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\\
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<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==

Latest revision as of 00:16, 2 November 2023

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:

$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}$

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 $ABC$ with $AB=BC=AC=s$. We will call the aforementioned point $P$. Call altitude from $P$ to $BC$ $PA'$. Similarly, we will name the other two altitudes $PB'$ and $PC'$. We can see that \[\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh\] Where h is the altitude. Multiplying both sides by $2$ and dividing both sides by $s$ gives us \[PA'+PB'+PC'=h\] The answer is $\textbf{(C)}$

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 47
Followed by
Problem 49
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All AHSME Problems and Solutions

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