Difference between revisions of "2015 AIME II Problems/Problem 15"
(→Solution 1) |
m (→Solution 11 (Trig)) |
||
(34 intermediate revisions by 13 users not shown) | |||
Line 19: | Line 19: | ||
==Hint== | ==Hint== | ||
− | + | <math>[ABC] = \frac{1}{2}ab \text{sin} C</math> is your friend for a quick solve. If you know about homotheties, go ahead, but you'll still need to do quite a bit of computation. If you're completely lost and you have a lot of time left in your mocking of this AIME, go ahead and use analytic geometry. | |
− | ==Solution 1== | + | ==Solution 1 (guys trig is fast)== |
+ | Let <math>M</math> be the intersection of <math>\overline{BC}</math> and the common internal tangent of <math>\mathcal P</math> and <math>\mathcal Q.</math> We claim that <math>M</math> is the circumcenter of right <math>\triangle{ABC}.</math> Indeed, we have <math>AM = BM</math> and <math>BM = CM</math> by equal tangents to circles, and since <math>BM = CM, M</math> is the midpoint of <math>\overline{BC},</math> implying that <math>\angle{BAC} = 90.</math> Now draw <math>\overline{PA}, \overline{PB}, \overline{PM},</math> where <math>P</math> is the center of circle <math>\mathcal P.</math> Quadrilateral <math>PAMB</math> is cyclic, and by Pythagorean Theorem <math>PM = \sqrt{5},</math> so by Ptolemy on <math>PAMB</math> we have <cmath>AB \sqrt{5} = 2 \cdot 1 + 2 \cdot 1 = 4 \iff AB = \dfrac{4 \sqrt{5}}{5}.</cmath> Do the same thing on cyclic quadrilateral <math>QAMC</math> (where <math>Q</math> is the center of circle <math>\mathcal Q</math> and get <math>AC = \frac{8 \sqrt{5}}{5}.</math> | ||
+ | |||
+ | Let <math>\angle A = \angle{DAB}.</math> By Law of Sines, <math>BD = 2R \sin A = 2 \sin A.</math> Note that <math>\angle{D} = \angle{ABC}</math> from inscribed angles, so | ||
+ | <cmath>\begin{align*} | ||
+ | [ABD] &= \dfrac{1}{2} BD \cdot AB \cdot \sin{\angle B} \ | ||
+ | &= \dfrac{1}{2} \cdot \dfrac{4 \sqrt{5}}{5} \cdot 2 \sin A \sin{\left(180 - \angle A - \angle D\right)} \ | ||
+ | &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \sin{\left(\angle A + \angle D\right)} \ | ||
+ | &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos D + \cos A \sin D\right) \ | ||
+ | &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos{\angle{ABC}} + \cos A \sin{\angle{ABC}}\right) \ | ||
+ | &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\dfrac{\sqrt{5} \sin A}{5} + \dfrac{2 \sqrt{5} \cos A}{5}\right) \ | ||
+ | &= \dfrac{4}{5} \cdot \sin A \left(\sin A + 2 \cos A\right) | ||
+ | \end{align*}</cmath> after angle addition identity. | ||
+ | |||
+ | Similarly, <math>\angle{EAC} = 90 - \angle A,</math> and by Law of Sines <math>CE = 8 \sin{\angle{EAC}} = 8 \cos A.</math> Note that <math>\angle{E} = \angle{ACB}</math> from inscribed angles, so | ||
+ | <cmath>\begin{align*} | ||
+ | [ACE] &= \dfrac{1}{2} AC \cdot CE \sin{\angle C} \ | ||
+ | &= \dfrac{1}{2} \cdot \dfrac{8 \sqrt{5}}{5} \cdot 8 \cos A \sin{\left[180 - \left(90 - \angle A\right) - \angle E\right]} \ | ||
+ | &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \sin{\left[\left(90 - \angle A\right) + \angle{ACB}\right]} \ | ||
+ | &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \left(\dfrac{2 \sqrt{5} \cos A}{5} + \dfrac{\sqrt{5} \sin A}{5}\right) \ | ||
