Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.8 Junior Find the answer

In right $\triangle ABC$ with legs $5$ and $12$ , arcs of circles are drawn, one with center $A$ and radius $12$ , the other with center $B$ and radius $5$ . They intersect the hypotenuse in $M$ and $N$ . Then, $MN$ has length

Pick one

Solution

First of all, from the Pythagorean Theorem, $AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$ . Also, since $AM$ and $AC$ are radii of the same circle, $AM=AC=12$ . Therefore, $MB=AB-AM=13-12=1$ . Also, since $BN$ and $BC$ are radii of the same circle, $BN=BC=5$ . We therefore have $MN=BN-BM=5-1=4, \boxed{\text{D}}$ .

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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.