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Difference between revisions of "2024 AMC 12A Problems/Problem 19"

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==Solution 2 (Law of Cosines + Law of Sines)==
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Draw diagonals <math>AC</math> and <math>BD</math>. By Law of Cosines,
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\begin{align*}
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AC^2&=3^2 + 5^2 - 2(3)(5)\cos \left(\frac{2\pi}{3} \right) \\
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&= 9+25 +15 \\
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&=49.
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\end{align*}
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Since <math>AC</math> is positive, taking the square root gives <math>AC=7.</math> Let <math>\angle BDC=\angle CBD=x</math>. Since <math>\triangle BCD</math> is isosceles, we have <math>\angle BCD=180-2x</math>. Notice we can eventually solve <math>BD</math> using the Extended Law of Sines: <cmath>\frac{BD}{\sin(180-2x)}=2r,</cmath> where <math>r</math> is the radius of the circumcircle <math>ABCD</math>. Since <math>\sin(180-2x)=\sin(2x)=2\sin(x)\cos(x)</math>, we simply our equation: <cmath>\frac{BD}{2\sin(x)\cos(x)}=2r.</cmath>
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Now we just have to find <math>\sin(x), \cos(x),</math> and <math>2r</math>. Since <math>ABCD</math> is cyclic, we have <math>\angle CBD = \angle CAD = x</math>. By Law of Cosines on <math>\triangle ADC</math>, we have  <cmath>3^2=7^2 + 5^2 - 70\cos(x).</cmath> Thus, <math>\cos(x)=\frac{13}{14}.</math> Similarly, by Law of Sines on <math>\triangle ACD</math>, we have <cmath>\frac{7}{\sin\left(\frac{2\pi}{3} \right)}=2r.</cmath> Hence, <math>2r=\frac{14\sqrt3}{3}</math>. Now, using Law of Sines on <math>\triangle BCD</math>, we have <math>\frac{3}{\sin(x)}=2r= \frac{14\sqrt3}{3},</math> so <math>\sin(x)=\frac{3\sqrt3}{14}.</math> Therefore, <cmath>\frac{BD}{2\left(\frac{3\sqrt3}{14}\right) \left(\frac{13}{14} \right)}=\frac{14\sqrt3}{3}.</cmath> Solving, <math>BD = \frac{39}{7},</math> so the answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
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~evanhliu2009
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2024|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:42, 8 November 2024

Problem

Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$. What is the length of the shorter diagonal of $ABCD$?

$\textbf{(A) }\frac{31}7 \qquad \textbf{(B) }\frac{33}7 \qquad \textbf{(C) }5 \qquad \textbf{(D) }\frac{39}7 \qquad \textbf{(E) }\frac{41}7 \qquad$

Solution 1

First, $\angle CBA=60 ^\circ$ by properties of cyclic quadrilaterals. Let $AC=u$. We apply the Law of Cosines on $\triangle ACD$: \[u^2=3^2+5^2-2(3)(5)\cos120^\circ\] \[u=7\]


Let $AB=v$. Apply the Law of Cosines on $\triangle ABC$: \[7^2=3^2+v^2-2(3)(v)\cos60^\circ\] \[v=\frac{3\pm13}{2}\] \[v=8\]


By Ptolemy’s Theorem, \[AB \cdot CD+AD \cdot BC=AC \cdot BD\] \[8 \cdot 3+5 \cdot 3=7BD\] \[BD=\frac{39}{7}\] Since $\frac{39}{7}<5$, The answer is $\boxed{\textbf{(D) }\frac{39}{7}}$.

~lptoggled, formatting by eevee9406


Solution 2 (Law of Cosines + Law of Sines)

Draw diagonals $AC$ and $BD$. By Law of Cosines, \begin{align*} AC^2&=3^2 + 5^2 - 2(3)(5)\cos \left(\frac{2\pi}{3} \right) \\ &= 9+25 +15 \\ &=49. \end{align*} Since $AC$ is positive, taking the square root gives $AC=7.$ Let $\angle BDC=\angle CBD=x$. Since $\triangle BCD$ is isosceles, we have $\angle BCD=180-2x$. Notice we can eventually solve $BD$ using the Extended Law of Sines: \[\frac{BD}{\sin(180-2x)}=2r,\] where $r$ is the radius of the circumcircle $ABCD$. Since $\sin(180-2x)=\sin(2x)=2\sin(x)\cos(x)$, we simply our equation: \[\frac{BD}{2\sin(x)\cos(x)}=2r.\] Now we just have to find $\sin(x), \cos(x),$ and $2r$. Since $ABCD$ is cyclic, we have $\angle CBD = \angle CAD = x$. By Law of Cosines on $\triangle ADC$, we have \[3^2=7^2 + 5^2 - 70\cos(x).\] Thus, $\cos(x)=\frac{13}{14}.$ Similarly, by Law of Sines on $\triangle ACD$, we have \[\frac{7}{\sin\left(\frac{2\pi}{3} \right)}=2r.\] Hence, $2r=\frac{14\sqrt3}{3}$. Now, using Law of Sines on $\triangle BCD$, we have $\frac{3}{\sin(x)}=2r= \frac{14\sqrt3}{3},$ so $\sin(x)=\frac{3\sqrt3}{14}.$ Therefore, \[\frac{BD}{2\left(\frac{3\sqrt3}{14}\right) \left(\frac{13}{14} \right)}=\frac{14\sqrt3}{3}.\] Solving, $BD = \frac{39}{7},$ so the answer is $\boxed{\textbf{(D) }\frac{39}{7}}$.

~evanhliu2009

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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