1966 AHSME Problems/Problem 38
Problem
In triangle 
the medians 
and 
to sides 
and 
, respectively, intersect in point 
. 
is the midpoint of side 
, and 
intersects in 
. If the area of triangle 
is 
, then the area of triangle is:
Solution
Construct triangle 
with points 
being the midpoints of sides 
, respectively. Proceed by drawing all medians. Then draw all medians (so draw 
). Next, draw line 
and label 's intersection with 
as the point . From the problem, the area of 
is , but by vertical angles we know that 
. Furthermore, since line is drawn from the midpoint of 
to the midpoint of 
, we know that is parallel to 
(via SAS similarity on triangles PCM and ABC). From these parallel lines we know that 
which indicates that 
. The linear ratio from 
to 
is 1:2 because line segment 
is one half of line segment 
since and make up the median 
. Thus the area ratio is 1:4. So has area 
. Since 
has the same height and base as we know that the area of 
. The medians form 3 triangles each with area 
of the total triangle (these triangles are 
). Thus since 
.
- LJ
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 37 |
Followed by Problem 39 | |
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