Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2024 AMC 12A Problems/Problem 19"
 
(11 intermediate revisions by 4 users not shown)
Line 1: Line 1:
−
{{duplicate|[[2024 AMC 12A Problems/Problem 19|2024 AMC 12A #19]] and [[2024 AMC 10A Problems/Problem 22|2024 AMC 10A #22]]}}
+
==Problem==
 +
Cyclic quadrilateral <math>ABCD</math> has lengths <math>BC=CD=3</math> and <math>DA=5</math> with <math>\angle CDA=120^\circ</math>. What is the length of the shorter diagonal of <math>ABCD</math>?
 +
 
 +
<math>\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</math>
 +
 
 +
==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, <math>\angle CBA=60 ^\circ</math> by properties of cyclic quadrilaterals.
 +
 
 +
Let <math>AC=u</math>. Apply the [[Law of Cosines]] on <math>\triangle ACD</math>:
 +
<cmath>u^2=3^2+5^2-2(3)(5)\cos120^\circ</cmath>
 +
<cmath>u=7</cmath>
 +
 
 +
Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</math>:
 +
<cmath>7^2=3^2+v^2-2(3)(v)\cos60^\circ</cmath>
 +
<cmath>v=\frac{3\pm13}{2}</cmath>
 +
<cmath>v=8</cmath>
 +
 
 +
By [[Ptolemy’s Theorem]],
 +
<cmath>AB \cdot CD+AD \cdot BC=AC \cdot BD</cmath>
 +
<cmath>8 \cdot 3+5 \cdot 3=7BD</cmath>
 +
<cmath>BD=\frac{39}{7}</cmath>
 +
Since <math>\frac{39}{7}<7</math>,
 +
The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
 +
 
 +
 
 +
~lptoggled,eevee9406, meh494
 +
 
 +
==Solution 2 (Law of Cosines + Law of Sines)==
 +
Draw diagonals <math>AC</math> and <math>BD</math>. 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 <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>
 +
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>.
  
−
==Problem==
+
~evanhliu2009
−
Let <math>a</math>, <math>b</math>, and <math>c</math> be pairwise relatively prime positive integers. Suppose one of these numbers is prime, and the other two are perfect squares. If <math>abc</math> has <math>15a</math> divisors and <math>a^{2}b^{2}c^{2}</math> has <math>15b</math> divisors, what is the least possible value of <math>a + b + c</math>?
+
 
 +
==Solution 3 (Law of Cosines + Cyclic Quadrilateral Property)==
 +
Draw diagonals <math>AC</math> and <math>BD</math>. 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, <math>AC=7</math>. Since <math>ABCD</math> is cyclic, <math>\angle CAD = \angle CBD</math>, so Law of Cosines once again with respect to <math>\angle CAD</math> on triangle <math>ACD</math> leads to
 +
\begin{align*}
 +
9&=5^2+7^2-2(7)(5)\cos\theta \\
 +
&= 74-70\cos\theta. \\
 +
\end{align*}
 +
Solving yields <math>\cos\theta=\frac{13}{14}</math>. Finally, in <math>\triangle CBD</math>, we have <math>BD=6\cos\theta \implies \boxed{\textbf{(D) }\frac{39}{7}}</math>.
 +
 
 +
~SirAppel
 +
 
 +
==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, Fast, Quick, Easy!==
  
−
<math>\textbf{(A)}~18\qquad\textbf{(B)}~44\qquad\textbf{(C)}~108\qquad\textbf{(D)}~141\qquad\textbf{(E)}~636</math>
+
https://youtu.be/g4xdfcFgwGo
  
−
==Solution==
+
https://youtu.be/RQucKqjdNv8
−
Define <math>\tau{(n)}</math> to be the number of divisors of <math>n</math> and let <math>\nu_{2}{(n)}</math> denote the <math>2</math>-adic valuation of <math>n</math>.
 
