Difference between revisions of "2019 AIME II Problems/Problem 11"
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label("$C$",C,dir(-75)); | label("$C$",C,dir(-75)); | ||
dot((2.68,2.25)); | dot((2.68,2.25)); | ||
− | label("$K$",(2.68,2.25), | + | label("$K$",(2.68,2.25),2*down); |
− | label("$\omega_1$",(- | + | label("$\omega_1$",(-4.5,1)); |
− | label("$\omega_2$",( | + | label("$\omega_2$",(12.75,6)); |
label("$7$",(A+B)/2,dir(140)); | label("$7$",(A+B)/2,dir(140)); | ||
label("$8$",(B+C)/2,dir(-90)); | label("$8$",(B+C)/2,dir(-90)); | ||
Line 25: | Line 25: | ||
− | Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> | + | Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math><math>^{(*)}</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> |
− | *Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}</math> | + | *Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=\angle AKB=-\frac{11}{21}</math> |
− | + | Giving us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>. | |
− | |||
− | |||
− | + | <math>^{(*)}</math> Let <math>O</math> be the center of <math>\omega_1</math>. Then <math>\angle KAC = 90 - \angle OAK = 90 - \frac{1}{2}(180 - \angle AOK) = \frac{\angle AOK}{2} = \angle ABK</math>. Thus, <math>\angle ABK = \angle KAC</math> | |
+ | |||
+ | -franchester; <math>^{(*)}</math> by firebolt360 | ||
+ | ===Supplement=== | ||
+ | *In order to get to the Law of Cosines first, we first apply the LOC to <math>\triangle{ABC},</math> obtaining <math>\angle{BAC}.</math> | ||
+ | *We angle chase before applying the law of cosines to <math>\angle{AKB}.</math> | ||
+ | |||
+ | Note that <math>\angle{ABK}=\angle{KAC}</math> and <math>\angle{KCA}=\angle{KAB}</math> from tangent-chord. | ||
+ | |||
+ | Thus, <math>\angle{AKC}=\angle{AKB}=180^{\circ}-(\angle{ABK}+\angle{KAB}).</math> | ||
+ | |||
+ | However from our tangent chord, note that: | ||
+ | <cmath>\angle{ABK}+\angle{KAB}=\angle{KAC}+\angle{KAB}=\angle{BAC}.</cmath> | ||
+ | Thus, <math>\angle{AKB}=180^\circ-\angle{BAC}.</math> | ||
+ | |||
+ | *As an alternative approach, note that the sum of the angles in quadrilateral <math>ABKC</math> is <math>360^{\circ}</math> and we can find <math>\angle{AKB}=\frac12</math> of convex <math>\angle{BKC},</math> which is just: | ||
+ | <cmath>\frac12 \left(360^{\circ}-2(\angle{KAB}+\angle{KBA}\right) = 180^\circ - \angle{BAC}.</cmath> | ||
+ | |||
+ | ~mathboy282 | ||
==Solution 2 (Inversion)== | ==Solution 2 (Inversion)== | ||
Consider an inversion with center <math>A</math> and radius <math>r=AK</math>. Then, we have <math>AB\cdot AB^*=AK^2</math>, or <math>AB^*=\frac{AK^2}{7}</math>. Similarly, <math>AC^*=\frac{AK^2}{9}</math>. Notice that <math>AB^*KC^*</math> is a parallelogram, since <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AC</math> and <math>AB</math>, respectively. Thus, <math>AC^*=B^*K</math>. Now, we get that | Consider an inversion with center <math>A</math> and radius <math>r=AK</math>. Then, we have <math>AB\cdot AB^*=AK^2</math>, or <math>AB^*=\frac{AK^2}{7}</math>. Similarly, <math>AC^*=\frac{AK^2}{9}</math>. Notice that <math>AB^*KC^*</math> is a parallelogram, since <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AC</math> and <math>AB</math>, respectively. Thus, <math>AC^*=B^*K</math>. Now, we get that | ||
− | <cmath>\cos(\angle AB^*K)= | + | <cmath>\cos(\angle AB^*K)=\cos(180-\angle BAC)=-\frac{11}{21}</cmath> |
so by Law of Cosines on <math>\triangle AB^*K</math> we have | so by Law of Cosines on <math>\triangle AB^*K</math> we have | ||
<cmath>(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)</cmath> | <cmath>(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)</cmath> | ||
<cmath>\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}</cmath> | <cmath>\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}</cmath> | ||
− | <cmath>\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{ | + | <cmath>\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{22AK^2}{63\cdot21}</cmath> |
<cmath>\Rightarrow AK=\frac{9}{2}</cmath> | <cmath>\Rightarrow AK=\frac{9}{2}</cmath> | ||
Then, our answer is <math>9+2=\boxed{11}</math>. | Then, our answer is <math>9+2=\boxed{11}</math>. | ||
-brianzjk | -brianzjk | ||
+ | == Solution 3 (Death By Trig Bash) == | ||
