2024 AMC 10B Problem 11

Attempt Problem 11 of the 2024 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10B solutions, or check the answer key.

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11.

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 vector area decomposition

Difficulty rating: 1500

Small hint:

Set

X = (0,0),

Y = (8,0),

W = (0,4),

Z = (8,4),

with

M = (m, 0)

and

A = (8, a)

Big hint:

Perpendicularity gives

m(8 - m) = 4a;

equal areas give

2m = 4(4 - a)

Solution:

Set

X = (0,0),

Y = (8,0),

W = (0,4),

Z = (8,4),

so

M = (m, 0)

and

A = (8, a).

The right angle means

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

which gives

-m(8 - m) + 4a = 0,

that is

m(8 - m) = 4a.

Equal areas

[WXM] = 2m

and

[WAZ] = 4(4 - a)

force

m = 8 - 2a,

so

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

Substitute back and

(8 - m)(2 - m) = 0.

Taking

M \ne Y

leaves

m = 2

and

a = 3.

Then

[WMA] = 32

- [WXM]

- [MYA]

- [AZW]

= 32 - 4 - 9 - 4

= 15.

Thus, C is the correct answer.