Difference between revisions of "1967 AHSME Problems/Problem 37"

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== Solution ==
 
== Solution ==
 
<math>\fbox{A}</math>
 
<math>\fbox{A}</math>
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WLOG let <math>RS</math> be the x-axis, or at least horizontal. The three lengths represent the y-coordinates of points <math>A,B,C</math>. As <math>G</math> is by definition the average of <math>A,B,C</math>, it's coordinates are the average of the coordinates of <math>A,B,</math> and <math>C</math>. Hence, the y-coordinate of <math>G</math>, which is also the distance from <math>G</math> to <math>RS</math>, is the average of the y-coordinates of <math>A,B,</math> and <math>C</math>, or <math>\frac{10+6+24}{3}</math>, hence the result.
  
 
== See also ==
 
== See also ==

Latest revision as of 01:04, 8 June 2018

Problem

Segments $AD=10$, $BE=6$, $CF=24$ are drawn from the vertices of triangle $ABC$, each perpendicular to a straight line $RS$, not intersecting the triangle. Points $D$, $E$, $F$ are the intersection points of $RS$ with the perpendiculars. If $x$ is the length of the perpendicular segment $GH$ drawn to $RS$ from the intersection point $G$ of the medians of the triangle, then $x$ is:

$\textbf{(A)}\ \frac{40}{3}\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ \frac{56}{3}\qquad \textbf{(D)}\ \frac{80}{3}\qquad \textbf{(E)}\ \text{undetermined}$

Solution

$\fbox{A}$

WLOG let $RS$ be the x-axis, or at least horizontal. The three lengths represent the y-coordinates of points $A,B,C$. As $G$ is by definition the average of $A,B,C$, it's coordinates are the average of the coordinates of $A,B,$ and $C$. Hence, the y-coordinate of $G$, which is also the distance from $G$ to $RS$, is the average of the y-coordinates of $A,B,$ and $C$, or $\frac{10+6+24}{3}$, hence the result.

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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