1963 AMC 12 Problem 35

Question

The lengths of the sides of a triangle are integers, and its area is also an integer. One side is

21

and the perimeter is

48.

The shortest side is:

8

10

12

14

16

Write the other sides as

x

and

27-x

, with semiperimeter

24

Heron’s formula makes the squared area

72(24-x)(x-3)

Let the other sides be

x

and

27-x,

with

x\leq27-x.

The semiperimeter is

24,

so Heron’s formula gives

\begin{aligned} K^2 &=24\cdot3(24-x)(x-3)\\ &=72(24-x)(x-3). \end{aligned}

The triangle inequalities give

4\leq x\leq13.

Checking these integers, the expression is not a square for

x=4,\ldots,9,

while at

x=10

it is

72\cdot14\cdot7=7056=84^2.

Thus the shortest side is

10.

Therefore, the correct answer is B.

Problem 35 in other years