Difference between revisions of "2019 AIME II Problems/Problem 11"
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*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>. This gives 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>. | *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>. This gives 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>. | ||
-franchester | -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 <math>BK</math> and <math>CK</math> yet we know <math>BC</math> 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)== | ==Solution 2 (Inversion)== | ||
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<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) == | ||
+ | 14. 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 AO_{1}B = \angle AO_{2}C = 2x</cmath> by the angle by by tangent. Then we also know that <cmath>\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x</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{OB_{1}}{\sin{90^{\circ}-x}}</cmath> <cmath>\implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}}</cmath> <cmath>\implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}</cmath> Using similar logic we obtain <math>OA_{1} =\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\sqrt{5}}{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 n+n = \boxed{11}.</math> | |
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− | |||
==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:2019 AIME II 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> | ||
+ | |||
+ | 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 00:44, 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 , 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.