+ | &= \dfrac{32}{5} \cdot \cos A \left(\sin A + 2 \cos A\right) | ||
+ | \end{align*}</cmath> after angle addition identity. | ||
+ | Setting the two areas equal, we get <cmath>\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \frac{1}{\sqrt{65}}</cmath> after Pythagorean Identity. Now plug back in and the common area is <math>\frac{64}{65} \iff \boxed{129}.</math> | ||
+ | |||
+ | ==Solution 2== | ||
<asy> | <asy> | ||
unitsize(35); | unitsize(35); | ||
Line 35: | Line 59: | ||
</asy> | </asy> | ||
− | Call <math>O_1</math> and <math>O_2</math> the centers of circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math>, respectively, and extend <math>CB</math> and <math>O_2O_1</math> to meet at point <math>N</math>. Call <math>K</math> and <math>L</math> the feet of the altitudes from <math>B</math> to <math>O_1N</math> and <math>C</math> to <math>O_2N</math>, respectively. Using the fact that <math>\triangle{O_1BN} \sim \triangle{O_2CN}</math> and setting <math>NO_1 = k</math>, we have that <math>\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}</math>. We can do some more length chasing using triangles similar to <math> | + | Call <math>O_1</math> and <math>O_2</math> the centers of circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math>, respectively, and extend <math>CB</math> and <math>O_2O_1</math> to meet at point <math>N</math>. Call <math>K</math> and <math>L</math> the feet of the altitudes from <math>B</math> to <math>O_1N</math> and <math>C</math> to <math>O_2N</math>, respectively. Using the fact that <math>\triangle{O_1BN} \sim \triangle{O_2CN}</math> and setting <math>NO_1 = k</math>, we have that <math>\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}</math>. We can do some more length chasing using triangles similar to <math>O_1BN</math> to get that <math>AK = AL = \frac{24}{15}</math>, <math>BK = \frac{12}{15}</math>, and <math>CL = \frac{48}{15}</math>. Now, consider the circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> on the coordinate plane, where <math>A</math> is the origin. If the line <math>\ell</math> through <math>A</math> intersects <math>\mathcal{P}</math> at <math>D</math> and <math>\mathcal{Q}</math> at <math>E</math> then <math>4 \cdot DA = AE</math>. To verify this, notice that <math>\triangle{AO_1D} \sim \triangle{EO_2A}</math> from the fact that both triangles are isosceles with <math>\angle{O_1AD} \cong \angle{O_2AE}</math>, which are corresponding angles. Since <math>O_2A = 4\cdot O_1A</math>, we can conclude that <math>4 \cdot DA = AE</math>. |
Line 43: | Line 67: | ||
Then we can find the coordinates of <math>D</math> by finding the point <math>(x,y)</math> other than <math>A = (0,0)</math> where the circle <math>\mathcal{P}</math> intersects <math>\ell</math>. <math>\mathcal{P}</math> can be represented with the equation <math>(x + 1)^2 + y^2 = 1</math>, and substituting <math>y = -\frac{3}{2}x</math> into this equation yields <math>x = 0, -\frac{8}{13}</math> as solutions. Discarding <math>x = 0</math>, the <math>y</math>-coordinate of <math>D</math> is <math>-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}</math>. The distance from <math>D</math> to <math>A</math> is then <math>\frac{4}{\sqrt{13}}.