  
−
Since <math>a, b, c</math> are pairwise relatively prime, the divisor function is multiplicative and <math>\tau(abc) = \tau(a)\tau(b)\tau(c)</math>. Two of these terms will be odd numbers, and the third one will be equal to <math>2</math>, since two of the numbers are perfect squares and one of them is prime. Therefore, <math>\nu_{2}{(\tau(abc))} = \nu_{2}{(15a)} = \nu_{2}{(a)} = 1</math>. If <math>a</math> is a perfect square then <math>\nu_{2}{(a)}</math> is even, and if <math>a</math> is an odd prime, then <math>\nu_{2}{(a)} = 0</math>, so we must have that <math>a = 2</math>. Therefore <math>a</math> is the prime, and <math>b</math> and <math>c</math> are perfect squares.
+
~MC
  
−
Now, <math>\tau(abc) = \tau(a)\tau(b)\tau(c) = 15a = 30 = 2\tau(b)\tau(c)</math>. This means that <math>\tau(b)\tau(c) = 15</math>. Since <math>b, c > 1</math>, (if either of them are one, then the other must be at least <math>3^{14}</math> which is clearly not minimum) the only way this is possible is if <math>\{\tau(a), \tau(b)\} = \{3, 5\}</math>. This means that one of them must be the square of a prime, and the other must be the fourth power of a prime. Thus <math>\tau(a^{2}b^{2}c^{2}) = 3 \cdot 5 \cdot 9 = 15b</math>, and <math>b = 9</math>.
+
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=f32mBtYTZp8
  
−
Thus <math>c</math> can be any fourth power of a prime that is relatively prime to both <math>2</math> and <math>9</math>, the least of which is <math>5^4 = 625</math>. The least possible value of <math>a + b + c</math> is <math>2 + 9 + 625 = \boxed{\textbf{(E)}~636}</math>.
+
== 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}}
−
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
+
{{MAA Notice}}
 +
==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:
 +
<math>AC^2 = 5^2 + 3^2 - 2(3)(5)(\cos 120^\circ)</math>
 +
<math>AC = 7</math>
 +
Now, our idea is to find side <math>AB</math> and then use Ptolemy's to find the other diagonal.
 +
LoS on <math>\triangle{ABC}</math> yields:
  
−
[[Category:Intermediate Number Theory Problems]]
+
<math>\frac{7}{\sin 60}</math> = <math>\frac{3}{\sin\angle{BAC}}</math> -> Note that <math>\angle{ABC} = 60^\circ</math>, due to opposite angles in a cyclic quadrilateral summing to <math>180^\circ</math>.
−
{{MAA Notice}}
+
<math>\sin\angle{BAC} = \frac{3\sqrt{3}}{14}</math>
 +
Now note that <math>\sin^2\theta + \cos^2\theta = 1</math>
 +
 
 +
This allows us to find <math>\cos\angle{BAC} = \sqrt{1- (\frac{3\sqrt{3}}{14})^2} = \frac{13}{14}</math>
 +
 
 +
Now, LoC again on <math>\triangle{ABC}</math> gives us <math>AB</math>:
 +
 
 +
Let <math>AB = x</math>:
 +
<math>x^2 + 7^2 - 2(7)(x)(\cos\angle{BAC} = 3^2</math>
 +
<math>x^2 -13x +40 = 0</math>
 +
<math>x = 5</math> and <math>x = 8</math>
 +
 
 +
Note that at this stage we can deduce that if we would have taken <math>\cos\angle{BAC} = -\frac{13}{14}</math>, then <math>AB</math> would have negative values.
 +
 
 +
Applying Ptolemy's:
 +
 
 +
<math>15 + 15 = 7(BD)</math> -> <math>BD = \frac{30}{7}</math>
 +
<math>24 + 15 = 7(BD)</math> -> <math>BD = \frac{39}{7}</math>
 +
 
 +
However, <math>\frac{30}{7}</math> isn't in the options, so we conclude <math>\boxed{D}</math>.
 +
~ cheltstudent
 +
I'm norz

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12A_Problems/Problem_19&oldid=264327"