Problem:
Isosceles triangles T and Tβ² 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 Tβ² 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 Tβ² from the sides with lengths 8 and b, respectively. The Pythagorean Theorem implies that g=52β42β=3, and so the area of T is 21ββ 8β 3=12, and the perimeter is 5+5+8=18. The Pythagorean Theorem implies that h=21β4a2βb2β. Thus 18=2a+b and
12=21βbβ 21β4a2βb2β=41βb4a2βb2β
Solving for a and substituting in the square of the second equation yields
Thus 64βb2(9βb)=b3β9b2+64=(bβ8)(b2βbβ8)=0. Because T and Tβ² are not congruent, it follows that bξ =8. Hence b2βbβ8=0 and the positive solution of this equation is 21β(33β+1). Because 25<33<36, the solution is between 21β(5+1)=3 and 21β(6+1)=3.5, so the closest integer is 3β .