed{\textbf{(E)}\text{ 500}}</math>.<br />
<br />
==Solution 9 (Cheap Solution - For when you are running out of time.)==<br />
<br />
WLOG, let <math>AB=BC=CD=200</math>, and let ABCD be inscribed in a clrcle with radius <math>200\sqrt2</math>. We draw perpendiculars from <math>B</math> and <math>C</math> to <math>AD</math>, and label the intersections <math>E</math> and <math>F</math>, respectively. We can see that <math>EF=200</math> (because BCFE is a rectangle), and since <math>AD</math> is clearly greater than 200, and and since <math>EF</math>, which is part of segment <math>AD</math>, is an integer, than we conclude that <math>AD</math> is also an integer or of the form <math>200+2*AE</math>. There is no reason for <math>AE</math> to be of the form <math>a\sqrt{b} - 100</math> because it seems too arbitrary. The only other integer choice is <math>\boxed{\textbf{(E)}\text{ 500}}</math>.<br />
<br />
==Solution 10 (Similar Triangles)==<br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations, L is used to write alpha= statement<br />
real RADIUS;<br />
pair A, B, C, D, E, F, O, L;<br />
RADIUS=3;<br />
<br />
//Variable Definitions<br />
A=RADIUS*dir(148.414);<br />
B=RADIUS*dir(109.471);<br />
C=RADIUS*dir(70.529);<br />
D=RADIUS*dir(31.586);<br />
E=extension(A,D,O,B);<br />
F=extension(A,D,O,C);<br />
L=midpoint(C--D);<br />
O=(0,0);<br />
<br />
//Path Definitions<br />
path quad = A -- B -- C -- D -- cycle;<br />
<br />
//Initial Diagram<br />
draw(circle(O, RADIUS), linewidth(0.8));<br />
draw(quad, linewidth(0.8));<br />
label("$A$",A,NW);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$D$",D,NE);<br />
label("$E$",E,SW);<br />
label("$F$",F,SE);<br />
label("$O$",O,SE);<br />
dot(O,linewidth(5));<br />
<br />
//Radii<br />
draw(O--A);<br />
draw(O--B);<br />
draw(O--C);<br />
draw(O--D);<br />
<br />
//Construction<br />
label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0));<br />
draw(anglemark(C,O,B));<br />
label("$\theta$",O,3N);<br />
draw(anglemark(E,F,O));<br />
label("$\alpha$",F,3SW);<br />
draw(anglemark(D,F,C));<br />
label("$\alpha$",F,3NE);<br />
draw(anglemark(F,C,D));<br />
label("$\alpha$",C,3SSE);<br />
draw(anglemark(C,D,F));<br />
label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW);<br />
</asy><br />
Label the points as shown, and let <math>\angle{EOF} = \theta</math>. Since <math>\overline{OB} = \overline{OC}</math>, and <math>\triangle{OFE} \sim \triangle{OCB}</math>, we get that <math>\angle{EFO} = 90-\frac{\theta}{2}</math>. We assign <math>\alpha</math> to <math>90-\frac{\theta}{2}</math> for simplicity. <br />
From here, by vertical angles <math>\angle{CFD} = \alpha</math>. Also, since <math>\triangle{OCB} \cong \triangle{ODC}</math>, <math>\angle{OCD} = \alpha</math>. This means that <math>\angle{CDF} = 180-2\alpha = \theta</math>, which leads to <math>\triangle{OCB} \sim \triangle{DCF}</math>. <br />
Since we know that <math>\overline{CD} = 200</math>, <math>\overline{DF} = 200</math>, and by similar reasoning <math>\overline{AE} = 200</math>. <br />
Finally, again using similar triangles, we get that <math>\overline{CF} = 100\sqrt{2}</math>, which means that <math>\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}</math>. We can again apply similar triangles (or use Power of a Point) to get <math>\overline{EF} = 100</math>, and finally <math>\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}</math> - ColtsFan10<br />
<br />
==Solution 11 (Parameshwara’s Formula for Circumradius) ==<br />
<br />
Scale down by <math>100</math>. We know that the semiperimeter of the quadrilateral is <math>\frac{(2 + 2 + 2 + x)}{2}</math> where <math>x = \overline {AD}</math>. Simplifying we get <math>\frac{6 + x}{2}</math>. Now, the radius is <math>2\sqrt {2}</math>, so <br />
<math>2\sqrt {2} = \frac{1}{4} \sqrt \frac {(4 + 2x)^{3}}{(\frac{2 + x}{2})^{3} (\frac {6 - x}{2})^{3}}</math>.<br />
<br />
Simplifying we get <math>x = 5</math>. So the answer is <math>500</math>.<br />
<br />
==Solution 12==<br />
<br />
Note that once we have drawn out the quadrilateral, the fourth side length looks similar to the diameter of the circle. Therefore, this leaves us with only D and E as viable options. With a bit of luck, choose E which is our final answer.<br />
<br />
~yk2007<br />
<br />
==Solution 13 (Complex Numbers)==<br />
<br />
We first scale down by a factor of <math>200\sqrt{2}</math>. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, so that <math>AD</math> is the length of the fourth side. We draw this in the complex plane so that <math>D</math> corresponds to the complex number <math>1</math>, and we let <math>C</math> correspond to the complex number <math>z</math>. Then, <math>A</math> corresponds to <math>z^3</math> and <math>B</math> corresponds to <math>z^2</math>. We are given that <math>\lvert z \rvert = 1</math> and <math>\lvert z-1 \rvert = 1/\sqrt{2}</math>, and we wish to find <math>\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert</math>. Let <math>z=a+bi</math>, where <math>a</math> and <math>b</math> are real numbers. Then, <math>a^2+b^2=1</math> and <math>a^2-2a+1+b^2=1/2</math>; solving for <math>a</math> and <math>b</math> yields <math>a=3/4</math> and <math>b=\sqrt{7}/4</math>. Thus, <math>AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}</math>. Scaling back up gives us a final answer of <math>\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{\textbf{(E)} 500}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
==Video Solution by AoPS (Deven Ware)==<br />
https://www.youtube.com/watch?v=hpSyHZwsteM<br />
<br />
==Video Solution by Walt S.==<br />
https://www.youtube.com/watch?v=3iDqR9YNNkU<br />
== Video Solution (Ptolemy’s Theorem) ==<br />
https://youtu.be/NsQbhYfGh1Q?t=5094<br />
<br />
~ pi_is_3.14<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/gCmQlaiEG5A<br />
<br />
~IceMatrix<br />
==See Also==<br />
<br />
{{AMC10 box|year=2016|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2016|ab=A|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_24&diff=1623322009 AMC 12B Problems/Problem 242021-09-16T19:33:00Z<p>Bobcats: </p>
<hr />
<div>== Problem ==<br />
For how many values of <math>x</math> in <math>[0,\pi]</math> is <math>\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)</math>?<br />
Note: The functions <math>\sin^{ - 1} = \arcsin</math> and <math>\cos^{ - 1} = \arccos</math> denote inverse trigonometric functions.<br />
<br />
<math>\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7</math><br />
<br />
== Solution ==<br />
<br />
First of all, we have to agree on the range of <math>\sin^{-1}</math> and <math>\cos^{-1}</math>. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: <math>\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2</math> and <math>\forall x: 0\leq \cos^{-1}(x) \leq \pi</math>.<br />
<br />
Hence we get that <math>\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x</math>, thus our equation simplifies to <math>\sin^{ - 1}(\sin 6x) = x</math>.<br />
<br />
Consider the function <math>f(x) = \sin^{ - 1}(\sin 6x) - x</math>. We are looking for roots of <math>f</math> on <math>[0,\pi]</math>.<br />
<br />
By analyzing properties of <math>\sin</math> and <math>\sin^{-1}</math> (or by computing the derivative of <math>f</math>) one can discover the following properties of <math>f</math>:<br />
<br />
* <math>f(0)=0</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[0,\pi/6]</math>.<br />
* <math>f</math> is decreasing and then increasing on <math>[\pi/6,2\pi/6]</math>.<br />
* <math>f</math> is increasing and then decreasing on <math>[2\pi/6,3\pi/6]</math>.<br />
<br />
For <math>x=\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0</math>. Hence <math>f</math> has exactly one root on <math>(0,\pi/6)</math>.<br />
<br />
For <math>x=2\pi/6</math> we have <math>f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0</math>. Hence <math>f</math> is negative on the entire interval <math>[\pi/6,2\pi/6]</math>.<br />
<br />
Now note that <math>\forall t: \sin^{-1}(t) \leq \pi/2</math>. Hence for <math>x > 3\pi/6</math> we have <math>f(x) < 0</math>, and we can easily check that <math>f(3\pi/6)<0</math> as well.<br />
<br />
Thus the only unknown part of <math>f</math> is the interval <math>(2\pi/6,3\pi/6)</math>. On this interval, <math>f</math> is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.<br />
<br />
To prove that there are two roots, it is enough to find any <math>x</math> from this interval such that <math>f(x)>0</math>.<br />
<br />
A good guess is its midpoint, <math>x=5\pi/12</math>, where the function <math>\sin^{-1}(\sin 6x)</math> has its local maximum. We can evaluate:<br />
<math>f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0</math>.<br />
<br />
Summary: The function <math>f</math> has <math>\boxed{\textbf{(B) }4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>.<br />
<br />
Actual solutions are <math>x=0</math>, <math>x=\pi/7</math>, <math>x=2\pi/5</math>, and <math>x=3\pi/7</math>.<br />
<br />
==Solution 2==<br />
Since <math>0\leq \cos^{-1}(a) \leq \pi</math> for all <math>a</math>, the equation reduces to <math>\sin^{-1}(\sin(6x)) = x</math>. Since <math>-\pi/2 \leq \sin^{-1}(a) \leq \pi/2</math> for all <math>a</math>, <math>0 \leq x \leq \pi/2</math>. To make the problem easier, we will measure angles in degrees. We will consider each sixth of the interval <math>[0, 90]</math>.<br />
<br />
For <math>0 \leq x \leq 15</math>, <math>6x</math> is in the first quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 6x</math>. Setting this equal to <math>x</math> yields the solution <math>x = 0</math>.<br />
<br />
For <math>15 \leq x \leq 30</math>, <math>6x</math> is in the second quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 180 - 6x</math>. This yields the solution <math>x = \frac{180}7</math>.<br />
<br />
For <math>30 \leq x \leq 45</math>, <math>6x</math> is in the third quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 180 - 6x</math>. As <math>\frac{180}{7}</math> is not on the interval <math>30 \leq x \leq 45</math>, this yields no solution.<br />
