1952 AHSME Problems/Problem 49
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:
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,
Alternative but very similar Solution
Let Then and hence Similarly, Then we can implement a similar but different area addition postulate to the first solution. It will be (PIE in action). Using transitive property Subtracting and adding on both sides gives: ~many credits to the first solution ~Lopkiloinm
Solution 2 (best solution)
We can force this triangle to be equilateral because the ratios are always no matter which rotation, and with the symmetry of the equilateral triangle, it is safe to assume that this may be the truth (symmetry is very important, kids). 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 (not nitrous oxide) by . 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 this coordinate with the coordinates of the center give you . Pythagorean theorem on that gives which simplifies to is vertically above by units. The scale factor is thus giving us an answer of: ~Lopkiloinm (Note: the presence of in the denominator gives hints on the answer, so when you see it, 1/7 looks like the obvious choice) (Another note: the question gives you the ratio of so I did not need to use that many steps to calculate the coordinate of , directly attaining it to be at After realizing this, this method should actually be pretty quick, may even be quicker than the non coordinate method. Regardless, if you don't get the ratio and you encounter a problem similar to this, the coordinate bash with all these calculations should get you the correct answer. Also, there is an even bashier way using all the points and shoelace formula. That takes too much time.)
Visual graphical proof. https://en.m.wikipedia.org/wiki/One-seventh_area_triangle#/media/File%3ATriangleOneSeventhAreaGraphicalSoln.png
Anybody down to write a solution using wooga looga theorem?
Solution 5 (Super fast)
We can use triangle area ratios. .
So let . Then .
Similarly, . So . Similarly, we can use this argument to find . So . So .
So therefore, since , select .
You can also use a similar idea (Principle of Inclusion-Exclusion) to find . Notice that the triangle we desire is the intersection set . So we just PIE it out.
|1952 AHSC (Problems • Answer Key • Resources)|
|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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50|
|All AHSME Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.