2015 AMC 12A Problem 20

Question

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?

3

4

5

6

8

Answer: A

Concepts:isosceles triangle system of equations polynomial

Difficulty rating: 2110

Small hint:

T

has perimeter

18

and area

12;

write those same two conditions for

T'

Big hint:

Eliminate

a

using

2a + b = 18;

you reach a cubic with the extraneous root

b = 8

to factor out

Solution:

The altitude of

T

to its base of length

8

is

\sqrt{5^2 - 4^2} = 3,

so

T

has area

\dfrac{1}{2}\cdot 8\cdot 3 = 12

and perimeter

18.

For

T'

we need

2a + b = 18

and area

\dfrac{1}{4}b\sqrt{4a^2 - b^2} = 12.

Substituting

a = \dfrac{18 - b}{2}

and squaring leads to

(b - 8)(b^2 - b - 8) = 0.

Since

T

and

T'

are not congruent,

b \ne 8,

so

b^2 - b - 8 = 0

and

b = \dfrac{1 + \sqrt{33}}{2}.

Because

25 \lt 33 \lt 36,

this is between

3

and

3.5,

so the closest integer is

3.

Thus, the correct answer is A.