<br />
For <math>45 \leq x \leq 60</math>, <math>6x</math> is in the fourth quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 6x - 360</math>. As <math>72</math> is not on the interval <math>45 \leq x \leq 60</math>, this yields no solution.<br />
<br />
For <math>60 \leq x \leq 75</math>, <math>6x</math> is in the first quadrant plus a full revolution. Thus, <math>\sin^{-1}(\sin(6x)) = 6x - 360</math>. This yields the solution <math>x = 72</math>.<br />
<br />
For <math>75 \leq x \leq 90</math>, <math>6x</math> is in the second quadrant plus a full revolution. Thus <math>\sin^{-1}(\sin(6x)) = 540 - 6x</math>. This yields the solution <math>x = \frac{540}7</math>.<br />
<br />
There are <math>\boxed{\textbf{(B) }4}</math> solutions, <math>x=0</math>, <math>x=\pi/7</math>, <math>x=2\pi/5</math>, and <math>x=3\pi/7</math>.<br />
<br />
==Solution 3==<br />
Algebraically, the inverse function of a function should just cancel out, leaving <math>6x=x</math>. However, upon inspection we find that the graphs of these "inverse function of the function" equations are also periodic, like their normal trig function counterparts, due to the fact that inverse trig functions will never return any angle value higher than <math>2\pi</math>. But instead of a smooth wave, these graphs are made up of zigzags with slope <math>1</math> and <math>-1</math>. Trying a few values, we see that<br />
<br />
<br />
<math>y = \arcsin{\sin{x}}</math><br />
<br />
<br />
peaks at <math>\frac{\pi}{2}</math> and<br />
<br />
<br />
<math>y = \arccos{\cos{x}}</math><br />
<br />
<br />
peaks at <math>\pi</math><br />
<br />
<br />
But we want the graph of <math>y = \arcsin{(\sin6{x})}</math>, which has a period of <math>\frac{\pi}{3}</math> instead of <math>2\pi</math>. So this means the interval <math>[0, \pi]</math> will show <math>3</math> periods instead of <math>\frac{1}{2}</math> of a period. Visually it would be lines with slopes <math>6</math> and <math>-6</math>. Using the graph paper given to us, we plot out the two equations according to the above and we see that they intersect <math>4</math> times <math>\Rightarrow \boxed{\text{B}}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_4&diff=1511992021 AIME I Problems/Problem 42021-04-07T20:48:00Z<p>Bobcats: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Find the number of ways <math>66</math> identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.<br />
<br />
==Solution 1==<br />
Suppose we have <math>1</math> coin in the first pile. Then <math>(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)</math> all work for a total of <math>31</math> piles. Suppose we have <math>2</math> coins in the first pile, then <math>(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)</math> all work, for a total of <math>29</math>. Continuing this pattern until <math>21</math> coins in the first pile, we have the sum <math>31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+</math><br />
<br />
<math>(29+26+23+\ldots+2)=176+155=\boxed{331}</math>.<br />
<br />
==Solution 2==<br />
Let the three piles have <math>a, b, c</math> coins respectively. If we disregard order, then we just need to divide by <math>3! = 6</math> at the end.<br />
<br />
We know <math>a + b + c = 66</math>. Since <math>a, b, c</math> are positive integers, there are <math>\binom{65}{2}</math> ways from Stars and Bars.<br />
<br />
However, we must discard the cases where <math>a = b</math> or <math>a = c</math> or <math>b = c</math>. The three cases are symmetric, so we just take the first case and multiply by 3. We have <math>2a + c = 66 \implies a = 1, 2, \dots 32</math> for 32 solutions. Multiplying by 3, we will subtract 96 from our total.<br />
<br />
But we undercounted where <math>a = b = c = 22</math>. This is first counted 1 time, then we subtract it 3 times, so we add it back twice. There is clearly only 1 way, for a total of 2.<br />
<br />
Hence, the answer is <math>\frac{\binom{65}{2} - 96 + 2}{6} = \boxed{331}.</math><br />
<br />
==Solution 3==<br />
Let the piles have <math>a, b</math> and <math>c</math> coins, with <math>0 < a < b < c</math>. Then, let <math>b = a + k_1</math>, and <math>c = b + k_2</math>, such that each <math>k_i \geq 1</math>. The sum is then <math>a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66</math>. This is simply the number of positive solutions to the equation <math>3x+2y+z = 66</math>. Now, we take cases on <math>a</math>. <br />
<br />
If <math>a = 1</math>, then <math>2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31</math>. Each value of <math>k_1</math> corresponds to a unique value of <math>k_2</math>, so there are <math>31</math> solutions in this case. Similarly, if <math>a = 2</math>, then <math>2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29</math>, for a total of <math>29</math> solutions in this case. If <math>a = 3</math>, then <math>2k_1 + k_1 = 57 \implies 1 \leq k_1 \leq 28</math>, for a total of <math>28</math> solutions. In general, the number of solutions is just all the numbers that aren't a multiple of <math>3</math>, that are less than or equal to <math>31</math>. <br />
<br />
We then add our cases to get <cmath>1 + 2 + 4 + 5 + \cdots + 31 = 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) = \frac{31(32)}{2} - 3(55) = 31 \cdot 16 - 165 = 496 - 165 = \boxed{331}</cmath> as our answer.<br />
<br />
==Solution 4==<br />
<br />
We make an equation for how this is true. a+b+c = 66, where a is less than b, is less than c. We don't have a clear solution, so we'll try complimentary counting. First, let's find where a doesn't have to be less than b, which doesn't have to be less than c. We have <math>\dbinom{65}{2}</math> =2080 by stars and bars. Now we need to subtract off the cases where it doesn't satisfy the condition.<br />
<br />
We first by starting out with a = b. We can write that as 2b + c = 66. We can find there are 32 integer solutions to this equation. There are 32 solutions for b=c, and a = c by symmetry. We also need to subtract 2 from the 96, because we counted 22, 22, 22 3 times. <br />
We then have to divide by 6 because there are 3! ways to order the a, b, and c. Therefore, we have <math>\dfrac{\dbinom{65}{2}-94}{6}</math> = <math>\dfrac{1986}{6} = </math><math>\boxed{331}</math><br />
<br />
~Arcticturn<br />
<br />
==Video Solution #1 ==<br />
https://youtu.be/M3DsERqhiDk?t=1073<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_13&diff=1509862021 AMC 10A Problems/Problem 132021-04-04T15:10:22Z<p>Bobcats: /* Solution */</p>
<hr />
<div>==Problem==<br />
What is the volume of tetrahedron <math>ABCD</math> with edge lengths <math>AB = 2</math>, <math>AC = 3</math>, <math>AD = 4</math>, <math>BC = \sqrt{13}</math>, <math>BD = 2\sqrt{5}</math>, and <math>CD = 5</math> ?<br />
<br />
<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math><br />
<br />
==Solution==<br />
Drawing the tetrahedron out and testing side lengths, we realize that the <math>\triangle ABD, \triangle ABC,</math> and <math>\triangle ABD</math> are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take <math>\triangle ADC</math> as the base, then <math>AB</math> must be the height. <math>\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2</math>, so we have an answer of <math>\boxed{\textbf{(C) } 4}</math>.<br />
<br />
==Similar Problem==<br />
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21<br />
<br />
==Video Solution (Simple & Quick)==<br />
https://youtu.be/bRrchiDCrKE<br />
<br />
~ Education, the Study of Everything<br />
<br />
== Video Solution (Using Pythagorean Theorem, 3D Geometry - Tetrahedron) ==<br />
https://youtu.be/i4yUaXVUWKE<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution by TheBeautyofMath==<br />
https://youtu.be/t-EEP2V4nAE?t=813<br />
<br />
~IceMatrix<br />
<br />
==See also==<br />
{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_7&diff=1417442008 AMC 12B Problems/Problem 72021-01-08T22:37:02Z<p>Bobcats: </p>
<hr />
<div>==Problem 7==<br />
For real numbers <math>a</math> and <math>b</math>, define <math>a\textdollar b = (a - b)^2</math>. What is <math>(x - y)^2\textdollar(y - x)^2</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy</math><br />
<br />
==Solution 1 ==<br />
<br />
<math>\left[ (x-y)^2 - (y-x)^2 \right]^2</math><br />
<br />
<math>\left[ (x-y)^2 - (x-y)^2 \right]^2</math><br />
<br />
<math>[0]^2</math><br />
<br />
<math>0 \Rightarrow \textbf{(A)}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_3&diff=1412732007 AMC 12B Problems/Problem 32021-01-01T21:05:31Z<p>Bobcats: /* Solution */</p>
<hr />
<div>==Problem==<br />
The point <math>O</math> is the center of the circle circumscribed about triangle <math>ABC</math>, with <math>\angle BOC = 120^{\circ}</math> and <math>\angle AOB = 140^{\circ}</math>, as shown. What is the degree measure of <math>\angle ABC</math>?<br />
<br />
[[Image:2007_12B_AMC-3.png]]<br />
<br />
<math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60</math><br />
==Solution==<br />
Since triangles <math>ABO</math> and <math>BOC</math> are isosceles, <math>\angle ABO=20^o</math> and <math>\angle OBC=30^o</math>. Therefore, <math>\angle ABC=50^o</math>, or <math>\mathrm{(D)}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2007|ab=B|num-b=2|num-a=4}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_3&diff=1412722007 AMC 12B Problems/Problem 32021-01-01T21:05:00Z<p>Bobcats: /* Solution */</p>
<hr />
<div>==Problem==<br />
The point <math>O</math> is the center of the circle circumscribed about triangle <math>ABC</math>, with <math>\angle BOC = 120^{\circ}</math> and <math>\angle AOB = 140^{\circ}</math>, as shown. What is the degree measure of <math>\angle ABC</math>?<br />
<br />
[[Image:2007_12B_AMC-3.png]]<br />
<br />
<math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60</math><br />
==Solution==<br />
