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

Latest revision as of 03:50, 14 August 2025

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

$[asy] import geometry; size(200); pair A = (-1.66, 0.33); pair B = (-9.61277, 1.19799); pair C = (-7.83974, 3.61798); pair D = (-4.88713, 4.14911); draw(circumcircle(A, B, C)); draw(A--C); draw(A--D); draw(C--D); draw(B--C); draw(A--B); label("$A$", A, E); label("$B$", B, W); label("$C$", C, NW); label("$D$", D, N); label("$7$", midpoint(A--C), SW); label("$5$", midpoint(A--D), NE); label("$3$", midpoint(C--D)+ dir(135)*0.3, N); label("$3$", midpoint(B--C)+dir(180)*0.3, NW); label("$8$", midpoint(A--B), S); markangle(Label("$60^\circ$", Relative(0.5)), A, B, C, radius=10); markangle(Label("$120^\circ$", Relative(0.5)), C, D, A, radius=10); [/asy]$ ~diagram by erics118

First, $\angle CBA=60 ^\circ$ by properties of cyclic quadrilaterals.

Let $AC=u$ . 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 $\frac{39}{7}<7$ , The answer is $\boxed{\textbf{(D) }\frac{39}{7}}$ .

~lptoggled,eevee9406, meh494

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

Solution 3 (Law of Cosines + Cyclic Quadrilateral Property)

Draw diagonals $AC$ and $BD$ . First, use Law of Cosines to get that \begin{align*} AC^2&=3^2 + 5^2 - 2(3)(5)\cos(120^{\circ}) \\ &= 9+25+15 \\ &=49. \end{align*} Thus, $AC=7$ . Since $ABCD$ is cyclic, $\angle CAD = \angle CBD$ , so Law of Cosines once again with respect to $\angle CAD$ on triangle $ACD$ leads to \begin{align*} 9&=5^2+7^2-2(7)(5)\cos\theta \\ &= 74-70\cos\theta. \\ \end{align*} Solving yields $\cos\theta=\frac{13}{14}$ . Finally, in $\triangle CBD$ , we have $BD=6\cos\theta \implies \boxed{\textbf{(D) }\frac{39}{7}}$ .

~SirAppel

Solution 4 (Law of Cosines+Law of Sines+Trig Identities)

Let $\angle BCA = x, \angle DCA = y$ . If we know $\cos(x+y)$ we can compute $BD$ . Notice that $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ . Now it remains to find all 4 terms in this equation. Applying Law of Cosines on triangle $ABC$ to find $\cos(x)$ , we find that $\cos(x)=-\frac{6}{42}=-\frac{1}{7}$ . Similarly we find that $\cos(y)=\frac{11}{14}$ . Now we compute $\sin(x)$ and $\sin(y)$ . Applying Law of Sines on triangle $ABC$ we see that $\frac{\sin(x)}{8}=\frac{\sin(\frac{\pi}{3})}{7}$ , or $\sin(x)=\frac{4\sqrt{3}}{7}$ . Similarly $\sin(y)=\frac{5\sqrt{3}}{14}$ . Now $\cos(x+y)=-\frac{71}{98}$ . Let $BD=k$ , we see that $k^2=3^2+3^2+2*3*3(\frac{71}{98})$ . Solving for $k$ yields $k=\frac{39}{7}$ .

~CreamyCream

Video Solution(Quick, fast, easy!)

https://youtu.be/RQucKqjdNv8

~MC

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=f32mBtYTZp8

@@ Line 90: / Line 90: @@
 ==Solution 4 (Law of Cosines+Law of Sines+Trig Identities)==
+Let <math>\angle BCA = x, \angle DCA = y</math>. If we know <math>\cos(x+y)</math> we can compute <math>BD</math>. Notice that <cmath>\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)</cmath>. Now it remains to find all 4 terms in this equation. Applying Law of Cosines on triangle <math>ABC</math> to find <math>\cos(x)</math>, we find that <math>\cos(x)=-\frac{6}{42}=-\frac{1}{7}</math>. Similarly we find that <math>\cos(y)=\frac{11}{14}</math>. Now we compute <math>\sin(x)</math> and <math>\sin(y)</math>. Applying Law of Sines on triangle <math>ABC</math> we see that <math>\frac{\sin(x)}{8}=\frac{\sin(\frac{\pi}{3})}{7}</math>, or <math>\sin(x)=\frac{4\sqrt{3}}{7}</math>. Similarly <math>\sin(y)=\frac{5\sqrt{3}}{14}</math>. Now <math>\cos(x+y)=-\frac{71}{98}</math>. Let <math>BD=k</math>, we see that <math>k^2=3^2+3^2+2*3*3(\frac{71}{98})</math>. Solving for <math>k</math> yields <math>k=\frac{39}{7}</math>.
+~CreamyCream
 ==Video Solution(Quick, fast, easy!)==

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

Page

Toolbox

Search