Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a triangle $ABC$, let $m_{a}$, $h_{a}$ be $A$-median, $A$-altitude.
Define $m_{b}$,$h_{b}$ and $m_{c}$,$h_{c}$ likewise.

Prove that $\dfrac{h_{a}}{m_{b}}+\dfrac{h_{b}}{m_{c}}+\dfrac{h_{c}}{m_{a}}\leq 3$

I have no solution.

share|cite|improve this question
    
The only reference I can find is: Walther Janous, "Further Inequalities of Erdos-Mordell Type", Forum Geometricorum 4 (2004), pp. 203–206 (PDF), citing p. 315 of D. S. Mitrinovic et al., Recent Advances in Geometric Inequalities, (Kluwer 1989; there's an expensive Springer reprint), which attributes to Klamkin and Meir the result that $\frac{\overline{h_1}}{m_1} + \frac{\overline{h_2}}{m_2} + \frac{\overline{h_3}}{m_3} \leqslant 3$, where $(\overline{h_1}, \overline{h_2}, \overline{h_3})$ is any permutation of $(h_1, h_2, h_3)$. – Calum Gilhooley Mar 25 at 0:08

Denote the side of the triangle opposite vertex $ A $ as $ a $. Now consider the points on side $ a $ where $ m_a,h_a $ intersect side $ a $. Label the intersection of $ a $ and $ h_a $ as the point $ X $, and label the intersection of $ a $ and $ m_a $ as the point $ Y $. Then the triangle $ XYA $ is a right triangle. The ratio $ \dfrac{h_a}{m_a} $ is actually the sine of the angle that $ m_a $ makes with side $ a $. Since the sine is always less than or equal to 1, we see that $ \dfrac{h_a}{m_a} \leq 1 $. The same argument shows that $ \dfrac{h_b}{m_b} \leq 1 $ and $ \dfrac{h_c}{m_c} \leq 1 $. Taking the sum of all three inequalities, you have your result.

share|cite|improve this answer
2  
Not $\dfrac{h_{a}}{m_{a}}$ but $\dfrac{h_{a}}{m_{b}}$ – chloe_shi Jun 11 '13 at 16:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.