Since triangles <math>ABO</math> and <math>BOC</math> are isosceles, <math>\angle ABO=20^o</math> and <math>\angle OBC=30^o</math>. Therefore, <math>\angle ABC=50^o</math>, or <math>\mathbf{(D)}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2007|ab=B|num-b=2|num-a=4}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems&diff=1185612005 AIME II Problems2020-02-29T01:12:26Z<p>Bobcats: /* Problem 6 */</p>
<hr />
<div>{{AIME Problems|year=2005|n=II}}<br />
<br />
== Problem 1 ==<br />
A game uses a deck of <math> n </math> different cards, where <math> n </math> is an integer and <math> n \geq 6. </math> The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math> n. </math><br />
<br />
[[2005 AIME II Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is <math> \frac mn, </math> where <math> m </math> and <math> n </math> are relatively prime integers, find <math> m+n. </math><br />
<br />
[[2005 AIME II Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is <math> \frac mn </math> where <math> m </math> and <math> n </math> are relatively prime integers. Find <math> m+n. </math><br />
<br />
[[2005 AIME II Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Find the number of positive integers that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math><br />
<br />
[[2005 AIME II Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Determine the number of ordered pairs <math> (a,b) </math> of integers such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math><br />
<br />
[[2005 AIME II Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
The cards in a stack of <math> 2n </math> cards are numbered consecutively from 1 through <math> 2n </math> from top to bottom. The top <math> n </math> cards are removed, kept in order, and form pile <math> A. </math> The remaining cards form pile <math> B. </math> The cards are then restacked by taking cards alternately from the tops of pile <math> B </math> and <math> A, </math> respectively. In this process, card number <math> (n+1) </math> becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles <math> A </math> and <math> B </math> are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. Find the number of cards in the magical stack in which card number 131 retains its original position.<br />
<br />
[[2005 AIME II Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math> (x+1)^{48}. </math><br />
<br />
[[2005 AIME II Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Circles <math> C_1 </math> and <math> C_2 </math> are externally tangent, and they are both internally tangent to circle <math> C_3. </math> The radii of <math> C_1 </math> and <math> C_2 </math> are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of <math> C_3 </math> is also a common external tangent of <math> C_1 </math> and <math> C_2. </math> Given that the length of the chord is <math> \frac{m\sqrt{n}}p </math> where <math> m,n, </math> and <math> p </math> are positive integers, <math> m </math> and <math> p </math> are relatively prime, and <math> n </math> is not divisible by the square of any prime, find <math> m+n+p. </math><br />
<br />
[[2005 AIME II Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
For how many positive integers <math> n </math> less than or equal to 1000 is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>?<br />
<br />
[[2005 AIME II Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Given that <math> O </math> is a regular octahedron, that <math> C </math> is the cube whose vertices are the centers of the faces of <math> O, </math> and that the ratio of the volume of <math> O </math> to that of <math> C </math> is <math> \frac mn, </math> where <math> m </math> and <math> n </math> are relatively prime integers, find <math> m+n. </math><br />
<br />
[[2005 AIME II Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Let <math> m </math> be a positive integer, and let <math> a_0, a_1,\ldots,a_m </math> be a sequence of reals such that <math> a_0 = 37, a_1 = 72, a_m = 0, </math> and <math> a_{k+1} = a_{k-1} - \frac 3{a_k} </math> for <math> k = 1,2,\ldots, m-1. </math> Find <math> m. </math><br />
<br />
[[2005 AIME II Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Square <math> ABCD </math> has center <math> O, AB=900, E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are positive integers and <math> r </math> is not divisible by the square of any prime, find <math> p+q+r. </math><br />
<br />
[[2005 AIME II Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Let <math> P(x) </math> be a polynomial with integer coefficients that satisfies <math> P(17)=10 </math> and <math> P(24)=17. </math> Given that <math> P(n)=n+3 </math> has two distinct integer solutions <math> n_1 </math> and <math> n_2, </math> find the product <math> n_1\cdot n_2. </math><br />
<br />
[[2005 AIME II Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
In triangle <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD. </math> Given that <math> BE=\frac pq </math> where <math> p </math> and <math> q </math> are relatively prime positive integers, find <math> q. </math><br />
<br />
[[2005 AIME II Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positive value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math> <br />
<br />
[[2005 AIME II Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
<br />
{{AIME box|year = 2005|n=II|before=[[2005 AIME I Problems]]|after=[[2006 AIME I Problems]]}}<br />
<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=51 2005 AIME II Math Jam Transcript]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_20&diff=1135882003 AMC 10B Problems/Problem 202019-12-28T16:51:32Z<p>Bobcats: /* Solution 4 */</p>
<hr />
<div>{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}<br />
<br />
==Problem==<br />
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.<br />
<center><asy><br />
unitsize(8mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=0;<br />
<br />
pair A=(0,0), B=(5,0), C=(5,3), D=(0,3);<br />
pair F=(1,3), G=(3,3);<br />
pair E=(5/3,5);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E,F,G}; dot(ps);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",E,N);<br />
label("$F$",F,SE);<br />
label("$G$",G,SW);<br />
label("$1$",midpoint(D--F),N);<br />
label("$2$",midpoint(G--C),N);<br />
label("$5$",midpoint(A--B),S);<br />
label("$3$",midpoint(A--D),W);<br />
</asy></center><br />
<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math><br />
<br />
==Solution 1==<br />
<br />
<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math><br />
<br />
<cmath>\frac{2}{5} = \frac{h-3}{h}</cmath><br />
<cmath>2h = 5h-15</cmath><br />
<cmath>3h = 15</cmath><br />
<cmath>h = 5</cmath><br />
<br />
The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>.<br />
<br />
==Solution 2==<br />
<br />
We can look at this diagram as if it were a coordinate plane with point <math>A</math> being <math>(0,0)</math>. This means that the equation of the line <math>AE</math> is <math>y=3x</math> and the equation of the line <math>EB</math> is <math>y=\frac{-3}{2}x+\frac{15}{2}</math>. From this we can set of the follow equation to find the <math>x</math> coordinate of point <math>E</math>:<br />
<br />
<cmath>3x=\frac{-3}{2}x+\frac{15}{2}</cmath><br />
<cmath>6x=-3x+15</cmath><br />
<cmath>9x=15</cmath><br />
<cmath>x=\frac{5}{3}</cmath><br />
<br />
We can plug this into one of our original equations to find that the <math>y</math> coordinate is <math>5</math>, meaning the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math><br />
<br />
==Solution 3==<br />
At points <math>A</math> and <math>B</math>, segment <math>AE</math> is 5 units from segment <math>BE</math>. At points <math>F</math> and <math>G</math>, the segments are 2 units from each other. This means that collectively, the two lines closed the distance between them by 3 units over a height of 3 units. Therefore, to close the next two units of distance, they will have to travel a height of 2 units.<br />
<br />
Then calculate the area of trapezoid <math>FGBA</math> and triangle <math>EGF</math> separately and add them. The area of the trapezoid is <math>\frac {2+5}{2}\cdot 3 = \frac {21}{2}</math> and the area of the triangle is <math>\frac{1}{2}\cdot 2 \cdot 2 = 2</math>. <math>\frac{21}{2}+2=\boxed{\textbf{(D)}\ \frac{25}{2}}</math><br />
<br />
==Solution 4==<br />
Since <math>\Delta{ABE}\sim{\Delta{FGE}}</math> then <math>[AFGB]\sim{[FXYG]}</math>, where <math>X</math> and <math>Y</math> are ponts on <math>EF</math> and <math>EG</math> respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of <math>\frac{FG}{AB}=\frac{2}{5}</math>, or something like this<br />
<cmath>[AEB]=[AFGB]+[FXYZ]+...</cmath><cmath>[AEB]=[AFGB]+\frac{2}{5}[AFGB]+(\frac{2}{5})^2[AFGB]+...</cmath>we have to find the ratio of the areas when the sides have shrunk by length <math>\frac{2}{5}l</math><br />
<asy><br />
unitsize(0.6 cm);<br />
<br />
pair A, B, C, D, E, F, G;<br />
<br />
A = (0,0);<br />
B = (5,0);<br />
C = (5,3);<br />
D = (0,3);<br />
F = (1,3);<br />
G = (3,3);<br />
E = extension(A,F,B,G);<br />
<br />
draw(A--B--C--D--cycle);<br />
draw(A--E--B);<br />
<br />
label("$A$", A, SW);<br />
label("$B$", B, SE);<br />
label("$C$", C, NE);<br />
label("$D$", D, NW);<br />
label("$E$", E, N);<br />
label("$F$", F, SE);<br />
label("$G$", G, SW);<br />
label("$2/5$", (D + F)/2, N);<br />
label("$4/5$", (G + C)/2, N);<br />
label("$6/5$", (B + C)/2, dir(0));<br />
label("$6/5$", (A + D)/2, W);<br />
label("$2$", (A + B)/2, S);<br />
</asy><br />
Let <math>[AFGB]'</math> be the area of the shape whose length is <math>\frac{2}{5}l</math><br />
<cmath>[AFGB]'=[ADCB]-[ADF]-[BCG]</cmath><cmath>[AFGB]'=12/5-6/25-12/25</cmath><cmath>[AFGB]'=42/25</cmath>Now comparing the ratios of <math>[AFGB]'</math> to <math>[AFGB]</math> we get<br />
<cmath>\frac{[AFGB]'}{[AFGB]}=\frac{42}{25}/\frac{21}{2}\implies \frac{[AFGB]'}{[AFGB]}=\frac{4}{25}</cmath>By applying an infinite summation<br />
