1969 Canadian MO Problems/Problem 4
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, .
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
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
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