2024 AMC 12A Problems/Problem 19: Difference between revisions

Latest revision as of 09:26, 28 October 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, Fast, Quick, Easy!

https://youtu.be/g4xdfcFgwGo

https://youtu.be/RQucKqjdNv8

~MC

Video Solution 1 by SpreadTheMathLove

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

Solution 5 (Law of Cosines+Law of Sines+Trig Identities+Ptolemy's)

First of all, we see that this is a cyclic quadrilateral problem. This makes us happy, as there are literally 2 things in a cyclic quadrilateral problem: Ptolemy's and the opposite angles sum to 180 degrees. These are useful theorems so we write them down beside our nicely drawn diagram. We now proceed to trig bash. LoC on triangle ABC yields: $AC^2 = 5^2 + 3^2 - 2(3)(5)(\cos 120^\circ)$ $AC = 7$ Now, our idea is to find side $AB$ and then use Ptolemy's to find the other diagonal. LoS on $\triangle{ABC}$ yields:

$\frac{7}{\sin 60}$ = $\frac{3}{\sin\angle{BAC}}$ -> Note that $\angle{ABC} = 60^\circ$ , due to opposite angles in a cyclic quadrilateral summing to $180^\circ$ . $\sin\angle{BAC} = \frac{3\sqrt{3}}{14}$ Now note that $\sin^2\theta + \cos^2\theta = 1$

This allows us to find $\cos\angle{BAC} = \sqrt{1- (\frac{3\sqrt{3}}{14})^2} = \frac{13}{14}$

Now, LoC again on $\triangle{ABC}$ gives us $AB$ :

Let $AB = x$ : $x^2 + 7^2 - 2(7)(x)(\cos\angle{BAC} = 3^2$ $x^2 -13x +40 = 0$ $x = 5$ and $x = 8$

Note that at this stage we can deduce that if we would have taken $\cos\angle{BAC} = -\frac{13}{14}$ , then $AB$ would have negative values.

Applying Ptolemy's:

$15 + 15 = 7(BD)$ -> $BD = \frac{30}{7}$ $24 + 15 = 7(BD)$ -> $BD = \frac{39}{7}$

However, $\frac{30}{7}$ isn't in the options, so we conclude $\boxed{D}$ . ~ cheltstudent I'm norz

2024 AMC 12A (Problems • Answer Key • Resources)
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