<cmath>[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{(\frac{4}{25})^n}</cmath><cmath>S=\frac{a_1}{1-r}</cmath><cmath>\boxed{[AEB]=\frac{25}{2}}</cmath><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}<br />
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_24&diff=1134832001 AMC 12 Problems/Problem 242019-12-26T23:03:15Z<p>Bobcats: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that <math>2\cdot BD=CD</math> and <math>\angle DAB=15^\circ</math>. Find <math>\angle ACB</math>.<br />
<br />
<math><br />
\text{(A) }54^\circ<br />
\qquad<br />
\text{(B) }60^\circ<br />
\qquad<br />
\text{(C) }72^\circ<br />
\qquad<br />
\text{(D) }75^\circ<br />
\qquad<br />
\text{(E) }90^\circ<br />
</math><br />
<br />
== Solution 1 ==<br />
<br />
<asy><br />
unitsize(2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B);<br />
pair ortho=rotate(-50)*(D-A);<br />
pair E=extension(C, ortho, A, D);<br />
draw(A--B);<br />
draw(B--C);<br />
draw(A--C);<br />
draw(C--E);<br />
draw(E--D);<br />
draw(A--D);<br />
<br />
<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$D$",D,NE);<br />
label("$E$",E,S);<br />
</asy><br />
<br />
We start with the observation that <math>\angle ADB = 180^\circ - 15^\circ - 45^\circ = 120^\circ</math>, and <math>\angle ADC = 15^\circ + 45^\circ = 60^\circ</math>.<br />
<br />
We can draw the height <math>CE</math> from <math>C</math> onto <math>AD</math>. In the triangle <math>CED</math>, we have <math>\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12</math>. Hence <math>ED = CD/2</math>.<br />
<br />
By the definition of <math>D</math>, we also have <math>BD=CD/2</math>, therefore <math>BD=DE</math>. This means that the triangle <math>BDE</math> is isosceles, and as <math>\angle BDE=120^\circ</math>, we must have <math>\angle BED = \angle EBD = 30^\circ</math>. <br />
<br />
Then we compute <math>\angle ABE = 45^\circ - 30^\circ = 15^\circ</math>, thus <math>\angle ABE = \angle BAE</math> and the triangle <math>ABE</math> is isosceles as well. Hence <math>AE=BE</math>.<br />
<br />
Now we can note that <math>\angle DCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ</math>, hence also the triangle <math>EBC</math> is isosceles and we have <math>BE=CE</math>.<br />
<br />
Combining the previous two observations we get that <math>AE=EC</math>, and as <math>\angle AEC=90^\circ</math>, this means that <math>\angle CAE = \angle ACE = 45^\circ</math>.<br />
<br />
Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
Draw a good diagram! Now, let's call <math>BD=t</math>, so <math>DC=2t</math>. Given the rather nice angles of <math>\angle ABD = 45^\circ</math> and <math>\angle ADC = 60^\circ</math> as you can see, let's do trig. Drop an altitude from <math>A</math> to <math>BC</math>; call this point <math>H</math>. We realize that there is no specific factor of <math>t</math> we can call this just yet, so let <math>AH=kt</math>. Notice that in <math>\triangle{ABH}</math> we get <math>BH=kt</math>. Using the 60-degree angle in <math>\triangle{ADH}</math>, we obtain <math>DH=\frac{\sqrt{3}}{3}kt</math>. The comparable ratio is that <math>BH-DH=t</math>. If we involve our <math>k</math>, we get:<br />
<br />
<math>kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t</math>. Eliminating <math>t</math> and removing radicals from the denominator, we get <math>k=\frac{3+\sqrt{3}}{2}</math>. From there, one can easily obtain <math>HC=3t-kt=\frac{3-\sqrt{3}}{2}t</math>. Now we finally have a desired ratio. Since <math>\tan\angle ACH = 2+\sqrt{3}</math> upon calculation, we know that <math>\angle ACH</math> can be simplified. Indeed, if you know that <math>\tan(75)=2+\sqrt{3}</math> or even take a minute or two to work out the sine and cosine using <math>\sin(x)^2+\cos(x)^2=1</math>, and perhaps the half- or double-angle formulas, you get <math>\boxed{75^\circ}</math>.<br />
<br />
== Solution 3==<br />
Without loss of generality, we can assume that <math>BD = 1</math> and <math>CD = 2</math>. As above, we are able to find that <math>\angle ADC = 60^\circ</math> and <math>\angle ADB = 120^\circ</math>.<br />
<br />
Using Law of Sines on triangle <math>ADB</math>, we find that <math>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}</math>. Since we know that <math>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}</math>, <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, and <math>\sin 120^\circ = \frac{\sqrt{3}}{2}</math>, we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>.<br />
<br />
Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <math>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ)</math>. Simplifying the right side, we get <math>AC^2 = 6</math>, so <math>AC = \sqrt{6}</math>.<br />
<br />
Now, we apply Law of Sines to triangle <math>ABC</math> to see that <math>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}</math>. After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <math>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}</math>.<br />
<br />
Dividing the right side by <math>\sqrt{3}</math>, we see that <math>\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4}</math>, so <math>\angle ACB</math> is either <math>75^\circ</math> or <math>105^\circ</math>. Since <math>105^\circ</math> is not a choice, we know <math>\angle ACB = \boxed{75^\circ}</math>.<br />
<br />
Note that we can also confirm that <math>\angle ACB \neq 105^\circ</math> by computing <math>\angle CAB</math> with Law of Sines.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2001|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_24&diff=1134822001 AMC 12 Problems/Problem 242019-12-26T23:00:52Z<p>Bobcats: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that <math>2\cdot BD=CD</math> and <math>\angle DAB=15^\circ</math>. Find <math>\angle ACB</math>.<br />
<br />
<math><br />
\text{(A) }54^\circ<br />
\qquad<br />
\text{(B) }60^\circ<br />
\qquad<br />
\text{(C) }72^\circ<br />
\qquad<br />
\text{(D) }75^\circ<br />
\qquad<br />
\text{(E) }90^\circ<br />
</math><br />
<br />
== Solution 1 ==<br />
<br />
<asy><br />
unitsize(2cm);<br />
defaultpen(0.8);<br />
pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B);<br />
pair ortho=rotate(-65)*(D-A);<br />
pair E=extension(C, ortho, A, D);<br />
draw(A--B);<br />
draw(B--C);<br />
draw(A--C);<br />
draw(C--E);<br />
draw(E--D);<br />
draw(A--D);<br />
<br />
\<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$D$",D,NE);<br />
label("$E$",E,S);<br />
</asy><br />
<br />
We start with the observation that <math>\angle ADB = 180^\circ - 15^\circ - 45^\circ = 120^\circ</math>, and <math>\angle ADC = 15^\circ + 45^\circ = 60^\circ</math>.<br />
<br />
We can draw the height <math>CE</math> from <math>C</math> onto <math>AD</math>. In the triangle <math>CED</math>, we have <math>\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12</math>. Hence <math>ED = CD/2</math>.<br />
<br />
By the definition of <math>D</math>, we also have <math>BD=CD/2</math>, therefore <math>BD=DE</math>. This means that the triangle <math>BDE</math> is isosceles, and as <math>\angle BDE=120^\circ</math>, we must have <math>\angle BED = \angle EBD = 30^\circ</math>. <br />
<br />
Then we compute <math>\angle ABE = 45^\circ - 30^\circ = 15^\circ</math>, thus <math>\angle ABE = \angle BAE</math> and the triangle <math>ABE</math> is isosceles as well. Hence <math>AE=BE</math>.<br />
<br />
Now we can note that <math>\angle DCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ</math>, hence also the triangle <math>EBC</math> is isosceles and we have <math>BE=CE</math>.<br />
<br />
Combining the previous two observations we get that <math>AE=EC</math>, and as <math>\angle AEC=90^\circ</math>, this means that <math>\angle CAE = \angle ACE = 45^\circ</math>.<br />
<br />
Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
Draw a good diagram! Now, let's call <math>BD=t</math>, so <math>DC=2t</math>. Given the rather nice angles of <math>\angle ABD = 45^\circ</math> and <math>\angle ADC = 60^\circ</math> as you can see, let's do trig. Drop an altitude from <math>A</math> to <math>BC</math>; call this point <math>H</math>. We realize that there is no specific factor of <math>t</math> we can call this just yet, so let <math>AH=kt</math>. Notice that in <math>\triangle{ABH}</math> we get <math>BH=kt</math>. Using the 60-degree angle in <math>\triangle{ADH}</math>, we obtain <math>DH=\frac{\sqrt{3}}{3}kt</math>. The comparable ratio is that <math>BH-DH=t</math>. If we involve our <math>k</math>, we get:<br />
<br />
<math>kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t</math>. Eliminating <math>t</math> and removing radicals from the denominator, we get <math>k=\frac{3+\sqrt{3}}{2}</math>. From there, one can easily obtain <math>HC=3t-kt=\frac{3-\sqrt{3}}{2}t</math>. Now we finally have a desired ratio. Since <math>\tan\angle ACH = 2+\sqrt{3}</math> upon calculation, we know that <math>\angle ACH</math> can be simplified. Indeed, if you know that <math>\tan(75)=2+\sqrt{3}</math> or even take a minute or two to work out the sine and cosine using <math>\sin(x)^2+\cos(x)^2=1</math>, and perhaps the half- or double-angle formulas, you get <math>\boxed{75^\circ}</math>.<br />
<br />
== Solution 3==<br />
Without loss of generality, we can assume that <math>BD = 1</math> and <math>CD = 2</math>. As above, we are able to find that <math>\angle ADC = 60^\circ</math> and <math>\angle ADB = 120^\circ</math>.<br />
<br />
Using Law of Sines on triangle <math>ADB</math>, we find that <math>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}</math>. Since we know that <math>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}</math>, <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, and <math>\sin 120^\circ = \frac{\sqrt{3}}{2}</math>, we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>.<br />
<br />
Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <math>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ)</math>. Simplifying the right side, we get <math>AC^2 = 6</math>, so <math>AC = \sqrt{6}</math>.<br />
