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A circle touches two sides of a triangle and two of its medians. Prove that the triangle is isosceles.

Figure

This problem came from the Mathematical Digest issue 62 (Jan 1986) which in turn cited a Russian mag called KVANT (meaning "Quantum").

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    $\begingroup$ Assume it's not true. Then you wouldn't ask the question. A contradiction. Therefore it's true. QED. :) $\endgroup$
    – SteveV
    Mar 13 '21 at 1:02
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As the triangles ADC and BEC have the same incircle and their areas are equal (half that of ABC), so are their perimeters
DC+DA+AC = EC+EB+BC
or, subtracting from both sides CD+CE+DA+EB
AE-EB = BD-DA.
This means that D and E lie on the same pair of hyperbolas with foci A and B. Since they also have the same distance to the base AB (half that of C) the triangle must be isosceles by symmetry.

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    $\begingroup$ It took me a moment to see why they have the same distance to the base, but that should have been obvious. This is a really clever proof. $\endgroup$ Mar 13 '21 at 8:38
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    $\begingroup$ @JaapScherphuis It's short ;-) I've added a half sentence explaining (if you can call it that) the distance to the base. $\endgroup$
    – loopy walt
    Mar 13 '21 at 9:50
  • $\begingroup$ That is clever. I don't think I've ever seen anyone use a hyperbola locus in a proof like this! $\endgroup$
    – Dr Xorile
    Mar 13 '21 at 17:34
  • $\begingroup$ Elegant, indeed. $\endgroup$ Mar 16 '21 at 6:16
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Alternative Proof

By Pitot Theorem we have $|CE|+|MD| = |CD|+|ME|$ and since the centroid divides each median in the ratio 2:1, this means that $$\frac{1}{2}|AC| + \frac{1}{3}|AD| = \frac{1}{2}|BC|+\frac{1}{3}|BE|$$ Now let $|AB| = c, |BC| = a, |CA|=b, |AD|=m_a, |BE|=m_b$.
Then Apollonius' Theorem tells us that $$m_a = \sqrt{\frac{2b^2+2c^2-a^2}{4}}\,\,\,\,,\,\,\,\,m_b = \sqrt{\frac{2a^2+2c^2-b^2}{4}}$$ Substituting this in above and multiplying across by $6$ yields $$ 3a + \sqrt{2b^2+2c^2-a^2} = 3b + \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 3(a-b) + \sqrt{2b^2+2c^2-a^2} = \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 9(a-b)^2 + 6(a-b)\sqrt{2b^2+2c^2-a^2} + 2b^2+2c^2-a^2 = 2a^2+2c^2-b^2$$ $$\Rightarrow (a-b)\left(9(a-b) + 6\sqrt{2b^2+2c^2-a^2} - 3(a+b)\right) = 0$$ which means either $a=b$ or $\sqrt{2b^2+2c^2-a^2} = 2b-a$.
If it is the latter then $$2b^2+2c^2-a^2 = 4b^2-4ab+a^2 $$ $$\Rightarrow c^2 = b^2-2ab+a^2 = (b-a)^2$$ $$\Rightarrow c = \pm(b-a) $$ which only happens for a degenerate triangle (does not satisfy the triangle inequality).
Hence $a=b$

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@loopwalt and @hexomino both gave excellent answers. I wanted to share the answer from the Mathematical Digest, because it's also elegant in its own way:

It starts the same way as @hexomino's proof (I didn't know it was called Pitot's Theorem!):

Since $CEMD$ circumscribes a circle, $CE+MD = CD + ME$ (Pitot's Theorem, but fun to prove). So (as in hexomino's proof):
$$\frac{1}{2}AC+\frac{1}{3}AD = \frac{1}{2}BC+\frac{1}{3}BE \tag1$$
Next, note that $\triangle ADC$ and $\triangle BEC$ have the same area ($\frac{1}{2}\triangle ABC$) and share a common incircle. This means that their perimeters are equal (another fun to prove and left as an exercise!). So $AD+\frac{1}{2}BC+AC = BE+\frac{1}{2}AC+BC$, which gives:
$$\frac{1}{2}AC+AD = \frac{1}{2}BC+BE \tag2$$
Now, $3\times(1)-(2)$ gives $AC=BC$

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