Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

8. In Rt ABC\triangle A B C, the altitude CDC D on the hypotenuse ABA B is 33, extend DCD C to point PP, such that CP=2C P=2, connect APA P, draw BFAPB F \perp A P, intersecting CDC D and APA P at points EE and FF respectively. Then the length of segment DED E is \qquad

A number or a short expression. Spacing and $ signs are ignored.

Solution

8. 95\frac{9}{5}.
As shown in Figure 5, let AD=aA D=a.
In the right triangle ABC\triangle A B C, given CD=3,CDABC D=3, C D \perp A B, we know BD=9aB D=\frac{9}{a}.
Notice that, in the right triangles APDEBD\triangle A P D \sim \triangle E B D,

then DEBD=ADPDDE\frac{D E}{B D}=\frac{A D}{P D} \Rightarrow D E =9a×a5=95=\frac{9}{a} \times \frac{a}{5}=\frac{9}{5}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.