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Difference between revisions of "2024 AMC 12A Problems/Problem 19"
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{{duplicate|[[2024 AMC 12A Problems/Problem 19|2024 AMC 12A #19]] and [[2024 AMC 10A Problems/Problem 22|2024 AMC 10A #22]]}}
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==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);
 +
 
 +
// from geogebra lol
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pair A = (-1.66, 0.33);
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pair B = (-9.61277, 1.19799);
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pair C = (-7.83974, 3.61798);
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pair D = (-4.88713, 4.14911);
 +
 
 +
draw(circumcircle(A, B, C));
 +
 
 +
draw(A--C);
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draw(A--D);
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draw(C--D);
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draw(B--C);
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draw(A--B);
 +
 
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label("$A$", A, E);
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label("$B$", B, W);
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label("$C$", C, NW);
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label("$D$", D, N);
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label("$7$", midpoint(A--C), SW);
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label("$5$", midpoint(A--D), NE);
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label("$3$", midpoint(C--D)+ dir(135)*0.3, N);
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label("$3$", midpoint(B--C)+dir(180)*0.3, NW);
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label("$8$", midpoint(A--B), S);
  
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==Problem==
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markangle(Label("$60^\circ$", Relative(0.5)), A, B, C, radius=10);
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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>?
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markangle(Label("$120^\circ$", Relative(0.5)), C, D, A, radius=10);
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</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>:
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<cmath>u^2=3^2+5^2-2(3)(5)\cos120^\circ</cmath>
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<cmath>u=7</cmath>
 +
 
 +
Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</math>:
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<cmath>7^2=3^2+v^2-2(3)(v)\cos60^\circ</cmath>
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<cmath>v=\frac{3\pm13}{2}</cmath>
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<cmath>v=8</cmath>
 +
 
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By [[Ptolemy’s Theorem]],
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<cmath>AB \cdot CD+AD \cdot BC=AC \cdot BD</cmath>
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<cmath>8 \cdot 3+5 \cdot 3=7BD</cmath>
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<cmath>BD=\frac{39}{7}</cmath>
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Since <math>\frac{39}{7}<7</math>,
 +
The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
 +
 
 +
 
 +
~lptoggled, formatting by eevee9406, typo fixed by meh494
  
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<math>\textbf{(A)}~18\qquad\textbf{(B)}~44\qquad\textbf{(C)}~108\qquad\textbf{(D)}~141\qquad\textbf{(E)}~636</math>
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==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 \\
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&=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>.
  
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==Solution==
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~evanhliu2009
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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>.
 
  
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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.
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==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*}
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9&=5^2+7^2-2(7)(5)\cos\theta \\
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&= 74-70\cos\theta. \\
 +
\end{align*}
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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>.
  
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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>.
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~SirAppel
  
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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>.
+
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=f32mBtYTZp8
  
 
==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}}
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{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
−
 
−
[[Category:Intermediate Number Theory Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:34, 21 March 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); // 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

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

Video Solution 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

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