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

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The answer is <math>3</math>. <math>\textbf{(A)}</math>
 
The answer is <math>3</math>. <math>\textbf{(A)}</math>
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 +
==Operation Descartes==
 +
For this new triangle, say its legs have length <math>d</math> and the base length <math>2c</math>. To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that <math>c+d=9</math> and <math>c*\sqrt{d^2-c^2}=12</math>. It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!!
 +
 +
Now, modify the square-root equation with <math>d=9-c</math>; you get <math>c^2*(81-18c)=144</math>, so <math>-18c^3+81c^2=144</math>. Divide by <math>-9</math> to get <math>2c^3-9c^2+16=0</math>. Obviously, <math>c=4</math> is a root as established by triangle <math>T</math>! So, use synthetic division to obtain <math>2c^2-c-4=0</math>, upon which <math>c=\frac{1+\sqrt{33}}{4}</math>, which is so close to <math>\frac{3}{2}</math> and even less than it. That's enough to confirm that the answer has to be <math>\textbf{A}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2015|ab=A|num-b=19|num-a=21}}

Revision as of 20:58, 2 February 2018

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

Solution 1

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 $2a = 18 - b$.

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 $\textbf{(A)}$.

Solution 2

Triangle $T$, being isosceles, has an area of $\frac{1}{2}(8)\sqrt{5^2-4^2}=12$ and a perimeter of $5+5+8=18$. Triangle $T'$ similarly has an area of $\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12$ and $2a+b=18$.

Now we apply our computational fortitude.

\[\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12\] \[(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=24\] \[(b)\sqrt{4a^2-b^2}=48\] \[b^2(4a^2-b^2)=48^2\] \[b^2(2a+b)(2a-b)=48^2\] Plug in $2a+b=18$ to obtain \[18b^2(2a-b)=48^2\] \[b^2(2a-b)=128\] Plug in $2a=18-b$ to obtain \[b^2(18-2b)=128\] \[2b^3-18b^2+128=0\] \[b^3-9b^2+64=0\] We know that $b=8$ is a valid solution by $T$. Factoring out $b-8$, we obtain \[(b-8)(b^2-b-8)=0 \Rightarrow b^2-b-8=0\] Utilizing the quadratic formula gives \[b=\frac{1\pm\sqrt{33}}{2}\] We clearly must pick the positive solution. Note that $5<\sqrt{33}<6$, and so ${3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}$, which clearly gives an answer of $\fbox{A}$, as desired.

Solution 3

Triangle T has perimeter $5 + 5 + 8 = 18$ so $18 = 2a + b$.

Using Heron's, we get $\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})(\frac{b}{2})^2(\frac{2a-b}{2})}$.

We know that $2a + b = 18$ from above so we plug that in, and we also know that then $2a - b = 18 - 2b$.

$12 = 3\frac{b}{2}\sqrt{9-b}$

$64 = 9b^2 - b^3$

We plug in 3 for $b$ in the LHS, and we get 54 which is too low. We plug in 4 for $b$ in the LHS, and we get 80 which is too high. We now know that b is some number between 3 and 4.

If $b \geq 3.5$, then we would round up to 4, but if $b < 3.5$, then we would round down to 3. So let us plug in 3.5 for b.

We get 67.375 which is too high, so we know that $b < 3.5$.

The answer is $3$. $\textbf{(A)}$

Operation Descartes

For this new triangle, say its legs have length $d$ and the base length $2c$. To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that $c+d=9$ and $c*\sqrt{d^2-c^2}=12$. It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!!

Now, modify the square-root equation with $d=9-c$; you get $c^2*(81-18c)=144$, so $-18c^3+81c^2=144$. Divide by $-9$ to get $2c^3-9c^2+16=0$. Obviously, $c=4$ is a root as established by triangle $T$! So, use synthetic division to obtain $2c^2-c-4=0$, upon which $c=\frac{1+\sqrt{33}}{4}$, which is so close to $\frac{3}{2}$ and even less than it. That's enough to confirm that the answer has to be $\textbf{A}$.

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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