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6000th Brilliant Problem Solved - Proof Problem

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 \(PCA'\).

Note by Sharky Kesa
1 year, 5 months ago

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Here'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 - 1 year, 5 months ago

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@Sharky Kesa

Do you have an elegant proof?

Deeparaj Bhat - 1 year, 5 months ago

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No, this is my proof.

Sharky Kesa - 1 year, 5 months ago

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@Sharky Kesa Ok. Why the equilateral triangle though?

Deeparaj Bhat - 1 year, 5 months ago

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@Deeparaj Bhat 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 - 1 year, 5 months ago

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Comment deleted May 29, 2016

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@Deeparaj Bhat 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.

Sharky Kesa - 1 year, 5 months ago

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@Sharky Kesa Ok. These transformations preserve ratio of lengths right? So, how are you going to make an arbitrary triangle an equilateral one?

Deeparaj Bhat - 1 year, 5 months ago

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@Deeparaj Bhat No, these transformations only preserve ratios of lengths of parallel segments.

Sharky Kesa - 1 year, 5 months ago

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@Sharky Kesa Ah! Ok. So, how do you know that the difference of the sum of the inradii is invariant under such transforms?

Deeparaj Bhat - 1 year, 5 months ago

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@Deeparaj Bhat It doesn't even seem likely.

Ameya Daigavane - 1 year, 5 months ago

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