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

(→‎Solution 1)
(→‎Solution 1)
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<cmath>u^2=3^2+5^2-2(3)(5)\cos120^\circ</cmath>
 
<cmath>u^2=3^2+5^2-2(3)(5)\cos120^\circ</cmath>
 
<cmath>u=7</cmath>
 
<cmath>u=7</cmath>
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Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</math>:
 
Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</math>:
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<cmath>v=\frac{3\pm13}{2}</cmath>
 
<cmath>v=\frac{3\pm13}{2}</cmath>
 
<cmath>v=8</cmath>
 
<cmath>v=8</cmath>
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By Ptolemy’s Theorem,
 
By Ptolemy’s Theorem,
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Since <math>\frac{39}{7}<5</math>,
 
Since <math>\frac{39}{7}<5</math>,
 
The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
 
The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
 +
 +
(someone please insert diagram)
  
 
~lptoggled, formatting by eevee9406
 
~lptoggled, formatting by eevee9406
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==Solution 2 (Law of Cosines + Law of Sines)==
 
==Solution 2 (Law of Cosines + Law of Sines)==

Revision as of 22:09, 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 $\boxed{\textbf{(D) }\frac{39}{7}}$.

(someone please insert diagram)

~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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