2024 AMC 12B Problem 7

Question

In the figure below

WXYZ

is a rectangle with

WX = 4

and

WZ = 8.

Point

M

lies on

\overline{XY},

point

A

lies on

\overline{YZ},

and

\angle WMA

is a right angle. The areas of

\triangle WXM

and

\triangle WAZ

are equal. What is the area of

\triangle WMA?

13

14

15

16

17

Answer: C

Concepts:coordinate geometry triangle area

Difficulty rating: 1420

Small hint:

Place

X = (0,0),

W = (0,4),

Y = (8,0),

Z = (8,4),

with

M = (m, 0)

and

A = (8, a)

Big hint:

The right angle gives

\overrightarrow{MW} \cdot \overrightarrow{MA} = 0,

and the equal-area condition gives a second equation; solve for

m

and

a

Solution:

Set

X = (0,0),

W = (0,4),

Y = (8,0),

Z = (8,4),

with

M = (m, 0)

on

\overline{XY}

and

A = (8, a)

on

\overline{YZ}.

Since

\angle WMA = 90^\circ,

\overrightarrow{MW} \cdot \overrightarrow{MA}

= (-m)(8-m)

+ 4a = 0,

so

4a = m(8-m).

The areas give

[\triangle WXM] = \tfrac12 \cdot 4 \cdot m = 2m

and

[\triangle WAZ]

= \tfrac12 \cdot 8 \cdot (4 - a)

= 4(4 - a).

Setting these equal yields

m = 8 - 2a.

Substituting

a = \tfrac{8-m}{2}

into

4a = m(8-m)

gives

2(8-m) = m(8-m),

so

m = 2

and

a = 3.

The other algebraic root,

m=8,

gives

M=A=Y

and no defined angle

\angle WMA,

so it is invalid. Then with

W = (0,4),

M = (2,0),

A = (8,3),

\begin{aligned} [\triangle WMA] &= \tfrac12\,\bigl|2(3 - 4) + 8(4 - 0)\bigr| \\ &= \tfrac12 (30) = 15. \end{aligned}

Thus, the correct answer is C.