Difference between revisions of "1952 AHSME Problems/Problem 49"
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== Alternative but very similar Solution == | == Alternative but very similar Solution == | ||
− | Let <math>[ABC]=K.</math> Then <math>[ADC] = \frac{1}{3}K,</math> and hence <math>[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.</math> Similarly, <math>[N_2EA]=[N_3FB]=[N_1DC] = \frac{1}{21}K.</math> <math>[ADC]+[BEA]+[CFB]=\frac{1}{3}K+\frac{1}{3}K+\frac{1}{3}K = K.</math> Then we can implement a similar but different area addition postulate to the first solution. It will be <math>[ADC]+[BEA]+[CFB]=K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3].</math> Using transitive property <cmath>K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3] = K.</cmath> Subtracting and adding on both sides gives: <cmath>[N_1N_2N_3] = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> ~many credits to the first solution | + | Let <math>[ABC]=K.</math> Then <math>[ADC] = \frac{1}{3}K,</math> and hence <math>[N_1DC] = \frac{1}{7} [ADC] = \frac{1}{21}K.</math> Similarly, <math>[N_2EA]=[N_3FB]=[N_1DC] = \frac{1}{21}K.</math> <math>[ADC]+[BEA]+[CFB]=\frac{1}{3}K+\frac{1}{3}K+\frac{1}{3}K = K.</math> Then we can implement a similar but different area addition postulate to the first solution. It will be <math>[ADC]+[BEA]+[CFB]=K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3].</math> Using transitive property <cmath>K-\frac{1}{21}K-\frac{1}{21}K-\frac{1}{21}K+[N_1N_2N_3] = K.</cmath> Subtracting and adding on both sides gives: <cmath>[N_1N_2N_3] = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> ~many credits to the first solution ~Lopkiloinm |
== Solution 2 (better solution)== | == Solution 2 (better solution)== |
Revision as of 16:18, 6 November 2020
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,
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 Using transitive property Subtracting and adding on both sides gives: ~many credits to the first solution ~Lopkiloinm
Solution 2 (better 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 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: ~Lopkiloinm (Note: the presence of in the denominator gives hints on the answer, so this solution still seems good)
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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 |
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