Problem 30 · 2010 Math Kangaroo
Stretch
Geometry & Measurement
reflectionpythagorean-triple
On the two legs of a right-angled triangle (with lengths a and b respectively) points P and Q respectively are chosen. Let K and H be the feet of the perpendiculars from P and Q respectively, to the hypotenuse of the triangle. How big is the smallest possible value of KP + PQ + QH?
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Answer: C — \(\frac{2ab}{\sqrt{a^2+b^2}}\)
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Hint 1 of 3
K and H sit on the hypotenuse; KP and QH are perpendicular to it, so the path KP–PQ–QH is a 'wall-bounce' route.
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Hint 2 of 3
Reflect to straighten a bouncing path: the shortest such path equals the straight-line distance between the reflected endpoints.
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Hint 3 of 3
The hypotenuse length is \(c=\sqrt{a^2+b^2}\) and the altitude to it is \(h=\tfrac{ab}{c}\) — the answer is built from h.
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Approach: straighten the bounce path by reflection
- Put the right angle at the origin with legs along the axes; the hypotenuse has length \(c=\sqrt{a^2+b^2}\) and the triangle's altitude to it is \(h=\frac{ab}{c}\).
- KP and QH are both perpendicular to the hypotenuse, so KP–PQ–QH is a path that leaves the hypotenuse, crosses, and returns — like light bouncing off two parallel walls a distance h apart.
- Reflecting the figure to straighten that bounce, the shortest total length is exactly twice the gap between the walls, i.e. \(2h\).
- Therefore the minimum is \(2h=\dfrac{2ab}{\sqrt{a^2+b^2}}\) — choice C.
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