A Property of Points on Incircle

Introduction and acknowledgment

Miguel Ochoa Sanchez (Peru) has posted an intriguing problem on top of a beautiful picture that I was forced to bespoil due to formatting requirements:

Point on incircle - problem

Miguel also posted an elegant solution that is reproduced below.


$P$ is a point on the incircle of $\Delta ABC,$ $x,y,z$ its distances to the sides $BC,$ $AC,$ and $AB,$ respectively. If vertex angles are denoted by the same letters as the vertices, and $x$ is the longest of the three segments, then



The proof depends on the following


Let $M$ be a point on a line tangent to circle $(O)$ with radius $R$ at point $N.$ $T$ a point on the circle such that $MT\perp MN.$ Denote $w=MT.$

Point on incircle - lemma

Then $TN=\sqrt{2Rw}.$

Proof of Lemma

Drop perpendicular $TS$ onto $ON.$

Point on incircle - lemma, solution

In $\Delta MNT,$ $TN^{2}=w^{2}+MN^{2}.$ In $\Delta OST,$ $TS^{2}=R^{2}-(R-w)^{2}=2Rw-w^{2}.$ Combining the two,

$\begin{align} TN^{2}&=w^{2}+MN^{2}\\ &=w^{2}+TS^{2}\\ &=w^{2}+(2Rw-w^{2})\\ &=2Rw, \end{align}$

as required.

Proof of Statement

According to Lemma, $FP=\sqrt{2Rz},$ $PE=\sqrt{2Ry},$ $PD=\sqrt{2Rx}.$

In $\Delta DOE,$ $\displaystyle DE=2R\cos\frac{C}{2}.$ Similarly, $\displaystyle FD=2R\cos\frac{B}{2}$ and $\displaystyle FE=2R\cos\frac{A}{2}.$

Point on incircle - solution>

<p>By <a href=Ptolemy's theorem in the quadrilateral $FPED,$ $PD\cdot FE=PE\cdot FD+FP\cdot DE.$ Substituting the above,

$\displaystyle\sqrt{2Rx}\cdot 2R\cos\frac{A}{2}=\sqrt{2Ry}\cdot 2R\cos\frac{B}{2}+\sqrt{2Rz}\cdot 2R\cos\frac{C}{2},$

and, after simplification


Ptolemy's Theorem

  1. Ptolemy's Theorem
  2. Sine, Cosine, and Ptolemy's Theorem
  3. Useful Identities Among Complex Numbers
  4. Ptolemy on Hinges
  5. Thébault's Problem III
  6. Van Schooten's and Pompeiu's Theorems
  7. Ptolemy by Inversion
  8. Brahmagupta-Mahavira Identities
  9. Casey's Theorem
  10. Three Points Casey's Theorem
  11. Ptolemy via Cross-Ratio
  12. Ptolemy Theorem - Proof Without Word
  13. Carnot's Theorem from Ptolemy's Theorem

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