6000th Brilliant Problem Solved - Proof Problem

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

Let PP be a point inside equilateral triangle ABCABC. Let A′A', B′B' and C′C' be the projections of PP onto sides BCBC, CACA and ABAB respectively. Prove that the sum of lengths of the inradii of triangles PAC′PAC', PBA′PBA' and PCB′PCB' equals the sum of lengths of the inradii of triangles PAB′PAB', PBC′PBC' and PCA′PCA'.

#Geometry

Note by Sharky Kesa
5 years ago

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Comments

Here's a proof, not very elegant though.

Let rΔXYZr_{\Delta XYZ} denote the inradius of ΔXYZ\Delta XYZ.

If ΔXYZ\Delta XYZ is right angled (∠XYZ=90∘\angle XYZ=90^\circ), then rΔXYZ=XY+YZ−ZX2r_{\Delta XYZ}=\frac{XY+YZ-ZX}{2}

The proof of this is given as a note below.

Using this, we have the following equations:

rΔPAC′=PC′+AC′−PA2rΔPAB′=PB′+AB′−PA2rΔPBA′=PA′+BA′−PB2rΔPBC′=PC′+BC′−PB2rΔPCB′=PB′+CB′−PC2rΔPCA′=PA′+CA′−PC2\begin{aligned} 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{aligned}

The conclusion now follows.

Note:

If Δ\Delta denotes the area of ΔXYZ\Delta XYZ and ss denotes it's semi-perimeter, then rΔXYZ=Δsr_{\Delta XYZ}=\frac{\Delta}{s} Where ∠XYZ=90∘\angle XYZ=90^\circ.

Now, we know that Δ=12⋅XY⋅ZY\Delta = \dfrac{1}{2} \cdot XY \cdot ZY

So, rΔXYZ=XY⋅ZYXY+YZ+ZX  ⟹  rΔXYZ=XY⋅ZY⋅(XY+YZ−ZX)(XY+YZ)2−ZX2  ⟹  rΔXYZ=XY+YZ−ZX2\begin{aligned} 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{aligned}

The third implication follows by Pythagoras' Theorem.

The proof is now complete.

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

A Former Brilliant Member - 5 years ago

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@Sharky Kesa

Do you have an elegant proof?

A Former Brilliant Member - 5 years ago

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No, this is my proof.

Sharky Kesa - 5 years ago

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@Sharky Kesa – Ok. Why the equilateral triangle though?

A Former Brilliant Member - 5 years ago

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@A Former Brilliant Member – I was hoping someone would have a proof specific for an equilateral triangle. Then, if it was asked for an arbitrary triangle, we could perform an affine transformation so it becomes an equilateral triangle, then use said proof.

Sharky Kesa - 5 years ago
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