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

Latest revision as of 17:31, 17 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

$[asy] import geometry; size(200); // from geogebra lol pair A = (-1.66, 0.33); pair B = (-9.61277, 1.19799); pair C = (-7.83974, 3.61798); pair D = (-4.88713, 4.14911); draw(circumcircle(A, B, C)); draw(A--C); draw(A--D); draw(C--D); draw(B--C); draw(A--B); label("$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, formatting by eevee9406, typo fixed by 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

Video Solution by SpreadTheMathLove

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

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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