2018 AMC 12A Problem #11

Geometry Level 2

A paper triangle with sides of lengths 3, 4, and 5 inches, as shown, is folded so that point A A falls on point B . B. What is the length in inches of the crease?

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15 8 \dfrac{15}{8} 2 1 + 1 2 2 1 + \dfrac{1}{2}\sqrt{2} 7 4 \dfrac{7}{4} 3 \sqrt{3}

Steven Yuan by Steven Yuan , 20, USA

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1 solution

Richard Costen
Feb 9, 2018

In △ A B C , tan ⁡ A = 3 4 . In △ A D E , tan ⁡ A = x 2.5 . ∴ 3 4 = x 2.5 . \text{In }\triangle ABC, \tan A=\frac34. \text{ In } \triangle ADE, \tan A=\frac{x}{2.5}. \quad\therefore \frac34 = \frac{x}{2.5}. Solving yields x = 15 8 x=\boxed{\frac{15}{8}}

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I've seen this exact problem on AMC10. I forgot which year.

genius kid - 1 year ago

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