Difference between revisions of "2019 AIME II Problems/Problem 11"
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By the definition of <math>K</math>, it is the spiral center mapping <math>BA\to AC</math>, which means that it is the midpoint of the <math>A</math>-symmedian chord. In particular, if <math>M</math> is the midpoint of <math>BC</math> and <math>M'</math> is the reflection of <math>A</math> across <math>K</math>, we have <math>\triangle ABM'\sim\triangle AMC</math>. By Stewart's Theorem, it then follows that | By the definition of <math>K</math>, it is the spiral center mapping <math>BA\to AC</math>, which means that it is the midpoint of the <math>A</math>-symmedian chord. In particular, if <math>M</math> is the midpoint of <math>BC</math> and <math>M'</math> is the reflection of <math>A</math> across <math>K</math>, we have <math>\triangle ABM'\sim\triangle AMC</math>. By Stewart's Theorem, it then follows that | ||
<cmath>AK = \frac{AM'}{2} = \frac{AC\cdot AB}{2AM} = \frac{7\cdot 9}{2\sqrt{\frac{9^2\cdot 4 + 7^2\cdot 4 - 4^2\cdot 8}{8}}} = \frac{7\cdot 9}{2\sqrt{49}} = \frac{9}{2}\implies m + n = \boxed{11}.</cmath> | <cmath>AK = \frac{AM'}{2} = \frac{AC\cdot AB}{2AM} = \frac{7\cdot 9}{2\sqrt{\frac{9^2\cdot 4 + 7^2\cdot 4 - 4^2\cdot 8}{8}}} = \frac{7\cdot 9}{2\sqrt{49}} = \frac{9}{2}\implies m + n = \boxed{11}.</cmath> | ||
+ | |||
+ | ==Solution (Inversion simplified)== | ||
+ | The median of <math>\triangle ABC</math> is <math>AM = \sqrt{\frac {AB^2 + AC^2 }{2} – \frac{BC^2}{4}} = 7.</math> | ||
+ | Consider an inversion with center <math>A</math> and radius <math>r=AK</math> (inversion with respect the red circle). | ||
+ | Let <math>K, B',</math> and <math>C'</math> be inverse points for <math>K, B,</math> and <math>C,</math> respectively. | ||
+ | Image of line <math>AB</math> is line <math>AB, B'</math> lies on this line. Image of <math>\omega_2</math> is line <math>KC'||AB</math> (circle <math>\omega_2</math> passes through K, C and is tangent to the line <math>AB</math> at point <math>A.</math> Diagram shows cicle and it’s image using same color). Similarly, <math>AC||B'K (B'K</math> is the image of the circle <math>\omega_1</math>). | ||
+ | Therefore AB’KC’ is a parallelogram, AF is median of treangle AB’C’ and AK = 2 AF. | ||
+ | Then, we have <math>AB’=\frac{r^2}{7}</math>, <math>AC’=\frac{r^2}{9}</math>.Triangle ABC \sim triangle AC’B’ with coefficient k = AB’ / AC = \frac{r^2}{7\cdot 9}. | ||
+ | So median AF = k AM, AK/2 = r/2 = AM k = 7 \frac{r^2}{7\cdot 9} \implies r = 9/2. | ||
+ | |||
+ | |||
==See Also== | ==See Also== |
Revision as of 11:24, 8 June 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 (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 cicle and it’s image using same color). Similarly,
is the image of the circle
).
Therefore AB’KC’ is a parallelogram, AF is median of treangle AB’C’ and AK = 2 AF.
Then, we have
,
.Triangle ABC \sim triangle AC’B’ with coefficient k = AB’ / AC = \frac{r^2}{7\cdot 9}.
So median AF = k AM, AK/2 = r/2 = AM k = 7 \frac{r^2}{7\cdot 9} \implies r = 9/2.
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 |
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