1969 Canadian MO Problems/Problem 4
Problem
Let be an equilateral triangle, and be an arbitrary point within the triangle. Perpendiculars are drawn to the three sides of the triangle. Show that, no matter where is chosen, .
Solution 1
Let a side of the triangle be and let denote the area of Note that because Dividing both sides by , the sum of the perpendiculars from equals (It is independant of point ) Because the sum of the sides is , the ratio is always
Solution 2
Let a be the side of the triangle. By Viviani's theorem (which has a similar proof to solution 1) PD + PE + PF = h where h is the height of the equilateral triangle. Now it is not hard to see that the ratio h : 3 * a = 1 / 2 root 3 from 30 - 60 - 90 triangles or trigonometry. ~AK2006
1969 Canadian MO (Problems) | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 5 |