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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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 \(\Delta ABC\). – Deeparaj Bhat · 6 months, 1 week ago

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@Sharky Kesa

Do you have an elegant proof? – Deeparaj Bhat · 6 months, 1 week ago

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– Sharky Kesa · 6 months, 1 week ago

No, this is my proof.Log in to reply

– Deeparaj Bhat · 6 months, 1 week ago

Ok. Why the equilateral triangle though?Log in to reply

– Sharky Kesa · 6 months, 1 week ago

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.Log in to reply

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– Sharky Kesa · 6 months, 1 week ago

An affine transformation can be thought of as stretching a plane in multiple manners. Basically, it is a stretch from \((x, y) \mapsto (ax+by+c, dx+ey+f)\) (with \(ad-bc \neq 0\)). There are many properties of an affine transformation, such as any triangle can be mapped to any other triangle, collinear points remain collinear, concurrent line remain concurrent and parallel lines remain parallel.Log in to reply

– Deeparaj Bhat · 6 months, 1 week ago

Ok. These transformations preserve ratio of lengths right? So, how are you going to make an arbitrary triangle an equilateral one?Log in to reply

– Sharky Kesa · 6 months, 1 week ago

No, these transformations only preserve ratios of lengths of parallel segments.Log in to reply

– Deeparaj Bhat · 6 months, 1 week ago

Ah! Ok. So, how do you know that the difference of the sum of the inradii is invariant under such transforms?Log in to reply

– Ameya Daigavane · 5 months, 4 weeks ago

It doesn't even seem likely.Log in to reply