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Difference between revisions of "2024 AMC 12B Problems/Problem 22"

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~tsun26
 
~tsun26
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==Solution 2 (Similar to Solution 1)==
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Let <math>\overline{BC}=a</math>, <math>\overline{AC}=b</math>, <math>\overline{AB}=c</math>. Extend <math>C</math> to point <math>D</math> on <math>\overline{AB}</math> such that <math>\angle ACD = \angle CAD</math>. This means <math>\triangle CDA</math> is isosceles, so <math>CD=DA</math>. Since <math>\angle CDB</math> is the exterior angle of <math>\triangle CDA</math>, we have <cmath>\angle CDB=m+m=2m=\angle CBD.</cmath> Thus, <math>\triangle CBD</math> is isosceles, so <math>CB=CD=DA=a.</math> Then, draw the altitude of <math>\triangle CBD</math>, from <math>C</math> to <math>\overline{BD}</math>, and let this point be <math>H</math>. Let <math>BH=HD=x</math>. Then, by Pythagorean Theorem,
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\begin{align*}
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CH^2&=a^2-x^2 \\
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CH^2&= b^2 - (c+x)^2.\\
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\end{align*}
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Thus, <cmath>a^2-x^2=b^2-(c-x)^2.</cmath> Solving for <math>x</math>, we have <math>x=\frac{a^2-b^2+c^2}{2c}.</math> Since <math>2x=c-a</math>, we have <cmath>c-a=\frac{a^2-b^2+c^2}{c},</cmath> and simplifying, we get <math>b^2=a^2+ac.</math> Now we can consider cases on what <math>a</math> is. (Note: Although there looks to be quite a few cases, they are just trivial and usually only take up to a few seconds max).
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Case <math>1</math>: <math>a=1</math>.
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This means <math>b^2=c+1</math>, so the least possible values are <math>b=2</math>, <math>c=3</math>, but this does not work as it does not satisfy the triangle inequality. Similarly, <math>b=3</math>, <math>c=8</math> also does not satisfy it. Anything larger goes beyond the answer choices, so we stop checking this case.
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Case <math>2</math>: <math>a=2</math>
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This means <math>b^2=2c+4</math>, so the least possible values for <math>b</math> and <math>c</math> are <math>b=4</math>,<math>c=6</math>, but this does not satisfy the triangle inequality, and anything larger does not satisfy the answer choices.
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Case <math>3</math>: <math>a=3</math>
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This means <math>b^2=3c+9</math>, and the least possible value for <math>b</math> is <math>b=6</math>, which occurs when <math>c=9</math>. Unfortunately, this also does not satisfy the triangle inequality, and similarly, any <math>b > 6</math> means the perimeter will get too big.
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Case <math>4</math>: <math>a=4</math>
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This means <math>b^2=4c+16</math>, so we have <math>b=6,c=5,a=4</math>, so the least possible perimeter so far is <math>4+5+6=15</math>.
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Case <math>5</math>: <math>a=5</math>
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We have <math>b^2=5c+25</math>, so least possible value for <math>b</math> is <math>b=10</math>, which already does not work as <math>a=5</math>, and the minimum perimeter is <math>15</math> already.
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Case <math>6</math>: <math>a=6</math>
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We have <math>b^2=6c+36</math>, so <math>b=10</math>, which already does not work.
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Then, notice that when <math>a\geq 7</math>, we also must have <math>b\geq8</math> and <math>c\geq1</math>, so <math>a+b+c \geq 16</math>, so the least possible perimeter is <math>\boxed{\textbf{(C) }15}.</math>
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~evanhliu2009

Revision as of 08:10, 14 November 2024

Problem 22

Let $\triangle{ABC}$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$. What is the least possible perimeter of such a triangle?

$\textbf{(A) }13 \qquad \textbf{(B) }14 \qquad \textbf{(C) }15 \qquad \textbf{(D) }16 \qquad \textbf{(E) }17 \qquad$

Solution 1

Let $AB=c$, $BC=a$, $AC=b$. According to the law of sines, \[\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}\] \[=2\cos \angle A\]

According to the law of cosines, \[\cos \angle A=\frac{b^2+c^2-a^2}{2bc}\]

Hence, \[\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}\]

This simplifies to $b^2=a(a+c)$. We want to find the positive integer solution $(a, b, c)$ to this equation such that $a, b, c$ forms a triangle, and $a+b+c$ is minimized. We proceed by casework on the value of $b$. Remember that $a<a+c$.

