Difference between revisions of "2020 AIME I Problems/Problem 13"
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Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{12}{\sqrt{7}}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11 | Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{12}{\sqrt{7}}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11 | ||
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+ | ==Solution 2== | ||
+ | Let <math>\overline{BC}</math> lie on the x-axis and <math>B</math> be the origin. <math>C</math> is <math>(5,0)</math>. Use Heron's formula to compute the area of triangle <math>ABC</math>. We have <math>s=\frac{15}{2}</math>. and <math>[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}</math>. We now find the altitude, which is <math>\frac{\frac{15\sqrt{7}}{2}}{5}=3\sqrt{7}{2}</math>, which is the y-coordinate of <math>A</math>. We now find the x-coordinate of <math>A</math>, which satisfies <math>x^2 + (3\sqrt{7}{2})^2=16</math>, which gives <math>x=\frac{1}{2}</math> since the triangle is acute. Now using the Angle Bisector Theorem, we have <math>\frac{4}{6}=\frac{BD}{CD}</math> and <math>BD+CD=5</math> to get <math>BD=2</math>. The coordinates of D are <math>(2,0)</math>. | ||
+ | Since we want the area of triangle <math>AEF</math>, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is <math>(\frac{5}{4}, \frac{3\sqrt{7}}{4})</math> and the slope of AD is <math>-\sqrt{7}</math>. The slope of the perpendicular bisector is <math>\frac{1}{\sqrt{7}}</math>. The equation is(in point slope form) <math>y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})</math>. | ||
+ | The slope of AB, or in trig words, the tangent of <math>\angle ABC</math> is <math>3\sqrt{7}</math>. | ||
+ | Finding <math>\sin{\angle ABC}=\frac{\frac{3\sqrt{7}}{2}}{4}=\frac{3\sqrt{7}}{8}</math> and <math>\cos{\angle ABC}=\frac{\frac{1}{2}}{4}=\frac{1}{8}</math>. Plugging this in to half angle tangent, it gives <math>\frac{\frac{3\sqrt{7}}{8}}{1+\frac{1}{8}}=\frac{\sqrt{7}}{3}</math> as the slope of the angle bisector, since it passes through <math>B</math>, the equation is <math>y=\frac{\sqrt{7}}{3}x</math>. | ||
+ | Similarly, the equation for the angle bisector of <math>C</math> will be $y=-\frac{1}{\sqrt{7}}(x-5). | ||
==See Also== | ==See Also== |
Revision as of 17:58, 12 March 2020
Contents
Problem
Point lies on side
of
so that
bisects
The perpendicular bisector of
intersects the bisectors of
and
in points
and
respectively. Given that
and
the area of
can be written as
where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
Solution 1
Points are defined as shown. It is pretty easy to show that by spiral similarity at
by some short angle chasing. Now, note that
is the altitude of
, as the altitude of
. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that
, the altitude of
. Similarly, the altitude of
is the altitude of
, or
. However, it's not too hard to see that
, and therefore
. From here, we get that the area of
is
, by similarity. ~awang11
Solution 2
Let lie on the x-axis and
be the origin.
is
. Use Heron's formula to compute the area of triangle
. We have
. and
. We now find the altitude, which is
, which is the y-coordinate of
. We now find the x-coordinate of
, which satisfies
, which gives
since the triangle is acute. Now using the Angle Bisector Theorem, we have
and
to get
. The coordinates of D are
.
Since we want the area of triangle
, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is
and the slope of AD is
. The slope of the perpendicular bisector is
. The equation is(in point slope form)
.
The slope of AB, or in trig words, the tangent of
is
.
Finding
and
. Plugging this in to half angle tangent, it gives
as the slope of the angle bisector, since it passes through
, the equation is
.
Similarly, the equation for the angle bisector of
will be $y=-\frac{1}{\sqrt{7}}(x-5).
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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