<br />
Now, we apply Law of Sines to triangle <math>ABC</math> to see that <math>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}</math>. After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <math>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}</math>.<br />
<br />
Dividing the right side by <math>\sqrt{3}</math>, we see that <math>\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4}</math>, so <math>\angle ACB</math> is either <math>75^\circ</math> or <math>105^\circ</math>. Since <math>105^\circ</math> is not a choice, we know <math>\angle ACB = \boxed{75^\circ}</math>.<br />
<br />
Note that we can also confirm that <math>\angle ACB \neq 105^\circ</math> by computing <math>\angle CAB</math> with Law of Sines.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2001|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_17&diff=1134422000 AMC 12 Problems/Problem 172019-12-26T15:59:50Z<p>Bobcats: /* Solution 3 (with minimal trig) */</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 />
== 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>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_10&diff=1051962000 AIME II Problems/Problem 102019-04-09T01:15:44Z<p>Bobcats: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the [[Perfect square|square]] of the [[radius]] of the circle.<br />
<br />
== Solution ==<br />
Call the [[center]] of the circle <math>O</math>. By drawing the lines from <math>O</math> tangent to the sides and from <math>O</math> to the vertices of the quadrilateral, four pairs of congruent [[right triangle]]s are formed.<br />
<br />
Thus, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180</math>.<br />
<br />
Take the <math>\tan</math> of both sides and use the identity for <math>\tan(A+B)</math> to get <math>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0</math>.<br />
<br />
Use the identity for <math>\tan(A+B)</math> again to get <math>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0</math>.<br />
<br />
Solving gives <math>r^2=\boxed{647}</math>.<br />
== Solution 2==<br />
Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on (<math>a, b, c,</math> and <math>d</math> are the tangent lengths, not the side lengths).<br />
<cmath>A = \sqrt{(a+b+c+d)(abc+bcd+cda+dac)} = 105\sqrt{647}</cmath><br />
<math>r^2=\frac{A}{a+b+c+d} = \boxed{647}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=9|num-a=11}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_11&diff=1042221991 AIME Problems/Problem 112019-03-10T19:47:31Z<p>Bobcats: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Twelve congruent disks are placed on a [[circle]] <math>C^{}_{}</math> of [[radius]] 1 in such a way that the twelve disks cover <math>C^{}_{}</math>, no two of the disks overlap, and so that each of the twelve disks is [[tangent (geometry)|tangent]] to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be written in the from <math>\pi(a-b\sqrt{c})</math>, where <math>a,b,c^{}_{}</math> are positive integers and <math>c^{}_{}</math> is not divisible by the square of any prime. Find <math>a+b+c^{}_{}</math>.<br />
<br />
<asy><br />
real r=2-sqrt(3);<br />
draw(Circle(origin, 1));<br />
int i;<br />
for(i=0; i<12; i=i+1) {<br />
draw(Circle(dir(30*i), r));<br />
dot(dir(30*i));<br />
}<br />
draw(origin--(1,0)--dir(30)--cycle);<br />
label("1", (0.5,0), S);</asy><br />
<br />
_Diagram by 1-1 is 3_<br />
Diagram is incorrect since the centers of the smaller circles are not on the larger circle<br />
<br />
== Solution ==<br />
We wish to find the radius of one circle, so that we can find the total area.<br />
<br />
Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length <math>2r</math> that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius <math>1</math>.<br />
<br />
We thus know that the [[apothem]] of the [[dodecagon]] is equal to <math>1</math>. To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote <math>A, M,</math> and <math>O</math> respectively. Notice that <math>OM=1</math>, and that <math>\triangle OMA</math> is a right triangle with [[hypotenuse]] <math>OA</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles. The area of one circle is thus <math>\pi(2 - \sqrt {3})^{2} = \pi (7 - 4 \sqrt {3})</math>, so the area of all <math>12</math> circles is <math>\pi (84 - 48 \sqrt {3})</math>, giving an answer of <math>84 + 48 + 3 = \boxed{135}</math>.<br />
<br />
-----<br />
Note that it is very very useful to memorize the side lengths of a 15-75-90 triangle, as these triangles often appear on higher level math contests. The side lengths are <math>\sqrt {3} - 1</math>, <math>\sqrt {3} + 1</math>, and <math>2\sqrt {2}</math><br />
<!--Let <math>R_{}^{}</math> and <math>r_{}^{}</math> denote the radii of the large and small circles (<math>R_{}^{}>r_{}^{}</math>), respectively. Suppose that there are <math>n_{}^{}</math> circles of radius <math>r_{}^{}</math> centered on the circumference of the circle having radius <math>R_{}^{}</math>. Let <math>O_{}^{}</math>, <math>P_{}^{}</math>, and <math>Q_{}^{}</math> label the vertices of the triangle with <math>O_{}^{}</math> being at the center of the large circle, whereas <math>P_{}^{}</math> and <math>Q_{}^{}</math> are the tangential points of one of the <math>n_{}^{}</math> small circles with its two other neighbours, and <math>S_{}^{}</math> its center. The angle subtended by <math>P_{}^{}OQ</math> is <math>2\pi/n_{}^{}</math>. The segments <math>O_{}^{}P</math> and <math>S_{}^{}P</math> are [[perpendicular]]. Therefore, triangle <math>P_{}^{}OS</math> is rectangular and the angle subtended by <math>P_{}^{}OS</math> equals <math>\pi/n_{}^{}</math>. Hence, the radius <math>r_{}^{}=R\sin(\pi/n)</math>. The total area <math>A_{n}^{}</math> of the <math>n_{}^{}</math> circles is thus given by <br />
<br />
<cmath><br />
A_{n}^{}=n\pi r^{2}=n\pi R^{2}\sin^{2}\left(\frac{\pi}{n}\right)=\frac{1}{2}n\pi R^{2}\left[1-\cos\left(\frac{2\pi}{n}\right)\right]\, .<br />
</cmath><br />
<br />
In the present problem, <math>n_{}^{}=12</math> and <math>R_{}^{}=1</math>. It follows that <math>A_{12}^{}=6\pi\left[1-\cos(\pi/6)\right]=\pi\left(6-3\sqrt{3}\right)\equiv \pi\left(a-b\sqrt{c}\right)</math>.<br />
<br />
In summary, <math>a_{}^{}+b+c=\boxed{012}</math>.--><br />
<br />
== See also ==<br />
{{AIME box|year=1991|num-b=10|num-a=12}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_12&diff=1037541986 AIME Problems/Problem 122019-02-23T00:16:18Z<p>Bobcats: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let the sum of a set of numbers be the sum of its elements. Let <math>S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>S</math> have the same sum. What is the largest sum a set <math>S</math> with these properties can have? <br />
== Solution ==<br />
By using the greedy algorithm, we obtain <math>\boxed{061}</math>, with <math>S=\{ 15,14,13,11,8\}</math>. We must now prove that no such set has sum greater than 61. Suppose such a set <math>S</math> existed. Then <math>S</math> must have more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>. <br />
<br />
<math>S</math> can't have more than 5 elements. To see why, note that at least <math>\dbinom{6}{0} + \dbinom{6}{1} + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57</math> of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element and so on), and each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum.<br />
<br />
<br />
Thus, <math>S</math> would have to have 5 elements. <math>S</math> contains both 15 and 14 (otherwise its sum is at most <math>10+11+12+13+15=61</math>). It follows that <math>S</math> cannot contain both <math>a</math> and <math>a-1</math> for any <math>a\leq 13</math>, or the subsets <math>\{a,14\}</math> and <math>\{a-1,15\}</math> would have the same sum. So now <math>S</math> must contain 13 (otherwise its sum is at most <math>15+14+12+10+8=59</math>), and <math>S</math> cannot contain 12, or the subsets <math>\{12,15\}</math> and <math>\{13,14\}</math> would have the same sum.<br />
<br />
Now the only way <math>S</math> could have sum at least <math>62=15+14+13+11+9</math> would be if <math>S=\{ 15,14,13,11,9\}</math>. But the subsets <math>\{9,15\}</math> and <math>\{11,13\}</math> have the same sum, so this set does not work. Therefore no <math>S</math> with sum greater than 61 is possible and 61 is indeed the maximum.<br />
<br />
== See also ==<br />
{{AIME box|year=1986|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_24&diff=990392018 AMC 8 Problems/Problem 242018-11-25T00:12:28Z<p>Bobcats: /* Note */</p>
<hr />
<div>==Problem 24==<br />
In the cube <math>ABCDEFGH</math> with opposite vertices <math>C</math> and <math>E,</math> <math>J</math> and <math>I</math> are the midpoints of edges <math>\overline{FB}</math> and <math>\overline{HD},</math> respectively. Let <math>R</math> be the ratio of the area of the cross-section <math>EJCI</math> to the area of one of the faces of the cube. What is <math>R^2?</math><br />
<br />
<asy><br />
size(6cm);<br />
pair A,B,C,D,EE,F,G,H,I,J;<br />
C = (0,0);<br />
B = (-1,1);<br />
D = (2,0.5);<br />
A = B+D;<br />
G = (0,2);<br />
F = B+G;<br />
H = G+D;<br />
EE = G+B+D;<br />
I = (D+H)/2; J = (B+F)/2;<br />
filldraw(C--I--EE--J--cycle,lightgray,black);<br />
draw(C--D--H--EE--F--B--cycle); <br />
draw(G--F--G--C--G--H);<br />
draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed);<br />
dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J);<br />
label("$A$",A,E);<br />
label("$B$",B,W);<br />
label("$C$",C,S);<br />
label("$D$",D,E);<br />
label("$E$",EE,N);<br />
label("$F$",F,W);<br />
label("$G$",G,N);<br />
label("$H$",H,E);<br />
label("$I$",I,E);<br />
label("$J$",J,W);<br />
</asy><br />
<br />
<math>\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}</math><br />
<br />
==Solution==<br />
<br />
Note that <math>EJCI</math> is a rhombus.<br />
Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt 3s</math> and <math>JI=\sqrt 2s</math>. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math><br />
