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Difference between revisions of "2015 AMC 12A Problems/Problem 20"

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<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math>
 
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math>
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==Solution==
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The area of <math>T</math> is <math>\dfrac{1}{2} \cdot 8 \cdot 3 = 12</math> and the perimeter is 18.
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The area of <math>T</math> is <math>\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math> and the perimeter is <math>2a + b</math>.
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Thus <math>2a + b = 18</math>, so <math>2a = 18 - a</math>.
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Thus <math>12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math>, so <math>48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}</math>.
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We square and divide 36 from both sides to obtain <math>64 = b^2 (9 - b)</math>, so <math>b^3 - 9b^2 + 64 = 0</math>. This factors as <math>(b - 8)(b^2 - b - 8) = 0</math>. Because clearly <math>b \neq 8</math> but <math>b > 0</math>, we have <math>b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.</math> The answer is <math>\textbf{(A)}</math>.

Revision as of 21:40, 4 February 2015

Problem

Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$, $5$, and $8$, while those of $T'$ have lengths $a$, $a$, and $b$. Which of the following numbers is closest to $b$?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$

Solution

The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.

The area of $T$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$.

Thus $2a + b = 18$, so $2a = 18 - a$.

Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$, so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}$.

We square and divide 36 from both sides to obtain $64 = b^2 (9 - b)$, so $b^3 - 9b^2 + 64 = 0$. This factors as $(b - 8)(b^2 - b - 8) = 0$. Because clearly $b \neq 8$ but $b > 0$, we have $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\textbf{(A)}$.

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