Geometry Problem 1624
Right Triangle, Incircle, Circumscribed Square, Inradius, Area
Problem Statement
In a right triangle $ABC$ with $\angle B = 90^\circ$, consider the square circumscribed about its incircle such that one of its sides lies on the hypotenuse $AC$.
Let $D$ and $E$ be the vertices of the square on the hypotenuse, with $D$ positioned between $A$ and $E$.
If $AD = m$ and $CE = n$:
To Prove:
The area of the square $S$ and the inradius $r$ satisfy:
$$S = 4r^2 = 2mn$$Strategic Synthetic Hints
- Idea 1 — Square Side and Inradius: Note that the distance from the incentro $I$ to the hypotenuse $AC$ is the inradius $r$. Since the circumscribed square has sides parallel and perpendicular to $AC$, its side length is $DE = 2r$, and its area is $S = (2r)^2 = 4r^2$.
- Idea 2 — Similar Triangles at Incenter: Consider the right isosceles triangle $\triangle DIE$ formed by the incenter $I$ and vertices $D, E$. The angles formed at $I$ with vertices $A$ and $C$ yield similar right triangles $\triangle ADI \sim \triangle IEC$.
- Idea 3 — Ratio of Outer Segments: Establish the side ratios from similarity $\frac{AD}{DI} = \frac{IE}{EC}$, which translates to $\frac{m}{r\sqrt{2}} = \frac{r\sqrt{2}}{n}$.
- Idea 4 — Geometric Mean & Area Identity: Cross-multiplying gives $mn = 2r^2$, from which it directly follows that $S = 4r^2 = 2mn$.
Foundation Theorems — Direct Tools
Pattern note: Projecting the circumscribed square onto the hypotenuse directly links the inradius $r$ to outer segments $m$ and $n$, yielding the fundamental geometric mean relation $DE = 2r = \sqrt{2mn}$ and area identity $S = 2mn$.
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