Geometry Problem 1618: Right Triangle, Altitude, Semicircle, and the Reciprocal Invariant Identity of Mixtilinear Incircles

Geometry Problem 1618: Right Triangle ABC with Altitude BH, Semicircle on AC, and Mixtilinear Incircles

Figure: The invariant relationship between the reciprocals of the inradii and the altitude BH in a mixtilinear configuration.

Problem Statement:

In a right triangle ABC ($\angle ABC = 90^\circ$), let BH be the altitude to the hypotenuse AC. A semicircle is constructed with diameter AC, and the segments AB and BC are circular arcs of this semicircle that form mixtilinear triangles AHB and BHC.

Definitions:

r₁ and R₁ are the radii of the incircles of right triangles AHB and BHC, respectively.
r₂ and R₂ are the radii of the incircles of the mixtilinear triangles AHB and BHC.

To Prove: The difference between the reciprocals of the radii of the incircle and the mixtilinear incircle is invariant for both triangles and equals the reciprocal of the altitude BH:

$$\frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{BH}$$

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Strategic Hints:

• The Scaling Bridge: Compare how the incircle and the mixtilinear incircle "fit" within the same corner of the triangle. What determines the ratio between their sizes?
• Self-Similarity: Observe that ▵AHB and ▵BHC are both similar to the parent ▵ABC. How can this shared structure help you express all radii using common segments?
• The Role of Tangency: Focus on the point where the mixtilinear incircle touches the semicircle. How does the distance from the vertex to this tangency point interact with the altitude BH?
• Discovering the Invariant: Combine the expressions for the two types of circles. What specific geometric property of the right triangle causes the side-length variables to vanish, leaving only the altitude in the final relation?