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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