+ | Let the centers of the circles be <math>O_{1}</math> and <math>O_{2}</math> where the <math>O_{1}</math> has the side length <math>7</math> contained in the circle. Now let <math>\angle BAC =x.</math> This implies <cmath>\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x</cmath> by the angle by by tangent. Then we also know that <cmath>\angle AO_{1}B = \angle AO_{2}C = 2x</cmath> Now we first find <math>\cos x.</math> We use law of cosines on <math>\bigtriangleup ABC</math> to obtain <cmath>64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}</cmath> <cmath>\implies \cos{x} =\frac{11}{21}</cmath> <cmath>\implies \sin{x} =\frac{8\sqrt{5}}{21}</cmath> Then applying law of sines on <math>\bigtriangleup AO_{1}B</math> we obtain <cmath>\frac{7}{\sin{2x}} =\frac{O_{1}B}{\sin{90^{\circ}-x}}</cmath> <cmath>\implies\frac{7}{2\sin{x}\cos{x}} =\frac{O_{1}B}{\cos{x}}</cmath> <cmath>\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}</cmath> Using similar logic we obtain <math>O_{2}A =\frac{189}{16\sqrt{5}}.</math> | ||
− | == | + | Now we know that <math>\angle O_{1}AO_{2}=180^{\circ}-x.</math> Thus using law of cosines on <math>\bigtriangleup O_{1}AO_{2}</math> yields <cmath>O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}</cmath> While this does look daunting we can write the above expression as <cmath>\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}</cmath> Then factoring yields <cmath>\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}</cmath> The area <cmath>[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}</cmath> Now <math>AK</math> is twice the length of the altitude of <math>\bigtriangleup O_{1}AO_{2}</math> so we let the altitude be <math>h</math> and we have <cmath>\frac{1}{2} \cdot h \cdot\frac{147}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}</cmath> <cmath>\implies h =\frac{9}{4}</cmath> Thus our desired length is <math>\frac{9}{2} \implies m+n = \boxed{11}.</math> |
− | + | -minor edits by faliure167 | |
− | |||
− | |||
==Solution 4 (Video)== | ==Solution 4 (Video)== | ||
Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI | Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI | ||
+ | |||
+ | ==Solution 5 (Olympiad Geometry)== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ==Solution 6 (Inversion simplified)== | ||
+ | [[File:AIME-II-2019-11.png|500px|right]] | ||
+ | The median of <math>\triangle ABC</math> is <math>AM = \sqrt{\frac {AB^2 + AC^2 }{2} – \frac{BC^2}{4}} = 7</math> (via Stewart's Theorem). | ||
+ | |||
+ | Consider an inversion with center <math>A</math> and radius <math>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 circle and its image using same color). | ||
+ | |||
+ | Similarly, <math>AC||B'K (B'K</math> is the image of the circle <math>\omega_1</math>). | ||
+ | |||
+ | Therefore <math>AB'KC'</math> is a parallelogram, <math>AF</math> is median of <math>\triangle AB'C'</math> and <math>AK = 2 AF.</math> | ||
+ | Then, we have <math>AB'=\frac{AK^2}{7}</math>. <math>\triangle ABC \sim \triangle AC'B'</math> with coefficient <math>k =\frac {AB'}{AC} = \frac{AK^2}{7\cdot 9}.</math> | ||
+ | |||
+ | 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''' | ||
+ | |||
+ | == Solution 7 (Heavy Bash) == | ||
+ | We start by assigning coordinates to point <math>A</math>, labeling it <math>(0,0)</math> and point <math>B</math> 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== |
Latest revision as of 19:52, 28 January 2024
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,
Giving us
so our answer is
.
Let
be the center of
. Then
. Thus,
-franchester; by firebolt360
Supplement
- In order to get to the Law of Cosines first, we first apply the LOC to
obtaining
- We angle chase before applying the law of cosines to
Note that and
from tangent-chord.
Thus,
However from our tangent chord, note that:
Thus,
- As an alternative approach, note that the sum of the angles in quadrilateral
is
and we can find
of convex
which is just:
~mathboy282
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)
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
-minor edits by faliure167
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
(via Stewart's Theorem).
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 , labeling it
and point
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 |
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