</math> The perpendicular distance from <math>B</math> to <math>AD</math> or the height of <math>\triangle{DBA}</math> is <math>\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.</math> Finally, the common area is <math>\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}</math>, and <math>m + n = 64 + 65 = \boxed{129}</math>. | Then we can find the coordinates of <math>D</math> by finding the point <math>(x,y)</math> other than <math>A = (0,0)</math> where the circle <math>\mathcal{P}</math> intersects <math>\ell</math>. <math>\mathcal{P}</math> can be represented with the equation <math>(x + 1)^2 + y^2 = 1</math>, and substituting <math>y = -\frac{3}{2}x</math> into this equation yields <math>x = 0, -\frac{8}{13}</math> as solutions. Discarding <math>x = 0</math>, the <math>y</math>-coordinate of <math>D</math> is <math>-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}</math>. The distance from <math>D</math> to <math>A</math> is then <math>\frac{4}{\sqrt{13}}.</math> The perpendicular distance from <math>B</math> to <math>AD</math> or the height of <math>\triangle{DBA}</math> is <math>\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.</math> Finally, the common area is <math>\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}</math>, and <math>m + n = 64 + 65 = \boxed{129}</math>. | ||
− | ==Solution | + | ==Solution 3== |
By [[homothety]], we deduce that <math>AE = 4 AD</math>. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of <math>P</math> and <math>Q</math> to <math>l</math>.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from <math>B</math> to <math>l</math> is four times that from <math>C</math> to <math>l</math>. Let the distance from <math>C</math> be <math>x</math> and the distance from <math>B</math> be <math>4x</math>. | By [[homothety]], we deduce that <math>AE = 4 AD</math>. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of <math>P</math> and <math>Q</math> to <math>l</math>.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from <math>B</math> to <math>l</math> is four times that from <math>C</math> to <math>l</math>. Let the distance from <math>C</math> be <math>x</math> and the distance from <math>B</math> be <math>4x</math>. | ||
Line 63: | Line 87: | ||
Call the intersection of lines <math>l</math> and <math>BC</math> <math>E</math>.You can use similar triangles to find that the distance from <math>B</math> to <math>E</math> is four times the distance from <math>C</math> to <math>E</math>. Then draw a perpendicular from <math>A</math> to <math>BC</math> and call the point <math>F</math>. <math>AF = \frac{8}{5}</math> and <math>FE = FC + CE = \frac{16}{5} + \frac{4}{3} = \frac{68}{15}</math>, so by the Pythagorean Theorem, <math>AE = \dfrac{20\sqrt{13}}{5}</math>. You can now use similar triangles to find that <math>x = \dfrac{8}{5\sqrt{13}}</math> and continue on like in solution 2. | Call the intersection of lines <math>l</math> and <math>BC</math> <math>E</math>.You can use similar triangles to find that the distance from <math>B</math> to <math>E</math> is four times the distance from <math>C</math> to <math>E</math>. Then draw a perpendicular from <math>A</math> to <math>BC</math> and call the point <math>F</math>. <math>AF = \frac{8}{5}</math> and <math>FE = FC + CE = \frac{16}{5} + \frac{4}{3} = \frac{68}{15}</math>, so by the Pythagorean Theorem, <math>AE = \dfrac{20\sqrt{13}}{5}</math>. You can now use similar triangles to find that <math>x = \dfrac{8}{5\sqrt{13}}</math> and continue on like in solution 2. | ||
