Problem:
Isosceles triangles T and T′ are not congruent but have the same area and the same perimeter. The sides of T have lengths of 5,5 , and 8 , while those of T′ have lengths a,a, and b. Which of the following numbers is closest to b ?
Answer Choices:
A. 3
B. 4
C. 5
D. 6
E. 8 Solution:
Let g and h be the lengths of the altitudes of T and T′ from the sides with lengths 8 and b, respectively. The Pythagorean Theorem implies that g=52−42​=3, and so the area of T is 21​⋅8⋅3=12, and the perimeter is 5+5+8=18. The Pythagorean Theorem implies that h=21​4a2−b2​. Thus 18=2a+b and
12=21​b⋅21​4a2−b2​=41​b4a2−b2​
Solving for a and substituting in the square of the second equation yields
Thus 64−b2(9−b)=b3−9b2+64=(b−8)(b2−b−8)=0. Because T and T′ are not congruent, it follows that bî€ =8. Hence b2−b−8=0 and the positive solution of this equation is 21​(33​+1). Because 25<33<36, the solution is between 21​(5+1)=3 and 21​(6+1)=3.5, so the closest integer is 3​ .