Geometry Problem 1625
Isosceles Right Triangle, Interior Point, Rotation, Invariant Angle, Altitude
Problem Statement
Let $ABC$ be an isosceles right triangle with $\angle B = 90^\circ$, and let $P$ be an interior point. Given that $BP = 4$ and $PA^2 - PC^2 = 32$, evaluate the length of the altitude $BH$ dropped from vertex $B$ to the line $PC$.
Strategic Synthetic Hints
- Idea 1 — Rotation of Triangle: Rotate the triangle $\triangle ABP$ by $90^\circ$ around vertex $B$ to form a congruent auxiliary triangle, placing segment $PA$ as $CP'$ and $BP$ as $BP' = 4$.
- Idea 2 — The Invariant Angle: Using the relation $PA^2 - PC^2 = 32$ alongside the auxiliary right-isosceles triangle $\triangle BPP'$ ($PP' = 4\sqrt{2}$), apply the Pythagorean theorem converse to deduce that $\angle BPC = 135^\circ$.
- Idea 3 — Notable Right Triangle for Altitude: The altitude $BH$ dropped to the extension of side $PC$ forms a $45^\circ$-$45^\circ$-$90^\circ$ right triangle $\triangle BPH$ with hypotenuse $BP = 4$.
- Idea 4 — Final Calculation: Directly compute the altitude length as $BH = \frac{BP}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Foundation Theorems — Direct Tools
Pattern note: Combining a $90^\circ$ rotation with algebraic distance invariants reveals a constant angle of $135^\circ$, reducing complex calculations to a straightforward $45^\circ$-$45^\circ$-$90^\circ$ triangle relation.
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