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The problem is from Kiselev's Geometry Exercise 587:

Prove that in a scalene triangle, the sum of the diameters of the inscribed and circumscribed circle is congruent to the sum of the segments of the altitudes from the orthocenter to the vertices.

Here is what I have tried: let $a, b, c$ be its sides, $r$ be the radius of the incircle, $R$ be that of the excircle, $h_a$ be the altitude perpendicular to $a$, $h_a'$ be the segment of $h_a$ from the orthocenter to $A$, the vertex on the other side of $a$. Let $S$ be the area of the triangle. Then

$\displaystyle\frac{a+b+c}{2}r = \frac{abc}{4R} = \frac{a}{2}h_a = \frac{b}{2}h_b = \frac{c}{2}h_c$

From the previous exercise 585, we also have $\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c} = \frac{1}{r}$, although I am not sure whether it will be useful.

Then I tried to prove it algebraically by using the formulas above, but it became very complex and I could not reduce the sum on the right side into the sum of the diameters of the incircle and the excircle.

Any help would be greatly appreciated.

Taxxi
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2 Answers2

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If you take the midpoints of $AB,BC$ and $AC$ to be $M_3,M_1$ and $M_2$, then $AH$ is twice as long as $OM_1$ and the same goes for the rest; $OM_1=R\cos A$.

Now, what you have to prove is that $R(\cos A+\cos B+\cos C)=r+R$, which is the same as $ \cos A+\cos B+\cos C=1+\frac{r}{R}$ and that is a well known equality.

SarGe
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Калин Върбанов
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  • Thank you for your help. I was able to understand the first part using Euler line (it is not in my book but I learned it through one of the exercises), but I am not sure if it is the standard way of deriving that $AH$ is twice as long as $OM1$ (is Euler line overkill?) As for the second one, I think understanding it as a corollary of Carnot's theorem as in here is the best idea. Do you have any opinion or idea on these topics to help me further? – Taxxi Jul 30 '20 at 12:58
  • Well, actually Carnot's theorem directly proves this. – Taxxi Jul 30 '20 at 13:04
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    Another way to solve it is to take the diameter AA0 and then to show that HBA0C is a parallelogram, you have many right angles there and once you have that it is indeed a parallelogram you know that M1 is the midpoint of HA0 and so OM1 is half as long as AH. – Калин Върбанов Jul 30 '20 at 13:05
  • Thank you so much; it took some time, but I was indeed able to prove it by that method as well. – Taxxi Jul 30 '20 at 13:52
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Hint: Let $H$ be the orthocenter of $\triangle ABC$, then $HA=2R\cos A$. Also, use $\displaystyle\cos A+\cos B+\cos C=1+\frac{r}{R}$.

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In $\triangle AHL$, we've $$\begin{align} \cos(90°-C)&=\frac{AL}{AH}\\ AH&=c\cos A\ \mathrm{cosec} C\tag{$\because$ In $\triangle ALB$, $\displaystyle\cos A=\frac{AL}{AB}$}\\ AH&=2R\cos A\tag{$\displaystyle\because R=\frac{c}{2\sin C}$}\end{align}$$

Symbols have their usual meaning.

SarGe
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  • I have found the proof of the second one, but I was not able to prove $HA=2R \cos A$. Could you help me a bit more on the first part? – Taxxi Jul 30 '20 at 12:22
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    @Taxxi, I've edited the solution. Let me know if there's still any confusion. – SarGe Jul 30 '20 at 14:02
  • Thank you; this method is really great since no other gadgetry is required. I think in the second line of the proof, $\cos C$ should be $\cos A$ and vice versa for cosec. I just have chosen the other answer while you were editing; hope you will receive more votes afterwards though. – Taxxi Jul 30 '20 at 14:19
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    @Taxxi, yes I corrected it. No worries regarding accepting the answer. :-) – SarGe Jul 30 '20 at 14:20
  • Thank you for understanding :) btw I think cosecA should be cosec $C$ as well. – Taxxi Jul 30 '20 at 14:22