Difference between revisions of "2019 AIME II Problems/Problem 11"
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So median <cmath>AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot \frac{AK^2}{7\cdot 9} \implies AK = \frac{9}{2}.</cmath> | So median <cmath>AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot \frac{AK^2}{7\cdot 9} \implies AK = \frac{9}{2}.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | == Solution 7 (Heavy Bash) == | ||
+ | We start by assigning coordinates to point A, labeling it <math>(0,0)</math> and point B at <math>(7,0)</math>, and letting point <math>C</math> be above the <math>x</math>-axis. Through an application of the Pythagorean Theorem and dropping an altitude to side <math>AB</math>, it is easy to see that <math>C</math> has coordinates <math>(33/7, 24\sqrt{5}/7)</math>. | ||
+ | |||
+ | Let <math>O1</math> be the center of circle <math>\omega_1</math> and <math>O2</math> be the center of circle <math>\omega_2</math>. Since circle <math>\omega_1</math> contains both points <math>A</math> and <math>B</math>, <math>O1</math> must lie on the perpendicular bisector of line <math>AB</math>, and similarly <math>O2</math> must lie on the perpendicular bisector of line <math>AC</math>. Through some calculations, we find that the perpendicular bisector of <math>AB</math> has equation <math>x = 3.5</math>, and the perpendicular bisector of <math>AC</math> has equation <math>y = {-11\sqrt{5}/40 \cdot x} + 189\sqrt{5}/80</math>. | ||
+ | |||
+ | Since circle <math>\omega_1</math> is tangent to line <math>AC</math> at <math>A</math>, its radius must be perpendicular to <math>AC</math> at <math>A</math>. | ||
+ | Therefore, the radius has equation <math>y = {{-11\cdot\sqrt{5}/40} \cdot x}</math>. Substituting the <math>x</math>-coordinate of <math>O1</math> into this, we find the y-coordinate of <math>O1 = {{-11 \cdot \sqrt{5}/40} \cdot 7/2} = {-77 \cdot \sqrt{5}/80}</math>. | ||
+ | |||
+ | Similarly, since circle <math>\omega_2</math> is tangent to line <math>AB</math> at <math>A</math>, its radius must be perpendicular to <math>AB</math> at <math>A</math>. Therefore, the radius has equation <math>x = 0</math> and combining with the previous result for <math>O2</math> we get that the coordinates of <math>O2</math> are <math>(0, 189\sqrt{5}/80)</math>. | ||
+ | |||
+ | We now find the slope of <math>O1O2</math>, the line joining the centers of circles <math>\omega_1</math> and <math>\omega_2</math>, which turns out to be <math>{({266 \cdot \sqrt{5} / 80}) \cdot -2/7} = {-19 \cdot \sqrt{5}/20}</math>. Since the <math>y</math>-intercept of that line is at <math>O2(0,189\sqrt{5}/80)</math>, the equation is <math>y = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}</math>. Since circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>A</math> and <math>K</math>, line <math>AK</math> is the radical axis of those circles, and since the radical axis is always perpendicular to the line joining the centers of the circles, <math>AK</math> has slope <math>{4 \cdot \sqrt{5}/19}</math>. Since point <math>A</math> is <math>(0,0)</math>, this line has a <math>y</math>-intercept of <math>0</math>, so it has equation <math>y</math> = <math>{{4 \cdot \sqrt{5}/19} \cdot x}</math>. | ||
+ | |||
+ | We set <math>{{4 \cdot \sqrt{5}/19} \cdot x} = {{-19 \cdot \sqrt{5}/20} \cdot x} + {189 \cdot \sqrt{5}/80}</math> in order to find the intersection <math>I</math> of the radical axis <math>AK</math> and <math>O1O2</math>. Through some moderate bashing, we find that the intersection point is <math>I(57/28, {3 \cdot \sqrt{5}/7})</math>. We know that either intersection point of two circles is the same distance from the intersection of radical axis and line joining the centers of those circles, so reflecting <math>A</math> over <math>I</math> yields <math>K</math> and <math>AK</math> = <math>2AI</math> = (This is the most tedious part of the bash) <math>{2 \cdot \sqrt{(57/28)^2 + ({3 \cdot \sqrt{5}/7)^2)}}} = {2 \cdot \sqrt{3969/784}} = {2 \cdot 63/28} = {2 \cdot 9/4} = 9/2</math>. Therefore the answer is <math>9 + 2 = \boxed{011}.</math> | ||
==See Also== | ==See Also== |
Revision as of 00:43, 4 September 2022
Contents
Problem
Triangle has side lengths
and
Circle
passes through
and is tangent to line
at
Circle
passes through
and is tangent to line
at
Let
be the intersection of circles
and
not equal to
Then
where
and
are relatively prime positive integers. Find
Solution 1
-Diagram by Brendanb4321
Note that from the tangency condition that the supplement of with respects to lines
and
are equal to
and
, respectively, so from tangent-chord,
Also note that
, so
. Using similarity ratios, we can easily find
However, since
and
, we can use similarity ratios to get
- Now we use Law of Cosines on
: From reverse Law of Cosines,
. This gives us
so our answer is
.
