2024 AMC 12B Problem 20

Problem 20 of 25HarderAlgebra Geometry

Suppose

A,

B,

and

C

are points in the plane with

AB = 40

and

AC = 42,

and let

x

be the length of the line segment from

A

to the midpoint of

\overline{BC}.

Define a function

f

by letting

f(x)

be the area of

\triangle ABC.

Then the domain of

f

is an open interval

(p, q),

and the maximum value

r

of

f(x)

occurs at

x = s.

What is

p + q + r + s?

Show solution

Solution

Let

a = BC.

The median length gives

x^2

= \dfrac{2\cdot 1600 + 2\cdot 1764 - a^2}{4}

= \dfrac{6728 - a^2}{4}.

The triangle inequality requires

2 \lt a \lt 82,

i.e.

4 \lt a^2 \lt 6724,

which translates to

1 \lt x \lt 41.

So

(p, q) = (1, 41).

With

AB = 40

and

AC = 42

fixed, the area

\tfrac12\cdot 40\cdot 42\sin A

is largest when

\angle A = 90^\circ,

giving

r = 840.

Then

a^2 = 40^2 + 42^2 = 3364,

so

x^2 = \dfrac{6728 - 3364}{4} = 841,

i.e.

s = 29.

Thus

p + q + r + s

= 1 + 41 + 840 + 29

= 911.

Thus, the correct answer is C.

More practice

Concepts: median (geometry) · triangle inequality · optimization

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.