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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:
  
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<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>
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\textbf{(B)}\ \text{Greater than the altitude of the triangle} \qquad\\
 
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\textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\
 
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\textbf{(D)}\ \text{One-half the sum of the sides of the triangle} \qquad\\
 
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\textbf{(E)}\ \text{Greatest when the point is the center of gravity}</math>
 
  
 
==Solution==
 
==Solution==
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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>.  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  
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<cmath>\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh)</cmath>
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<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>
 
The answer is <math>\textbf{(C)}</math>
 
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

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 (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

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