-franchester
- The motivation for using the Law of Cosines ("LoC") is after finding the similar triangles it's hard to figure out what to do with
and
yet we know
which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealing with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got stuck). To anyone in the future reading this, math is beautiful.
-First
11111111:L)xiexie
Solution 2 (Inversion)
Consider an inversion with center and radius
. Then, we have
, or
. Similarly,
. Notice that
is a parallelogram, since
and
are tangent to
and
, respectively. Thus,
. Now, we get that
so by Law of Cosines on
we have
Then, our answer is
.
-brianzjk
Solution 3 (Death By Trig Bash)
14. Let the centers of the circles be and
where the
has the side length
contained in the circle. Now let
This implies
by the angle by by tangent. Then we also know that
Now we first find
We use law of cosines on
to obtain
Then applying law of sines on
we obtain
Using similar logic we obtain
Now we know that Thus using law of cosines on
yields
While this does look daunting we can write the above expression as
Then factoring yields
The area
Now
is twice the length of the altitude of
so we let the altitude be
and we have
Thus our desired length is
Solution 4 (Video)
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI
Solution 5 (Olympiad Geometry)
By the definition of , it is the spiral center mapping
, which means that it is the midpoint of the
-symmedian chord. In particular, if
is the midpoint of
and
is the reflection of
across
, we have
. By Stewart's Theorem, it then follows that
Solution 6 (Inversion simplified)
The median of is
Consider an inversion with center and radius
(inversion with respect the red circle).
Let
and
be inverse points for
and
respectively.
Image of line is line
lies on this line.
Image of is line
(circle
passes through K, C and is tangent to the line
at point
Diagram shows circle and its image using same color).
Similarly, is the image of the circle
).
Therefore is a parallelogram,
is median of
and
Then, we have
.
with coefficient
So median
vladimir.shelomovskii@gmail.com, vvsss
Solution 7 (Heavy Bash)
We start by assigning coordinates to point A, labeling it and point B at
, and letting point
be above the
-axis. Through an application of the Pythagorean Theorem and dropping an altitude to side
, it is easy to see that
has coordinates
.
Let be the center of circle
and
be the center of circle
. Since circle
contains both points
and
,
must lie on the perpendicular bisector of line
, and similarly
must lie on the perpendicular bisector of line
. Through some calculations, we find that the perpendicular bisector of
has equation
, and the perpendicular bisector of
has equation
.
Since circle is tangent to line
at
, its radius must be perpendicular to
at
.
Therefore, the radius has equation
. Substituting the
-coordinate of
into this, we find the y-coordinate of
.
Similarly, since circle is tangent to line
at
, its radius must be perpendicular to
at
. Therefore, the radius has equation
and combining with the previous result for
we get that the coordinates of
are
.
We now find the slope of , the line joining the centers of circles
and
, which turns out to be
. Since the
-intercept of that line is at
, the equation is
. Since circles
and
intersect at points
and
, line
is the radical axis of those circles, and since the radical axis is always perpendicular to the line joining the centers of the circles,
has slope
. Since point
is
, this line has a
-intercept of
, so it has equation
=
.
We set in order to find the intersection
of the radical axis
and
. Through some moderate bashing, we find that the intersection point is
. We know that either intersection point of two circles is the same distance from the intersection of radical axis and line joining the centers of those circles, so reflecting
over
yields
and
=
= (This is the most tedious part of the bash)
. Therefore the answer is
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.