1969 Canadian MO Problems/Problem 4
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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
Let a side of the triangle be 
and let ![[ABC]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/b/7/8b78d271c3d197e48980793383e67d5fd4a27dfc.gif)
denote the area of 
Note that because ![2[\triangle ABC]=2[\triangle APB]+2[\triangle APC]+2[\triangle BPC],](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/9/a/3/9a38b9ebc5fc1043a9e05de496915f9238d6fb04.gif)
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 
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 3 | 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | Followed by Problem 5 |
