Difference between revisions of "1950 AHSME Problems/Problem 48"
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begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be <math>ABC</math> with <math>AB=BC=AC=s</math>. Name the point <math>P</math>. Name the altitude from <math>P</math> to <math>BC</math> <math>PA'</math>. Similarly, we will name the other two altitudes <math>PB'</math> and <math>PC'</math>. We can see that | begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be <math>ABC</math> with <math>AB=BC=AC=s</math>. Name the point <math>P</math>. Name the altitude from <math>P</math> to <math>BC</math> <math>PA'</math>. Similarly, we will name the other two altitudes <math>PB'</math> and <math>PC'</math>. We can see that | ||
| − | <cmath>\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh | + | <cmath>\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh</cmath> |
Where h is the altitude. Multiplying both sides by <math>2</math> and dividing both sides by <math>s</math> gives us | Where h is the altitude. Multiplying both sides by <math>2</math> and dividing both sides by <math>s</math> gives us | ||
<cmath>PA'+PB'+PC'=h</cmath> | <cmath>PA'+PB'+PC'=h</cmath> | ||
Revision as of 16:03, 9 May 2015
Problem
A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
Solution
begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be 
with 
. Name the point 
. Name the altitude from to 
. Similarly, we will name the other two altitudes 
and 
. We can see that
![\[\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh\]](http://latex.artofproblemsolving.com/f/7/6/f76759b8329b2e9dd1a58e22ad29072a74ac2215.png)
Where h is the altitude. Multiplying both sides by 
and dividing both sides by 
gives us
![\[PA'+PB'+PC'=h\]](http://latex.artofproblemsolving.com/6/e/a/6ea60170e4d145d40d38631dcc455414c64f2eab.png)
The answer is 
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 47 |
Followed by Problem 49 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
