2024 AMC 12B Problem 21

Question

The measures of the smallest angles of three different right triangles sum to

90^\circ.

All three triangles have side lengths that are primitive Pythagorean triples. Two of them are

3

-

4

-

5

and

5

-

12

-

13.

What is the perimeter of the third triangle?

40

126

154

176

208

The smallest angles have tangents

\tfrac34

and

\tfrac{5}{12};

the third must satisfy

\alpha + \beta + \gamma = 90^\circ,

so

\tan\gamma = \cot(\alpha + \beta)

Compute

\tan(\alpha + \beta) = \dfrac{56}{33},

so

\tan\gamma = \dfrac{33}{56};

find the primitive triple with legs

33

and

56

The smallest angles

\alpha, \beta

of the

3

-

4

-

5

and

5

-

12

-

13

triangles have

\tan\alpha = \tfrac34

and

\tan\beta = \tfrac{5}{12}.

By the tangent addition formula,

\begin{aligned} &\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} \\ &= \frac{\tfrac{14}{12}}{\tfrac{33}{48}} \\ &= \frac{56}{33}. \end{aligned}

The third smallest angle

\gamma

satisfies

\gamma = 90^\circ - (\alpha + \beta),

so

\tan\gamma = \dfrac{33}{56}.

The right triangle with legs

33

and

56

has hypotenuse

\sqrt{33^2 + 56^2} = \sqrt{4225} = 65,

a primitive triple. Its perimeter is

33 + 56 + 65 = 154.

Thus, the correct answer is C.

Problem 21 in other years