Difference between revisions of "1952 AHSME Problems/Problem 49"
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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 | 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 ( | + | == Solution 2 (best solution)== |
We can force this triangle to be equilateral because the ratios are always <math>3:3:1</math> 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 <math>B</math> be at <math>(0,0)</math>, <math>A</math> be at <math>(3,3\sqrt{3})</math>, and <math>C</math> be at <math>(6,0)</math>. We then create a new point <math>O</math> at the center of everything. It should be noted because of similarity between <math>\triangle N_{1}N_{2}N_{3}</math> and <math>\triangle ABC</math>, we can find the scale factor between the two triangle by simply dividing <math>AO</math> by <math>N_{2}O</math> (nitrous oxide). First, we need to find the coordinates of <math>O</math> and <math>N_{2}</math>. <math>O</math> is easily found at <math>(3,\sqrt{3})</math> and <math>N_{2}</math> be found by calculating equation of <math>BE</math> and <math>AD</math>.<math>E</math> is located <math>(4,2\sqrt{3})</math> so <math>BE</math> is <math>y=\frac{x\sqrt{3}}{2}</math>. <math>D</math> be at <math>(4,0)</math> and the slope is <math>-3\sqrt{3}</math>. We see that they be at the same <math>x</math>-value. Quick maths calculate the x value to be <math>4-\frac{2\sqrt{3}}{3\sqrt{3}+\frac{\sqrt{3}}{2}}</math> which be <math>3\frac{3}{7}</math>. Another quick maths caculation of the <math>y</math>-value lead it be equal <math>2\sqrt{3}*\frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}+3\sqrt{3}}</math> which be <math>1\frac{5}{7}\sqrt{3}</math>. Peferct, so now <math>N_{2}</math> be at <math>(3\frac{3}{7},1\frac{5}{7}\sqrt{3})</math>. Subtracting the coordinate with the center give you <math>(\frac{3}{7}, \frac{5}{7}\sqrt{3})</math>. I don't even want to do this anymore so here is the answer: ~Lopkiloinm <cmath>\boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> (Note: the presence of <math>7</math> in the denominator gives hints on the answer, so this solution still seems good) | We can force this triangle to be equilateral because the ratios are always <math>3:3:1</math> 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 <math>B</math> be at <math>(0,0)</math>, <math>A</math> be at <math>(3,3\sqrt{3})</math>, and <math>C</math> be at <math>(6,0)</math>. We then create a new point <math>O</math> at the center of everything. It should be noted because of similarity between <math>\triangle N_{1}N_{2}N_{3}</math> and <math>\triangle ABC</math>, we can find the scale factor between the two triangle by simply dividing <math>AO</math> by <math>N_{2}O</math> (nitrous oxide). First, we need to find the coordinates of <math>O</math> and <math>N_{2}</math>. <math>O</math> is easily found at <math>(3,\sqrt{3})</math> and <math>N_{2}</math> be found by calculating equation of <math>BE</math> and <math>AD</math>.<math>E</math> is located <math>(4,2\sqrt{3})</math> so <math>BE</math> is <math>y=\frac{x\sqrt{3}}{2}</math>. <math>D</math> be at <math>(4,0)</math> and the slope is <math>-3\sqrt{3}</math>. We see that they be at the same <math>x</math>-value. Quick maths calculate the x value to be <math>4-\frac{2\sqrt{3}}{3\sqrt{3}+\frac{\sqrt{3}}{2}}</math> which be <math>3\frac{3}{7}</math>. Another quick maths caculation of the <math>y</math>-value lead it be equal <math>2\sqrt{3}*\frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}+3\sqrt{3}}</math> which be <math>1\frac{5}{7}\sqrt{3}</math>. Peferct, so now <math>N_{2}</math> be at <math>(3\frac{3}{7},1\frac{5}{7}\sqrt{3})</math>. Subtracting the coordinate with the center give you <math>(\frac{3}{7}, \frac{5}{7}\sqrt{3})</math>. I don't even want to do this anymore so here is the answer: ~Lopkiloinm <cmath>\boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> (Note: the presence of <math>7</math> in the denominator gives hints on the answer, so this solution still seems good) | ||
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 (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
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 | ||
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