$\textbf{Case 1: b=1}$ Clearly, this case yields no valid solutions.

$\textbf{Case 2: b=2}$ For this case, we must have $a=1$ and $c=3$. However, $(1, 2, 3)$ does not form a triangle. Hence this case yields no valid solutions.

$\textbf{Case 3: b=3}$ For this case, we must have $a=1$ and $c=9$. However, $(1, 3, 9)$ does not form a triangle. Hence this case yields no valid solutions.

$\textbf{Case 4: b=4}$ For this case, $a=1$ and $c=15$, or $a=2$ and $c=6$. As one can check, this case also yields no valid solutions

$\textbf{Case 5: b=5}$ For this case, we must have $a=1$ and $c=24$. There are no valid solutions

$\textbf{Case 6: b=6}$ For this case, $a=2$ and $c=16$, or $a=4$ and $c=5$, or $a=3$ and $c=9$. The only valid solution for this case is $(4, 6, 5)$.

It is safe to assume that $(4, 5, 6)$ will be the solution with least perimeter. Hence, the answer is $\fbox{\textbf{(C) }15}$

~tsun26

Solution 2 (Similar to Solution 1)

Let $\overline{BC}=a$, $\overline{AC}=b$, $\overline{AB}=c$. Extend $C$ to point $D$ on $\overline{AB}$ such that $\angle ACD = \angle CAD$. This means $\triangle CDA$ is isosceles, so $CD=DA$. Since $\angle CDB$ is the exterior angle of $\triangle CDA$, we have \[\angle CDB=m+m=2m=\angle CBD.\] Thus, $\triangle CBD$ is isosceles, so $CB=CD=DA=a.$ Then, draw the altitude of $\triangle CBD$, from $C$ to $\overline{BD}$, and let this point be $H$. Let $BH=HD=x$. Then, by Pythagorean Theorem, \begin{align*} CH^2&=a^2-x^2 \\ CH^2&= b^2 - (c+x)^2.\\ \end{align*} Thus, \[a^2-x^2=b^2-(c-x)^2.\] Solving for $x$, we have $x=\frac{a^2-b^2+c^2}{2c}.$ Since $2x=c-a$, we have \[c-a=\frac{a^2-b^2+c^2}{c},\] and simplifying, we get $b^2=a^2+ac.$ Now we can consider cases on what $a$ is. (Note: Although there looks to be quite a few cases, they are just trivial and usually only take up to a few seconds max).

Case $1$: $a=1$.

This means $b^2=c+1$, so the least possible values are $b=2$, $c=3$, but this does not work as it does not satisfy the triangle inequality. Similarly, $b=3$, $c=8$ also does not satisfy it. Anything larger goes beyond the answer choices, so we stop checking this case.

Case $2$: $a=2$ This means $b^2=2c+4$, so the least possible values for $b$ and $c$ are $b=4$,$c=6$, but this does not satisfy the triangle inequality, and anything larger does not satisfy the answer choices.

Case $3$: $a=3$ This means $b^2=3c+9$, and the least possible value for $b$ is $b=6$, which occurs when $c=9$. Unfortunately, this also does not satisfy the triangle inequality, and similarly, any $b > 6$ means the perimeter will get too big.

Case $4$: $a=4$ This means $b^2=4c+16$, so we have $b=6,c=5,a=4$, so the least possible perimeter so far is $4+5+6=15$.

Case $5$: $a=5$ We have $b^2=5c+25$, so least possible value for $b$ is $b=10$, which already does not work as $a=5$, and the minimum perimeter is $15$ already.

Case $6$: $a=6$ We have $b^2=6c+36$, so $b=10$, which already does not work.

Then, notice that when $a\geq 7$, we also must have $b\geq8$ and $c\geq1$, so $a+b+c \geq 16$, so the least possible perimeter is $\boxed{\textbf{(C) }15}.$

~evanhliu2009

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