− | ==Solution | + | ==Solution 4== |
<math>DE</math> goes through <math>A</math>, the point of tangency of both circles. So <math>DE</math> intercepts equal arcs in circle <math>P</math> and <math>Q</math>: [[homothety]]. Hence, <math>AE=4AD</math>. We will use such similarity later. | <math>DE</math> goes through <math>A</math>, the point of tangency of both circles. So <math>DE</math> intercepts equal arcs in circle <math>P</math> and <math>Q</math>: [[homothety]]. Hence, <math>AE=4AD</math>. We will use such similarity later. | ||
Line 87: | Line 111: | ||
Thus, <math>[\triangle CEG]=64/65\cdot 8\div 2=256/65</math>. The area of the whole cyclic quadrilateral is <math>64/5+256/65=(832+256)/65=1088/65</math>. Lastly, the common area is <math>1/17</math> the area of the quadrilateral, or <math>64/65</math>. So <math>64+65=\boxed{129}</math>. | Thus, <math>[\triangle CEG]=64/65\cdot 8\div 2=256/65</math>. The area of the whole cyclic quadrilateral is <math>64/5+256/65=(832+256)/65=1088/65</math>. Lastly, the common area is <math>1/17</math> the area of the quadrilateral, or <math>64/65</math>. So <math>64+65=\boxed{129}</math>. | ||
− | ==Solution 4== | + | ==Solution 5 (HARD computation)== |
− | Let the center of | + | <asy> |
+ | unitsize(35); | ||
+ | draw(Circle((-1,0),1)); | ||
+ | draw(Circle((4,0),4)); | ||
+ | pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; | ||
+ | A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); | ||
+ | label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); | ||
+ | draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); | ||
+ | dot(O_1);dot(O_2); | ||
+ | draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5)); | ||
+ | //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); | ||
+ | path circle2 = Circle((4,0),4); | ||
+ | N = (-8/3,0); | ||
+ | pair X =rotate(180,O_2)*E; | ||
+ | pair Y = (8,0); | ||
+ | draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y); | ||
+ | dot("$X$", X, NE);dot("$Y$", Y, NE); | ||
+ | </asy> | ||
+ | |||
+ | Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because <math>AE=XY</math> and <math>AE \parallel XY</math>, <math>XYE</math> is right angle. | ||
+ | |||
+ | First, <math>\frac{NO_1}{NO_1+5} = \frac{1}{4}</math>, so <math>NO_1=\frac{5}{3}</math>. And, | ||
+ | <cmath>\cos{\angle{AO_2C}}=\cos{2\angle{AYC}} = \frac{O_2C}{NO_1+5} = \frac{3}{5}</cmath> | ||
+ | <cmath>\sin{\angle{AYC}} = \sqrt{\dfrac{1-\cos{2\angle{AYC}}}{2}}=\frac{1}{\sqrt{5}}</cmath> | ||
+ | <cmath> \cos{\angle{AYC}} = \frac{2}{\sqrt{5}}</cmath> | ||
+ | Then, | ||
+ | <cmath>[AEC] = \frac{1}{2}AE*CE*\sin{\angle{ACE}}=\frac{1}{2}AE*8\sin{\angle{CYE}}*\frac{1}{\sqrt{5}}</cmath> | ||
+ | <cmath>[CXY] = \frac{1}{2}CX*XY*\sin{\angle{CXY}}=\frac{1}{2}*8\sin{\angle{XYC}}*XY*\sin{\angle{CAY}}</cmath> | ||
+ | Since <math>\angle{CAY} = 90 - \angle{AYC}</math>, <math>\angle{XYC} = 90 - \angle{CYE}</math>, <math>XY = AE</math>, we have | ||
+ | <cmath>[CXY] = \frac{1}{2}*8AE\cos{\angle{CYE}}*\cos{\angle{AYC}}=\frac{1}{2}*8AE\cos{\angle{CYE}}*\frac{2}{\sqrt{5}}</cmath> | ||
