Problem 24 · 2016 Math Kangaroo
Stretch
Geometry & Measurement
pythagorean-triple
In the right-angled triangle ABC (with the right angle at A) the angle bisectors of the acute angles intersect at point P. The distance of P to the hypotenuse is \(\sqrt{8}\). What is the distance of P to A?
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Answer: E — 4
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Hint 1 of 3
Two angle bisectors meeting is the incentre, so its distance to every side is the same inradius.
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Hint 2 of 3
At the right angle A, the incentre sits on the bisector of a \(90^\circ\) angle, a \(45^\circ\) line from each leg.
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Hint 3 of 3
Relate AP to the inradius using that \(45^\circ\) geometry.
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Approach: incentre geometry at the right angle
- The two acute-angle bisectors meet at the incentre P, so its distance to the hypotenuse is the inradius \(r = \sqrt{8}\).
- P is also distance \(r\) from each leg, so from the right-angle vertex A it lies along the \(45^\circ\) bisector at distance \(r\sqrt{2}\).
- \(AP = \sqrt{8}\cdot\sqrt{2} = \sqrt{16} = 4\), answer E.
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