perplexus.info :: Geometry : Product of Ratios

Product of Ratios (Posted on 2008-09-08)

Let ABC be an arbitrary triangle with circumradius R, incenter I, and inradius r.
Let I_A, I_B, and I_C be the excenters of ABC lying on the bisectors of interior angles A, B, and C respectively.

What is the value of the following product of ratios ?



    |II_A|     |II_B|     |II_C|      r
   ------- x ------- x ------- x ---
    |IA|      |IB|      |IC|      R

Submitted by Bractals

Rating: 4.0000 (1 votes)

Solution: (Hide)

Let a, b, and c be the length of the sides opposite vertices A, B, and C respectively,
Let s be the semiperimeter. Let X and X_A be the projections of I and I_A on ray AB.
From similar right triangles we have
|II_A| |XX_A| a ------- = ------- = ----- |IA| |XA| s-a Similarly, |II_B| b |II_C| c ------- = ----- and ------- = ----- |IB| s-b |IC| s-c Therefore, |II_A| |II_B| |II_C| r abc r ------- x ------- x ------- x --- = ------------------- |IA| |IB| |IC| R (s-a)(s-b)(s-c) R rs abc = ------------------ x ----- (1) s(s-a)(s-b)(s-c) R
From equations (7), (8), and (9) for the area of a triangle at this link we have
abc Δ = ----- = rs = √(s[s-a][s-b][s-c]) (2) 4 R
Applying equation (2) to equation (1) we get
|II_A| |II_B| |II_C| r Δ ------- x ------- x ------- x --- = ---- x 4 Δ |IA| |IB| |IC| R Δ² = 4

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Answer	K Sengupta	2022-03-21 01:16:10
The Geometer's Sketchpad Answer (spoiler)	Charlie	2008-09-08 14:58:29

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