Difference between revisions of "1950 AHSME Problems/Problem 48"
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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: | 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: | ||
| − | <math>\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ | + | <math> \textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity} </math> |
| − | \textbf{(B)}\ \text{Greater than the altitude of the triangle} \qquad\\ | ||
| − | \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ | ||
| − | \textbf{(D)}\ \text{One-half the sum of the sides of the triangle} \qquad\\ | ||
| − | \textbf{(E)}\ \text{Greatest when the point is the center of gravity}</math> | ||
==Solution== | ==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 <math>ABC</math> with <math>AB=BC=AC=s</math>. We will call the aforementioned point <math>P</math>. Call 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> | ||
| + | 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> | ||
| + | The answer is <math>\textbf{(C)}</math> | ||
| + | ==Note== | ||
| + | Thie result is exactly the Viviani theorem. | ||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=l4lAvs2P_YA&t=251s | ||
| + | |||
| + | ~MathProblemSolvingSkills.com | ||
| + | |||
==See Also== | ==See Also== | ||
Latest revision as of 00:16, 2 November 2023
Contents
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 
. We will call the aforementioned point 
. Call 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 
Note
Thie result is exactly the Viviani theorem.
Video Solution
https://www.youtube.com/watch?v=l4lAvs2P_YA&t=251s
~MathProblemSolvingSkills.com
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 47 |
Followed by Problem 49 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
