Difference between revisions of "1952 AHSME Problems/Problem 49"
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== Solution 2 == | == Solution 2 == | ||
− | We can force this triangle to be equilateral because the ratios are always <math>3:3:1</math>, 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 <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: <cmath>\boxed{\textbf{(C) }\frac{1}{7}[ABC]}</cmath> | + | We can force this triangle to be equilateral because the ratios are always <math>3:3:1</math>, 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 <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> |
== See also == | == See also == |
Revision as of 15:59, 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,
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: ~Lopkiloinm
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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