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

(→‎Solution 3 (Law of Cosines + Cyclic Quadrilateral Property))
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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==
 
==Problem==
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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>?
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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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<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>
 
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==Solution 1==
 
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<asy>
 
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import geometry;
 
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size(200);
 
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// 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);
 
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draw(circumcircle(A, B, C));
 
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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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markangle(Label("$60^\circ$", Relative(0.5)), A, B, C, radius=10);
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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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markangle(Label("$120^\circ$", Relative(0.5)), C, D, A, radius=10);
 
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</asy>
 
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~diagram by erics118
 
  
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First, <math>\angle CBA=60 ^\circ</math> by properties of cyclic quadrilaterals.
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==Solution==
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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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Let <math>AC=u</math>. Apply the [[Law of Cosines]] on <math>\triangle ACD</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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<cmath>u^2=3^2+5^2-2(3)(5)\cos120^\circ</cmath>
 
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<cmath>u=7</cmath>
 
  
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Let <math>AB=v</math>. Apply the Law of Cosines on <math>\triangle ABC</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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<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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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>.
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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>,
 
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The answer is <math>\boxed{\textbf{(D) }\frac{39}{7}}</math>.
 
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~lptoggled, formatting by eevee9406, typo fixed by meh494
 
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==Solution 2 (Law of Cosines + Law of Sines)==
 
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Draw diagonals <math>AC</math> and <math>BD</math>. By Law of Cosines,
 
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\begin{align*}
 
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AC^2&=3^2 + 5^2 - 2(3)(5)\cos \left(\frac{2\pi}{3} \right) \\
 
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&= 9+25 +15 \\
 
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&=49.
 
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\end{align*}
 
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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>
 
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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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~evanhliu2009
 
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==Solution 3 (Law of Cosines + Cyclic Quadrilateral Property)==
 
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Draw diagonals <math>AC</math> and <math>BD</math>. First, use Law of Cosines to get that
 
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\begin{align*}
 
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AC^2&=3^2 + 5^2 - 2(3)(5)\cos(120^{\circ}) \\
 
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&= 9+25+15 \\
 
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&=49.
 
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\end{align*}
 
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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
 
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\begin{align*}
 
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9&=5^2+7^2-2(7)(5)\cos\theta \\
 
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&= 74-70\cos\theta. \\
 
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\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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~SirAppel
 
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==Video Solution by SpreadTheMathLove==
 
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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}}
 +
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 +
 +
[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:23, 20 March 2025

The following problem is from both the 2024 AMC 12A #19 and 2024 AMC 10A #22, so both problems redirect to this page.

Problem

Let $a$, $b$, and $c$ be pairwise relatively prime positive integers. Suppose one of these numbers is prime, and the other two are perfect squares. If $abc$ has $15a$ divisors and $a^{2}b^{2}c^{2}$ has $15b$ divisors, what is the least possible value of $a + b + c$?

$\textbf{(A)}~18\qquad\textbf{(B)}~44\qquad\textbf{(C)}~108\qquad\textbf{(D)}~141\qquad\textbf{(E)}~636$

Solution

Define $\tau{(n)}$ to be the number of divisors of $n$ and let $\nu_{2}{(n)}$ denote the $2$-adic valuation of $n$.

Since $a, b, c$ are pairwise relatively prime, the divisor function is multiplicative and $\tau(abc) = \tau(a)\tau(b)\tau(c)$. Two of these terms will be odd numbers, and the third one will be equal to $2$, since two of the numbers are perfect squares and one of them is prime. Therefore, $\nu_{2}{(\tau(abc))} = \nu_{2}{(15a)} = \nu_{2}{(a)} = 1$. If $a$ is a perfect square then $\nu_{2}{(a)}$ is even, and if $a$ is an odd prime, then $\nu_{2}{(a)} = 0$, so we must have that $a = 2$. Therefore $a$ is the prime, and $b$ and $c$ are perfect squares.

Now, $\tau(abc) = \tau(a)\tau(b)\tau(c) = 15a = 30 = 2\tau(b)\tau(c)$. This means that $\tau(b)\tau(c) = 15$. Since $b, c > 1$, (if either of them are one, then the other must be at least $3^{14}$ which is clearly not minimum) the only way this is possible is if $\{\tau(a), \tau(b)\} = \{3, 5\}$. 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 $\tau(a^{2}b^{2}c^{2}) = 3 \cdot 5 \cdot 9 = 15b$, and $b = 9$.

Thus $c$ can be any fourth power of a prime that is relatively prime to both $2$ and $9$, the least of which is $5^4 = 625$. The least possible value of $a + b + c$ is $2 + 9 + 625 = \boxed{\textbf{(E)}~636}$.

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
2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 10 Problems and Solutions

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