Difference between revisions of "1966 AHSME Problems/Problem 38"
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== Solution == | == Solution == | ||
+ | Construct triangle <math>\triangle ABC</math> with points <math>M,N,P</math> being the midpoints of sides <math>\overline{CB}, \overline{AB}, \overline{AC}</math>, respectively. Proceed by drawing all medians. Then draw all medians (so draw <math>\overline{AM}, \overline{BP}, \overline{CN}</math>). Next, draw line <math>\overline{PM}</math> and label <math>\overline{PM}</math>'s intersection with <math>\overline{CN}</math> as the point <math>Q</math>. From the problem, the area of <math>\triangle QMO</math> is <math>n</math>, but by vertical angles we know that <math>\angle QOM = \angle AOM</math>. Furthermore, since line <math>\overline{PM}</math> is drawn from the midpoint of <math>\overline{AC}</math> to the midpoint of <math>\overline{CB}</math>, we know that <math>\overline{PM}</math> is parallel to <math>\overline{AB}</math> (via SAS similarity on triangles PCM and ABC). From these parallel lines we know that <math>\angle PMO = \angle OAB</math> which indicates that <math>\triangle QOM \sim \triangle ANO</math>. The linear ratio from <math>\triangle QOM</math> to <math>\triangle ANO</math> is 1:2 because line segment <math>\overline{OM}</math> is one half of line segment <math>\overline{AO}</math> since <math>\overline{AO}</math> and <math>\overline{OM}</math> make up the median <math>\overline{AM}</math>. Thus the area ratio is 1:4. So <math>\triangle ANO</math> has area <math>4n</math>. Since <math>\triangle ONB</math> has the same height and base as <math>\triangle ANO</math> we know that the area of <math>\triangle AOB = 8n</math>. The medians form 3 triangles each with area <math>1/3</math> of the total triangle (these triangles are <math>\triangle AOB, \triangle COB, \triangle COA</math>). Thus since <math>\triangle AOB = 4n \underset{\text{multiply by 3}}{\implies} \triangle ABC = 24n</math> <math>\fbox{D}</math>. | ||
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+ | - LJ | ||
== See also == | == See also == |
Latest revision as of 19:21, 3 April 2023
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 | |
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 | ||
All AHSME Problems and Solutions |
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