<br />
==Note==<br />
In the 2008 AMC 10A, Question 21 (https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_21) was nearly identical to this question, it's the same but for this question, you have to look for the square of the area, not the actual area.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2018|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_24&diff=990382018 AMC 8 Problems/Problem 242018-11-25T00:09:36Z<p>Bobcats: /* Note */</p>
<hr />
<div>==Problem 24==<br />
In the cube <math>ABCDEFGH</math> with opposite vertices <math>C</math> and <math>E,</math> <math>J</math> and <math>I</math> are the midpoints of edges <math>\overline{FB}</math> and <math>\overline{HD},</math> respectively. Let <math>R</math> be the ratio of the area of the cross-section <math>EJCI</math> to the area of one of the faces of the cube. What is <math>R^2?</math><br />
<br />
<asy><br />
size(6cm);<br />
pair A,B,C,D,EE,F,G,H,I,J;<br />
C = (0,0);<br />
B = (-1,1);<br />
D = (2,0.5);<br />
A = B+D;<br />
G = (0,2);<br />
F = B+G;<br />
H = G+D;<br />
EE = G+B+D;<br />
I = (D+H)/2; J = (B+F)/2;<br />
filldraw(C--I--EE--J--cycle,lightgray,black);<br />
draw(C--D--H--EE--F--B--cycle); <br />
draw(G--F--G--C--G--H);<br />
draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed);<br />
dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J);<br />
label("$A$",A,E);<br />
label("$B$",B,W);<br />
label("$C$",C,S);<br />
label("$D$",D,E);<br />
label("$E$",EE,N);<br />
label("$F$",F,W);<br />
label("$G$",G,N);<br />
label("$H$",H,E);<br />
label("$I$",I,E);<br />
label("$J$",J,W);<br />
</asy><br />
<br />
<math>\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}</math><br />
<br />
==Solution==<br />
<br />
Note that <math>EJCI</math> is a rhombus.<br />
Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt 3s</math> and <math>JI=\sqrt 2s</math>. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math><br />
<br />
==Note==<br />
In the 2008 AMC 10A, Question 21 was nearly identical to this question, it's the same but for this question, you have to look for the square of the area, not the actual area.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2018|num-b=23|num-a=25}}<br />
<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_7&diff=989282016 AIME II Problems/Problem 72018-11-22T16:17:03Z<p>Bobcats: /* Solution */</p>
<hr />
<div>==Problem==<br />
Squares <math>ABCD</math> and <math>EFGH</math> have a common center and <math>\overline{AB} || \overline{EF}</math>. The area of <math>ABCD</math> is 2016, and the area of <math>EFGH</math> is a smaller positive integer. Square <math>IJKL</math> is constructed so that each of its vertices lies on a side of <math>ABCD</math> and each vertex of <math>EFGH</math> lies on a side of <math>IJKL</math>. Find the difference between the largest and smallest positive integer values for the area of <math>IJKL</math>.<br />
<br />
==Solution==<br />
Letting <math>AI=a</math> and <math>IB=b</math>, we have <math>IJ^{2}=a^{2}+b^{2} \geq 1008</math> by Cauchy-Schwarz inequality. Also, since <math>EFGH||ABCD</math>, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and <math>2</math> adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since <math>2016=12^{2} \cdot 14</math>, we have the maximum area is <math>2016 \cdot \dfrac{11}{12} = 1848</math> (the areas of the squares from largest to smallest are <math>12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14</math> forming a geometric progression). <br />
<br />
<br />
The minimum area is <math>1008</math> (every square is half the area of the square whose sides its vertices touch), so the desired answer is <math>1848-1008=\boxed{840}</math>.<br />
<br />
Solution by Shaddoll (edited by ppiittaattoo)<br />
<br />
== See also ==<br />
{{AIME box|year=2016|n=II|num-b=6|num-a=8}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_22&diff=967452011 AMC 10A Problems/Problem 222018-08-04T00:23:40Z<p>Bobcats: /* Solution 2 */</p>
<hr />
<div>== Problem 22 ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math><br />
<br />
== Solution 1 ==<br />
<br />
Let vertex <math>A</math> be any vertex, then vertex <math>B</math> be one of the diagonal vertices to <math>A</math>, <math>C</math> be one of the diagonal vertices to <math>B</math>, and so on. We consider cases for this problem.<br />
<br />
In the case that <math>C</math> has the same color as <math>A</math>, <math>D</math> has a different color from <math>A</math> and so <math>E</math> has a different color from <math>A</math> and <math>D</math>. In this case, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices (any color but the color of <math>A</math>), <math>C</math> has <math>1</math> choice, <math>D</math> has <math>5</math> choices, and <math>E</math> has <math>4</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600</math> combinations.<br />
<br />
In the case that <math>C</math> has a different color from <math>A</math> and <math>D</math> has a different color from <math>A</math>, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices, <math>C</math> has <math>4</math> choices (since <math>A</math> and <math>B</math> necessarily have different colors), <math>D</math> has <math>4</math> choices, and <math>E</math> has <math>4</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920</math> combinations.<br />
<br />
In the case that <math>C</math> has a different color from <math>A</math> and <math>D</math> has the same color as <math>A</math>, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices, <math>C</math> has <math>4</math> choices, <math>D</math> has <math>1</math> choice, and <math>E</math> has <math>5</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600</math> combinations.<br />
<br />
Adding all those combinations up, we get <math>600+1920+600=\boxed{3120 \ \mathbf{(C)}}</math>.<br />
<br />
==Solution 2==<br />
<br />
First, notice that there can be <math>3</math> cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.<br />
<br />
Case <math>1</math>: There are <math>6!</math> ways of assigning each vertex a different color. <math>6! = 720</math><br />
<br />
Case <math>2</math>: There are <math>\frac {6!}{2!} * 5</math> ways. After picking four colors, we can rotate our pentagon in <math>5</math> ways to get different outcomes. <math>\frac {6!}{2} * 5 = 1800</math><br />
<br />
Case <math>3</math>: There are <math>\frac {\frac {6!}{3!} * 10}{2!}</math> ways of arranging the final case. We can pick <math>3</math> colors for our pentagon. There are <math>5</math> spots for the first pair of colors. Then, there are <math>2</math> possible ways we can put the final pair in the last <math>3</math> spaces. But because the two pairs are indistinguishable, we divide by <math>2!</math>. <math>\frac {\frac {6!}{3!} * 10}{2!} = 600</math><br />
<br />
Adding all the possibilities up, we get <math>720+1800+600=\boxed{3120 \ \mathbf{(C)}}</math><br />
<br />
~ZericHang<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2011|ab=A|num-b=21|num-a=23}}<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems/Problem_6&diff=964972007 AMC 10B Problems/Problem 62018-07-26T01:15:48Z<p>Bobcats: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The <math>2007 \text{ AMC }10</math> will be scored by awarding <math>6</math> points for each correct response, <math>0</math> points for each incorrect response, and <math>1.5</math> points for each problem left unanswered. After looking over the <math>25</math> problems, Sarah has decided to attempt the first <math>22</math> and leave only the last <math>3</math> unanswered. How many of the first <math>22</math> problems must she solve correctly in order to score at least <math>100</math> points?<br />
<br />
<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17</math><br />
<br />
==Solution==<br />
<br />
Sarah is leaving <math>3</math> questions unanswered, guaranteeing her <math>3 \times 1.5 = 4.5</math> points. She will either get <math>6</math> points or <math>0</math> points for the rest of the questions. Let <math>x</math> be the number of questions Sarah answers correctly.<br />
<cmath>\begin{align*}<br />
6x+4.5 &\ge 100\\<br />
6x &\ge 95.5\\<br />
x &\ge 15.92<br />
\end{align*}</cmath><br />
The number of questions she answers correctly has to be a whole number, so round up to get <math>\boxed{\textbf{(D) }16}</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2007|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_13&diff=964672006 AMC 10A Problems/Problem 132018-07-23T23:05:37Z<p>Bobcats: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A player pays <math>\textdollar 5</math> to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)<br />
<br />
<math>\mathrm{(A) \ } \textdollar12\qquad\mathrm{(B) \ } \textdollar30\qquad\mathrm{(C) \ } \textdollar50\qquad\mathrm{(D) \ } \textdollar60\qquad\mathrm{(E) \ } \textdollar 100\qquad</math><br />
<br />
== Solution ==<br />
<br />
The probability of rolling an even number on the first turn is <math>\frac{1}{2}</math> and the probability of rolling the same number on the next turn is <math>\frac{1}{6}</math>. The probability of winning is <math>\frac{1}{12}</math>. If the game is to be fair, the amount paid, <math>5</math> dollars, must be <math>\frac{1}{12}</math> the amount of prize money, so the answer is<br />
<math>\boxed{\text{(D) } 60}.</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_13&diff=964662006 AMC 10A Problems/Problem 132018-07-23T23:05:18Z<p>Bobcats: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A player pays <math>\textdollar 5</math> to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)<br />
<br />
<math>\mathrm{(A) \ } \textdollar12\qquad\mathrm{(B) \ } \textdollar30\qquad\mathrm{(C) \ } \textdollar50\qquad\mathrm{(D) \ } \textdollar60\qquad\mathrm{(E) \ } \textdollar 100\qquad</math><br />
<br />
== Solution ==<br />
<br />
