Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
Line 5: Line 5:
 
<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>
  
−
==Solution 1 ==
+
==Solution==
 +
===Solution 1 ===
 
The area of <math>T</math> is <math>\dfrac{1}{2} \cdot 8 \cdot 3 = 12</math> and the perimeter is 18.
 
The area of <math>T</math> is <math>\dfrac{1}{2} \cdot 8 \cdot 3 = 12</math> and the perimeter is 18.
  
Line 16: Line 17:
 
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>.
 
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>.
  
−
==Solution 2==
+
===Solution 2===
  
 
Triangle <math>T</math>, being isosceles, has an area of <math>\frac{1}{2}(8)\sqrt{5^2-4^2}=12</math> and a perimeter of <math>5+5+8=18</math>.
 
Triangle <math>T</math>, being isosceles, has an area of <math>\frac{1}{2}(8)\sqrt{5^2-4^2}=12</math> and a perimeter of <math>5+5+8=18</math>.
Line 40: Line 41:
 
<cmath>b=\frac{1\pm\sqrt{33}}{2}</cmath>
 
<cmath>b=\frac{1\pm\sqrt{33}}{2}</cmath>
 
We clearly must pick the positive solution. Note that <math>5<\sqrt{33}<6</math>, and so <math>{3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}</math>, which clearly gives an answer of <math>\fbox{A}</math>, as desired.
 
We clearly must pick the positive solution. Note that <math>5<\sqrt{33}<6</math>, and so <math>{3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}</math>, which clearly gives an answer of <math>\fbox{A}</math>, as desired.
 +
 +
== See Also ==
 +
{{AMC12 box|year=2015|ab=A|num-b=19|num-a=21}}

Revision as of 00:54, 5 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

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 $2a = 18 - a$.

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 $\textbf{(A)}$.

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.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_20&oldid=67583"