1952 AHSME Problems/Problem 49
Contents
[hide]Problem
In the figure, ,
and
are one-third of their respective sides. It follows that
, and similarly for lines BE and CF. Then the area of triangle
is:
Solution
Let Then
and hence
Similarly,
Then
and same for the other quadrilaterals. Then
is just
minus all the other regions we just computed. That is,
Solution 2
We can force this triangle to be equilateral because the ratios are always , and with the symmetry of the equilateral triangle, it is safe to assume that this may be the truth. Then, we can do a simple coordinate bash. Let
be at
,
be at
, and
be at
. We then create a new point
at the center of everything. It should be noted because of similarity between
and
, we can find the scale factor between the two triangle by simply dividing
by
(nitrous oxide). First, we need to find the coordinates of
and
.
is easily found at
and
be found by calculating equation of
and
.
is located
so
is
.
be at
and the slope is
. We see that they be at the same
-value. Quick maths calculate the x value to be
which be
. Another quick maths caculation of the
-value lead it be equal
which be
. Peferct, so now
be at
. Subtracting the coordinate with the center give you
. I don't even want to do this anymore so here is the answer:
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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