A geometry problem for the Regional Mathematical Olympiad

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Problem:

Prove that in any acute angled triangle of sides a, b, c, semi perimeter p, in-radius r, and circumradius R, the following inequalities hold:

\frac{2}{5} \leq \frac{Rp}{2aR+bc} < \frac{1}{2}

Proof:

Let D be the foot of the altitude from A, and D will be on the side BC since the triangle has only acute angles. Now, by summing up AD+BD>AB and CD+AD>AC, we get

2h_{a}+a>b+c \Longrightarrow h_{a}>p-a \Longrightarrow \frac{2S}{a}>p-a \Longrightarrow 2S > a(p-a) \Longrightarrow \frac{abc}{2R} > a(p-a) \Longrightarrow bc+2aR > 2pR \Longrightarrow \frac{pR}{bc+2aR}<\frac{1}{2}

For the other part, we have the following equivalences:

\frac{2}{5} \leq \frac{Rp}{2aR+bc} \Longleftrightarrow 2aR+bc \leq \frac{5pR}{2} \Longleftrightarrow 4aR + \frac{8SR}{a} \leq 5pR \Longleftrightarrow 4a^{2}+4ah_{a} \leq 5ap \Longleftrightarrow (a+h_{a}) \leq \frac{5p}{4}

—– call the above as relation I

But, b=\sqrt{h_{a}^{2}+BD^{2}} and c=\sqrt{h_{a}^{2}+CD^{2}} and by Minkowski’s inequality, we have b+c \geq \sqrt{4h_{a}^{2}+a^{2}}. Then, we will have

\frac{5p}{4} = \frac{5(a+b+c)}{8} \geq \frac{5(a+\sqrt{4h_{a}^{2}+a^{2}})}{8} \geq a+h_{a} \Longleftrightarrow 5\sqrt{4h_{a}^{2}+a^{2}} \geq 3a +8h_{a} \Longleftrightarrow 100h_{a}^{2}+25a^{2} \geq 64h_{a}^{2}+9a^{2}+48ah_{a} \Longleftrightarrow 36h_{a}^{2}+16a^{2} \geq 48ah_{a}

which holds for all positive a and h_{a} because of the AM-GM inequality. This tells us that (1) is true, and thus so is our conclusion.

Ref: Problems for the Mathematical Olympiads (from the First Team Selection Test to the IMO) by Andrei Negut

More geometry later !

Nalin Pithwa

 

 

 

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