To commemorate my 6000th solved problem, which was this, I have decided to post this proof problem.

Let \(P\) be a point inside equilateral triangle \(ABC\). Let \(A'\), \(B'\) and \(C'\) be the projections of \(P\) onto sides \(BC\), \(CA\) and \(AB\) respectively. Prove that the sum of lengths of the inradii of triangles \(PAC'\), \(PBA'\) and \(PCB'\) equals the sum of lengths of the inradii of triangles \(PAB'\), \(PBC'\) and \(PCA'\).

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## Comments

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TopNewestHere's a proof, not very elegant though.

Let \(r_{\Delta XYZ}\) denote the inradius of \(\Delta XYZ\).

If \(\Delta XYZ\) is right angled (\(\angle XYZ=90^\circ\)), then \[r_{\Delta XYZ}=\frac{XY+YZ-ZX}{2}\]

The proof of this is given as a note below.

Using this, we have the following equations:

\[\begin{align} r_{\Delta PAC'}&=\frac{PC'+AC'-PA}{2} \\r_{\Delta PAB'}&=\frac{PB'+AB'-PA}{2} \\r_{\Delta PBA'}&=\frac{PA'+BA'-PB}{2} \\r_{\Delta PBC'}&=\frac{PC'+BC'-PB}{2} \\r_{\Delta PCB'}&=\frac{PB'+CB'-PC}{2} \\r_{\Delta PCA'}&=\frac{PA'+CA'-PC}{2}\end{align}\]

The conclusion now follows.

Note:If \(\Delta\) denotes the area of \(\Delta XYZ\) and \(s\) denotes it's semi-perimeter, then \[r_{\Delta XYZ}=\frac{\Delta}{s}\] Where \(\angle XYZ=90^\circ\).

Now, we know that \(\Delta = \dfrac{1}{2} \cdot XY \cdot ZY\)

So, \[\begin{align} r_{\Delta XYZ}&=\frac{XY \cdot ZY}{XY+YZ+ZX} \\\implies r_{\Delta XYZ}&=\frac{XY \cdot ZY \cdot (XY+YZ-ZX)}{(XY+YZ)^2-ZX^2} \\\implies r_{\Delta XYZ}&=\frac{XY+YZ-ZX}{2} \end{align}\]

The third implication follows by Pythagoras' Theorem.

The proof is now complete.

Also, notice that the fact that \(\Delta ABC\) was equilateral was not required. The result holds for any arbitrary triangle \(\Delta ABC\).

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@Sharky Kesa

Do you have an elegant proof?

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No, this is my proof.

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Comment deleted May 29, 2016

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