1953 AMC 12 Problem 50

Question

One of the sides of a triangle is divided into segments of

6

and

8

units by the point of tangency of the inscribed circle. If the radius of the circle is

4,

then the length of the shortest side of the triangle is:

12

units

13

units

14

units

15

units

16

units

Equal tangent segments from a vertex make the three sides

14,

6+z,

and

8+z

for some

z

Use both

K=rs

and Heron’s formula with semiperimeter

s=14+z

Let the two tangent segments from the third vertex each have length

z.

Equal tangents from a common vertex make the side lengths

14,\qquad 6+z,\qquad 8+z,

with semiperimeter

s=14+z.

Since the inradius is

4,

K=rs=4(14+z).

Heron’s formula gives

\begin{aligned} K^2 &=(14+z)z\cdot8\cdot6\\ &=48z(14+z). \end{aligned}

Equating the squares yields

16(14+z)=48z,

so

z=7.

The sides are

14,

13,

and

15,

and the shortest is

13.

Thus, the correct answer is B.

Problem 50 in other years