Difference between revisions of "1952 AHSME Problems/Problem 49"
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== Solution == | == Solution == | ||
− | <math>C</ | + | 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] = \frac{1}{21}K.</math> Then <math>[N_2N_1CE] = [ADC] - [N_1DC]-[N_2EA] = \frac{5}{21}K,</math> and same for the other quadrilaterals. Then <math>[N_1N_2N_3]</math> is just <math>[ABC]</math> minus all the other regions we just computed. That is, <cmath>[N_1N_2N_3] = K - 3\left(\frac{1}{21}K\right) - 3\left(\frac{5}{21}\right)K = K - \frac{6}{7}K = \boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> |
== See also == | == See also == |
Revision as of 03:09, 6 August 2016
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,
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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All AHSME Problems and Solutions |
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