Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Find the answer

5. As shown in Figure 1, in the right triangle $\triangle ABC$ , $AC=1, BC=$ $2, P$ is a moving point on the hypotenuse $AB$ . Draw $PE \perp BC, PF$ $\perp AC$ , with the feet of the perpendiculars being $E$ and $F$ , respectively, and connect $EF$ . Then the minimum length of the line segment $EF$ is $\qquad$

Your answer

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

Solution

5. $\frac{2}{\sqrt{5}}$ .

From the problem, we know that the hypotenuse $A B=\sqrt{5}$ .
Since $E F=C P$ , therefore,
$E F$ is shortest $\Leftrightarrow C P$ is shortest $\Leftrightarrow C P \perp A B$ . Hence, the minimum length of $E F$ is $\frac{1 \times 2}{\sqrt{5}}=\frac{2}{\sqrt{5}}$ .

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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.