The probability of rolling an even number on the first turn is <math>\frac{1}{2}</math> and the probability of rolling the same number on the next turn is <math>\frac{1}{6}</math>. The probability of winning is <math>\frac{1}{12}</math>. If the game is to be fair, the amount paid, <math>5</math> dollars, must be <math>\frac{1}{12}</math> the amount of prize money, so the answer is<br />
<math>\text{(D) } 60.</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_8&diff=964652006 AMC 10A Problems/Problem 82018-07-23T23:03:36Z<p>Bobcats: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
A [[parabola]] with equation <math>y=x^2+bx+c</math> passes through the points <math> (2,3) </math> and <math> (4,3) </math>. What is <math>c</math>?<br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>.<br />
<br />
Then we get a system of two equations:<br />
<br />
<math>3=4+2b+c</math><br />
<br />
<math>3=16+4b+c</math><br />
<br />
Subtracting the first equation from the second we have:<br />
<br />
<math>0=12+2b</math><br />
<br />
<math>b=-6</math><br />
<br />
Then using <math>b=-6</math> in the first equation:<br />
<br />
<math>0=1+-12+c</math><br />
<br />
<math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer.<br />
<br />
=== Solution 2 ===<br />
<br />
Alternatively, notice that since the equation is that of a monic parabola, the vertex is likely <math>(3,2)</math>. Thus, the form of the equation of the parabola is <math>y - 2 = (x - 3)^2</math>. Expanding this out, we find that <math>c = 11</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
The points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>.<br />
<br />
The <math>x</math>-coordinate of the vertex is also equal to <math>\frac{-b}{2a}</math>, so set this equal to <math>3</math> and solve for <math>b</math>, given that <math>a=1</math>:<br />
<br />
<math>x=\frac{-b}{2a}</math><br />
<br />
<math>3=\frac{-b}{2}</math><br />
<br />
<math>6=-b</math><br />
<br />
<math>b=-6</math><br />
<br />
Now the equation is of the form <math>y=x^2-6x+c</math>. Now plug in the point <math>(2,3)</math> and solve for <math>c</math>:<br />
<br />
<math>y=x^2-6x+c</math><br />
<br />
<math>3=2^2-6(2)+c</math><br />
<br />
<math>3=4-12+c</math><br />
<br />
<math>3=-8+c</math><br />
<br />
<math>\boxed{ \text{(E) }c=11}</math><br />
<br />
=== Solution 4 ===<br />
Substituting y into the two equations, we get:<br />
<br />
<math>3=x^2+bx+c</math><br />
<br />
Which can be written as:<br />
<br />
<math>x^2+bx+c-3=0</math><br />
<br />
4, 2, are the solutions to the quadratic. Thus:<br />
<br />
<math>c-3=4\times2</math><br />
<br />
<math>c-3=8</math><br />
<br />
<math>c=11</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=7|num-a=9}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_8&diff=964642006 AMC 10A Problems/Problem 82018-07-23T23:03:21Z<p>Bobcats: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
A [[parabola]] with equation <math>y=x^2+bx+c</math> passes through the points <math> (2,3) </math> and <math> (4,3) </math>. What is <math>c</math>?<br />
<br />
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11 </math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Substitute the points <math> (2,3) </math> and <math> (4,3) </math> into the given equation for <math> (x,y) </math>.<br />
<br />
Then we get a system of two equations:<br />
<br />
<math>3=4+2b+c</math><br />
<br />
<math>3=16+4b+c</math><br />
<br />
Subtracting the first equation from the second we have:<br />
<br />
<math>0=12+2b</math><br />
<br />
<math>b=-6</math><br />
<br />
Then using <math>b=-6</math> in the first equation:<br />
<br />
<math>0=1+-12+c</math><br />
<br />
<math>c=11 \Longrightarrow \mathrm{(E)}</math> is the answer.<br />
<br />
=== Solution 2 ===<br />
<br />
Alternatively, notice that since the equation is that of a monic parabola, the vertex is likely <math>(3,2)</math>. Thus, the form of the equation of the parabola is <math>y - 2 = (x - 3)^2</math>. Expanding this out, we find that <math>c = 11</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
The points given have the same <math>y</math>-value, so the vertex lies on the line <math>x=\frac{2+4}{2}=3</math>.<br />
<br />
The <math>x</math>-coordinate of the vertex is also equal to <math>\frac{-b}{2a}</math>, so set this equal to <math>3</math> and solve for <math>b</math>, given that <math>a=1</math>:<br />
<br />
<math>x=\frac{-b}{2a}</math><br />
<br />
<math>3=\frac{-b}{2}</math><br />
<br />
<math>6=-b</math><br />
<br />
<math>b=-6</math><br />
<br />
Now the equation is of the form <math>y=x^2-6x+c</math>. Now plug in the point <math>(2,3)</math> and solve for <math>c</math>:<br />
<br />
<math>y=x^2-6x+c</math><br />
<br />
<math>3=2^2-6(2)+c</math><br />
<br />
<math>3=4-12+c</math><br />
<br />
<math>3=-8+c</math><br />
<br />
<math>\boxed{ \text{(E)}c=11}</math><br />
<br />
=== Solution 4 ===<br />
Substituting y into the two equations, we get:<br />
<br />
<math>3=x^2+bx+c</math><br />
<br />
Which can be written as:<br />
<br />
<math>x^2+bx+c-3=0</math><br />
<br />
4, 2, are the solutions to the quadratic. Thus:<br />
<br />
<math>c-3=4\times2</math><br />
<br />
<math>c-3=8</math><br />
<br />
<math>c=11</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=7|num-a=9}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_4&diff=964632006 AMC 12A Problems/Problem 42018-07-23T22:58:41Z<p>Bobcats: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2008 AMC 10A #4]]}}<br />
== Problem ==<br />
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?<br />
<br />
<math>\mathrm{(A)}\ 17\qquad\mathrm{(B)}\ 19\qquad\mathrm{(C)}\ 21\qquad\mathrm{(D)}\ 22\qquad\mathrm{(E)}\ 23</math><br />
<br />
== Solution 1 ==<br />
From the [[greedy algorithm]], we have <math>9</math> in the hours section and <math>59</math> in the minutes section. <math>9+5+9=23\Rightarrow\mathrm{(E)}</math><br />
<br />
==Solution 2==<br />
<br />
With a matrix we can see<br />
<math><br />
\begin{bmatrix}<br />
1+2&9&6&3\\<br />
1+11&8&5&2\\<br />
1+0&7&4&2<br />
\end{bmatrix}<br />
</math><br />
The largest digit sum we can see is <math>9</math>.<br />
For the minutes digits, we can combine the largest <math>2</math> digits, which are <math>9,5=9+5=14</math> which we can then do <math>14+9=23</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{AMC10 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_4&diff=964622006 AMC 12A Problems/Problem 42018-07-23T22:58:11Z<p>Bobcats: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2008 AMC 10A #4]]}}<br />
== Problem ==<br />
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?<br />
<br />
<math>\mathrm{(A)}\ 17\qquad\mathrm{(B)}\ 19\qquad\mathrm{(C)}\ 21\qquad\mathrm{(D)}\ 22\qquad\mathrm{(E)}\ 23</math><br />
<br />
== Solution 1 ==<br />
From the [[greedy algorithm]], we have <math>9</math> in the hours section and <math>59</math> in the minutes section. <math>9+5+9=23\Rightarrow\mathrm{(E)}</math><br />
<br />
==Solution 2==<br />
<br />
With a matrix we can see<br />
<math><br />
\begin{bmatrix}<br />
1+2&9&6&3\\<br />
1+11&8&5&2\\<br />
1+0&7&4&2<br />
\end{bmatrix}<br />
</math><br />
The largest digit sum we can see is <math>9</math><br />
For the minutes digits, we can combine the largest <math>2</math> digits, which are <math>9,5=9+5=14</math> which we can then do <math>14+9=23</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{AMC10 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_4&diff=964612006 AMC 12A Problems/Problem 42018-07-23T22:55:54Z<p>Bobcats: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2008 AMC 10A #4]]}}<br />
== Problem ==<br />
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?<br />
<br />
<math>\mathrm{(A)}\ 17\qquad\mathrm{(B)}\ 19\qquad\mathrm{(C)}\ 21\qquad\mathrm{(D)}\ 22\qquad\mathrm{(E)}\ 23</math><br />
<br />
== Solution 1 ==<br />
From the [[greedy algorithm]], we have <math>9</math> in the hours section and <math>59</math> in the minutes section. <math>9+5+9=23\Rightarrow\mathrm{(E)}</math><br />
<br />
==Solution 2==<br />
<br />
With a matrix we can see<br />
<math><br />
\[\begin{bmatrix}<br />
1+2&9&6&3\\<br />
1+11&8&5&2\\<br />
1+0&7&4&2<br />
\end{bmatrix}\]<br />
</math><br />
The largest digit sum we can see is <math>9</math><br />
For the minutes digits, we can combine the largest <math>2</math> digits, which are <math>9,5=9+5=14</math> which we can then do <math>14+9=23</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{AMC10 box|year=2006|ab=A|num-b=3|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_9&diff=964602006 AMC 10B Problems/Problem 92018-07-23T22:29:22Z<p>Bobcats: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Francesca uses 100 grams of lemon juce, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade? <br />
<br />
<math> \mathrm{(A) \ } 129\qquad \mathrm{(B) \ } 137\qquad \mathrm{(C) \ } 174\qquad \mathrm{(D) \ } 233\qquad \mathrm{(E) \ } 411 </math><br />
<br />
== Solution ==<br />
The calorie to gram ratio of Francesca's lemonade is <math>\frac{25+386+0}{100+100+400}=\frac{411\textrm{ calories}}{600\textrm{ grams}}=\frac{137\textrm{ calories}}{200\textrm{ grams}}</math> <br />
<br />
So in <math>200\textrm{ grams}</math> of Francesca's lemonade there are <math>200\textrm{ grams}\cdot\frac{137\textrm{ calories}}{200\textrm{ grams}}=137\textrm{ calories} \Rightarrow B</math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2006|ab=B|num-b=8|num-a=10}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_25&diff=963822005 AMC 10B Problems/Problem 252018-07-21T00:31:02Z<p>Bobcats: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A subset <math>B</math> of the set of integers from <math>1</math> to <math>100</math>, inclusive, has the property that no two elements of <math>B</math> sum to <math>125</math>. What is the maximum possible number of elements in <math>B</math>?<br />
<br />
<math>\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68 </math><br />
<br />
==Solution 1==<br />
