# Geometric inequality (I.S.I. B.Stat 2005 Problem 5 solution)

Consider an acute angled triangle PQR such that C,I and O are the circumcentre, incentre and orthocentre respectively. Suppose $$\displaystyle{ \angle QCR, \angle QIR }$$ and $$\displaystyle{ \angle QOR }$$, measured in degrees, are $$\displaystyle{ \alpha, \beta }$$ and $$\gamma$$ respectively. Show that $$\displaystyle { \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} > \frac{1}{45} }$$

Discussion:

Teacher:

Use this diagram. Angle chasing should suffice. Assume the that the measure of $$\angle PQR = Q , \angle RPQ = P , \angle PRQ = R$$

Student: $$\alpha = 2P$$ since it is the angle at the center (twice the angle at circumference). Again $$\displaystyle { \angle IQR = \frac{Q}{2} , \angle IRQ = \frac {R}{2} \implies \beta = 180^o – \frac{Q+R}{2} = 90^o + \frac{P}{2} }$$
Finally $$\angle BQR = \angle OQR = 90^o -R, \angle ORQ = 90^o – Q \implies \gamma = R+Q$$

Hence $$\displaystyle { \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{1}{2P}+\frac{1}{90 + \frac{P}{2}}+\frac{1}{180-P} }$$

But here is a problem. If I maximize P (by replacing it 90), it adversely affects 180 – P as Pis negative there. Somehow we must balance the value of P.

Teacher: Right. If you replace P by 90, 1/2P will remain greater than 1/180, but 1/90 will be less than 1/180-P.
Here we need to use Arithmetic Mean > Harmonic Mean inequality which state for positive numbers a, b, c we have $$\displaystyle { \frac{3}{\frac{1}{a}+ \frac{1}{b} + \frac{1}{c} } \le \frac{a+b+c}{3} \implies \frac{9}{a+b+c} \le \frac{1}{a}+ \frac{1}{b} + \frac{1}{c} }$$

$$\displaystyle { \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{1}{2P}+\frac{1}{90 + \frac{P}{2}}+\frac{1}{180-P} \ge \frac{9} { 2P +90 + \frac{P}{2} + 180 – P } }$$

Student: Ok. I think I can do it from here.

Since it is an acute angled triangle, $$P \le 90^o$$
Hence replacing P by 90, we get

$$\displaystyle { \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} > \frac{9} { 90 +90 + \frac{90}{2} + 180 } = \frac{9}{9 \times 45} = \frac{1}{45} }$$

Teacher: Just one point of caution; all numbers are in degrees. Also mention that ‘increasing the denominator decreases a number’.