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 WXYZWXYZ is a rectangle with WX=4WX = 4 and WZ=8.WZ = 8. Point MM lies on XY,\overline{XY}, point AA lies on YZ,\overline{YZ}, and WMA\angle WMA is a right angle. The areas of WXM\triangle WXM and WAZ\triangle WAZ are equal. What is the area of WMA?\triangle WMA?

1313

1414

1515

1616

1717

Answer: C
Concepts:coordinate geometryvectorarea decomposition
Difficulty rating: 1500
Solution:

Set X=(0,0),X = (0,0), Y=(8,0),Y = (8,0), W=(0,4),W = (0,4), Z=(8,4),Z = (8,4), so M=(m,0)M = (m, 0) and A=(8,a).A = (8, a). The right angle means MWMA=0,\overrightarrow{MW} \cdot \overrightarrow{MA} = 0, which gives m(8m)+4a=0,-m(8 - m) + 4a = 0, that is m(8m)=4a.m(8 - m) = 4a. Equal areas [WXM]=2m[WXM] = 2m and [WAZ]=4(4a)[WAZ] = 4(4 - a) force m=82a,m = 8 - 2a, so a=8m2.a = \tfrac{8 - m}{2}. Substitute back and (8m)(2m)=0.(8 - m)(2 - m) = 0. Taking MYM \ne Y leaves m=2m = 2 and a=3.a = 3. Then [WMA]=32[WMA] = 32 [WXM]- [WXM] [MYA]- [MYA] [AZW]- [AZW] =32494= 32 - 4 - 9 - 4 =15.= 15. Thus, C is the correct answer.

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