The question asks for the maximum possible number of elements. The integers from <math>1</math> to <math>24</math> can be included because you cannot make <math>125</math> with integers from <math>1</math> to <math>24</math> without the other number being greater than <math>100</math>. The integers from <math>25</math> to <math>100</math> are left. They can be paired so the sum is <math>125</math>: <math>25+100</math>, <math>26+99</math>, <math>27+98</math>, <math>\ldots</math>, <math>62+63</math>. That is <math>38</math> pairs, and at most one number from each pair can be included in the set. The total is <math>24 + 38 = \boxed{\mathrm{(C)}\ 62}</math>.<br />
Also, it is possible to see that since the numbers <math>1</math> to <math>24</math> are in the set there are only the numbers <math>25</math> to <math>100</math> to consider. As <math>62+63</math> gives <math>125</math>, the numbers <math>25</math> to <math>62</math> can be put in subset <math>B</math> without having two numbers add up to <math>125</math>. In this way, subset <math>B</math> will have the numbers <math>1</math> to <math>62</math>, and so <math>\boxed{\mathrm{(C)}\ 62}</math>.<br />
<br />
==Solution 2==<br />
"Cut" <math>125</math> into half. The maximum integer value in the smaller half is <math>62</math>. Thus the answer is <math>\boxed{\mathrm{(C)}\ 62}</math>.<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_14&diff=963812005 AMC 10B Problems/Problem 142018-07-21T00:20:23Z<p>Bobcats: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Equilateral <math> \triangle ABC</math> has side length <math>2</math>, <math> M</math> is the midpoint of <math> \overline{AC}</math>, and <math> C</math> is the midpoint of <math> \overline{BD}</math>. What is the area of <math> \triangle CDM</math>?<br />
<asy>defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
pair B = (0,0);<br />
pair A = 2*dir(60);<br />
pair C = (2,0);<br />
pair D = (4,0);<br />
pair M = midpoint(A--C);<br />
<br />
label("$A$",A,NW);label("$B$",B,SW);label("$C$",C, SE);label("$M$",M,NE);label("$D$",D,SE);<br />
<br />
draw(A--B--C--cycle);<br />
draw(C--D--M--cycle);</asy><math> \textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad \textrm{(B)}\ \frac {3}{4}\qquad \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad \textrm{(D)}\ 1\qquad \textrm{(E)}\ \sqrt {2}</math><br />
== Solution ==<br />
===Solution 1===<br />
The area of a triangle can be given by <math>\frac12 ab \sin C</math>. <math>MC=1</math> because it is the midpoint of a side, and <math>CD=2</math> because it is the same length as <math>BC</math>. Each angle of an equilateral triangle is <math>60^\circ</math> so <math>\angle MCD = 120^\circ</math>. The area is <math>\frac12 (1)(2) \sin<br />
120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}</math>.<br />
<br />
===Solution 2===<br />
In order to calculate the area of <math>\triangle CDM</math>, we can use the formula <math>A=\dfrac{1}{2}bh</math>, where <math>\overline{CD}</math> is the base. We already know that <math>\overline{CD}=2</math>, so the formula now becomes <math>A=h</math>. We can drop verticals down from <math>A</math> and <math>M</math> to points <math>E</math> and <math>F</math>, respectively. We can see that <math>\triangle AEC \sim \triangle MFC</math>. Now, we establish the relationship that <math>\dfrac{AE}{MF}=\dfrac{AC}{MC}</math>. We are given that <math>\overline{AC}=2</math>, and <math>M</math> is the midpoint of <math>\overline{AC}</math>, so <math>\overline{MC}=1</math>. Because <math>\triangle AEB</math> is a <math>30-60-90</math> triangle and the ratio of the sides opposite the angles are <math>1-\sqrt{3}-2</math> <math>\overline{AE}</math> is <math>\sqrt{3}</math>. Plugging those numbers in, we have <math>\dfrac{\sqrt{3}}{MF}=\dfrac{2}{1}</math>. Cross-multiplying, we see that <math>2\times\overline{MF}=\sqrt{3}\times1\implies \overline{MF}=\dfrac{\sqrt{3}}{2}</math> Since <math>\overline{MF}</math> is the height <math>\triangle CDM</math>, the area is <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}</math>.<br />
<br />
===Solution 3===<br />
Draw a line from <math>M</math> to the midpoint of <math>\overline{BC}</math>. Call the midpoint of <math>\overline{BC}</math> <math>P</math>. This is an equilateral triangle, since the two segments <math>\overline{PC}</math> and <math>\overline{MC}</math> are identical, and <math>\angle C</math> is 60°. Using the Pythagorean Theorem, point <math>M</math> to <math>\overline{BC}</math> is <math>\dfrac{\sqrt{3}}{2}</math>. Also, the length of <math>\overline{CD}</math> is 2, since <math>C</math> is the midpoint of <math>\overline{BD}</math>. So, our final equation is <math>\dfrac{\sqrt{3}}{2}\times2\over2</math>, which just leaves us with <math>\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}</math>.<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2005|ab=B|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_8&diff=963612005 AMC 10A Problems/Problem 82018-07-20T01:34:04Z<p>Bobcats: /* Problem */</p>
<hr />
<div>== Problem ==<br />
In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE=1</math>. What is the area of the inner square <math>EFGH</math>?<br />
<br />
[[File:AMC102005Aq.png]]<br />
<br />
<math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 </math><br />
<br />
==Solution==<br />
We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So:<br />
<br />
<cmath>1^2 + (HE+1)^2=\sqrt{50}^2</cmath><br />
<br />
<cmath>1 + (HE+1)^2=50</cmath><br />
<br />
<cmath>(HE+1)^2=49</cmath><br />
<br />
<cmath>HE+1=7</cmath><br />
<br />
<cmath>HE=6</cmath> <br />
So, the area of the square is <math>6^2=\boxed{36} \Rightarrow \mathrm{(C)}</math>.<br />
<br />
==See Also==<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
<br />
*[[2005 AMC 10A Problems]]<br />
<br />
*[[2005 AMC 10A Problems/Problem 7|Previous Problem]]<br />
<br />
*[[2005 AMC 10A Problems/Problem 9|Next Problem]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=1995_AJHSME_Problems/Problem_18&diff=882261995 AJHSME Problems/Problem 182017-11-12T16:53:44Z<p>Bobcats: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The area of each of the four congruent L-shaped regions of this 100-inch by 100-inch square is 3/16 of the total area. How many inches long is the side of the center square?<br />
<br />
<asy><br />
draw((2,2)--(2,-2)--(-2,-2)--(-2,2)--cycle);<br />
draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle);<br />
draw((0,1)--(0,2));<br />
draw((1,0)--(2,0));<br />
draw((0,-1)--(0,-2));<br />
draw((-1,0)--(-2,0));<br />
</asy><br />
<br />
<math>\text{(A)}\ 25 \qquad \text{(B)}\ 44 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 62 \qquad \text{(E)}\ 75</math><br />
<br />
==Solution==<br />
<br />
The area taken up by the L's is <math>4*\frac{3}{16}=\frac{3}{4}</math> of the area of the whole square. What remains has <math>\frac{1}{4}</math> of the area of the larger square. <math>\frac{100*100}{4}=\frac{100}{2}*\frac{100}{2}=50*50</math> is the area of the smaller square, so its side length is 50. <math>\text{(C)}</math><br />
<br />
==See Also==<br />
{{AJHSME box|year=1995|num-b=17|num-a=19}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_15&diff=881541986 AJHSME Problems/Problem 152017-11-09T19:32:37Z<p>Bobcats: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Sale prices at the Ajax Outlet Store are <math>50\% </math> below original prices. On Saturdays an additional discount of <math>20\% </math> off the sale price is given. What is the Saturday price of a coat whose original price is <math>\textdollar 180</math>?<br />
<br />
<math>\text{(A)}</math> <math>\textdollar 54</math><br />
<br />
<math>\text{(B)}</math> <math>\textdollar 72</math><br />
<br />
<math>\text{(C)}</math> <math>\textdollar 90</math><br />
<br />
<math>\text{(D)}</math> <math>\textdollar 108</math><br />
<br />
<math>\text{(D)}</math> <math>\textdollar 110</math><br />
<br />
==Solution==<br />
<br />
First we need to do the first discount, which, at Ajax Outlet Store, would be there any day of the week.<br />
<cmath>\begin{align*}<br />
180 \times 50\% &= 180 \times \frac{1}{2} \\<br />
&= 90 \\<br />
\end{align*}</cmath><br />
<br />
If we discount <math>20\% </math>, then <math>80 \% </math> will be left, so after the second discount, we get<br />
<cmath>\begin{align*}<br />
90 \times 80\% &= 90 \times \frac{8}{10} \\<br />
&= 9 \times 8 \\<br />
&= 72 \\<br />
\end{align*}</cmath><br />
<br />
<math>\boxed{\text{B}}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=14|num-a=16}}<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bobcatshttps://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_15&diff=881531986 AJHSME Problems/Problem 152017-11-09T19:30:19Z<p>Bobcats: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Sale prices at the Ajax Outlet Store are <math>50\% </math> below original prices. On Saturdays an additional discount of <math>20\% </math> off the sale price is given. What is the Saturday price of a coat whose original price is <math>\dollar 180</math>?<br />
<br />
<math>\text{(A)}</math> <dollar/><math>54</math><br />
<br />
<math>\text{(B)}</math> <dollar/><math>72</math><br />
<br />
<math>\text{(C)}</math> <dollar/><math>90</math><br />
<br />
<math>\text{(D)}</math> <dollar/><math>108</math><br />
<br />
<math>\text{(D)}</math> <dollar/><math>110</math><br />
<br />
==Solution==<br />
<br />
First we need to do the first discount, which, at Ajax Outlet Store, would be there any day of the week.<br />
<cmath>\begin{align*}<br />
180 \times 50\% &= 180 \times \frac{1}{2} \\<br />
&= 90 \\<br />
\end{align*}</cmath><br />
<br />
If we discount <math>20\% </math>, then <math>80 \% </math> will be left, so after the second discount, we get<br />
<cmath>\begin{align*}<br />
90 \times 80\% &= 90 \times \frac{8}{10} \\<br />
&= 9 \times 8 \\<br />
&= 72 \\<br />
\end{align*}</cmath><br />
<br />
<math>\boxed{\text{B}}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=14|num-a=16}}<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bobcats