+ | Since <math>\triangle{CXY}</math> is four times in scale to <math>\triangle{AEC}</math>, their area ratio is 16. Divide the two equations for the two areas, we have | ||
+ | <cmath>\tan{\angle{CYE}} = \frac{1}{8}</cmath> | ||
+ | With this angle found, everything else just follows. | ||
+ | <cmath>\sin{\angle{CYE}} = \dfrac{\tan{\angle{CYE}}}{\sqrt{1+\tan^2{\angle{CYE}}}}=\dfrac{1}{\sqrt{65}}</cmath> | ||
+ | <cmath>\cos{\angle{CYE}} = \dfrac{8}{\sqrt{65}}</cmath> | ||
+ | <cmath>\sin{\angle{AYE}} = \sin({\angle{AYC}+\angle{CYE}} )= \frac{1}{\sqrt{5}}*\frac{8}{\sqrt{65}} + \frac{2}{\sqrt{5}}*\frac{1}{\sqrt{65}} = \frac{2}{\sqrt{13}}</cmath> | ||
+ | <cmath>AE = 8\sin{\angle{AYE}} = \frac{16}{\sqrt{13}}</cmath> | ||
+ | <cmath>[AEC] = \frac{1}{2}*8*\frac{16}{\sqrt{13}}*\dfrac{1}{\sqrt{65}}*\frac{1}{\sqrt{5}}=\frac{64}{65}</cmath> | ||
+ | |||
+ | ==Solution 6 (Simple computation)== | ||
+ | |||
+ | Let <math>K</math> be the intersection of <math>BC</math> and <math>AE</math>. Since the radii of the two circles are 1:4, so we have <math>AD:AE=1:4</math>, and the distance from <math>B</math> to line <math>l</math> and the distance from <math>C</math> to line <math>l</math> are in a ratio of 4:1, so <math>BK:CK=4:1</math>. We can easily calculate the length of <math>BC</math> to be 4, so <math>CK=\frac{4}{3}</math>. Let <math>J</math> be the foot of perpendicular line from <math>A</math> to <math>BC</math>, we can know that <math>BJ:CJ=1:4</math>, so <math>BJ = 0.8</math>, <math>CJ=3.2</math>, <math>AJ=1.6</math>, and <math>AK=\sqrt{1.6^2+\left(3.2+\frac{4}{3}\right)^2}=\frac{4}{3}\sqrt{13}</math>. Since <math>CK^2 = EK\cdot AK</math>, so <math>EK=\frac{4}{39}\sqrt{13}</math>, and <math>AE = \frac{4}{3}\sqrt{13} - \frac{4}{39}\sqrt{13}= \frac{16}{13}\sqrt{13}</math>. <math>\sin\angle AKB=\frac{AJ}{AK} = \frac{1.6}{\frac{4}{3}\sqrt{13}}=\frac{1.2}{\sqrt{13}}</math>, so the distance from <math>C</math> to line <math>l</math> is <math>d=CK\cdot \sin\angle AKB = \frac{4}{3}\cdot \frac{1.2}{\sqrt{13}}=\frac{1.6}{\sqrt{13}}</math>. so the area is | ||
+ | <cmath> | ||
+ | [ACE] = \frac{1}{2}\cdot AE\cdot d = \frac{1}{2}\cdot\frac{16}{13}\sqrt{13}\frac{1.6}{\sqrt{13}} = \frac{64}{65} | ||
+ | </cmath> | ||
+ | The final answer is <math>\boxed{129}</math>. | ||
+ | |||
+ | --- by Dan Li | ||
+ | |||
+ | ==Solution 7== | ||
+ | Consider the common tangent from <math>A</math> to both circles. Let this intersect <math>BC</math> at point <math>K</math>. From equal tangents, we have <math>BK=AK=CK</math>, which implies that <math>\angle BAC = 90^\circ</math>. | ||
+ | |||
+ | Let the center of <math>\mathcal{P}</math> be <math>O_1</math>, and the center of <math>\mathcal{Q}</math> be <math>O_2</math>. Angle chasing, we find that <math>\triangle O_1DA \sim \triangle O_2EA</math> with a ratio of <math>1:4</math>. Hence <math>4AD = AE</math>. | ||
+ | |||
+ | We can easily deduce that <math>BC=4</math> by dropping an altitude from <math>O_1</math> to <math>O_2C</math>. Let <math>\angle ABC = \theta</math>. By some simple angle chasing, we obtain that <math>\angle BO_1A = 2\angle BDA = 2\angle ABC = 2\theta,</math> and similarly <math>\angle CO_2A = 180 - 2\theta</math>. | ||
+ | |||
+ | Using LoC, we get that <math>AB = \sqrt{2-2\cos2\theta}</math> and <math>AC = \sqrt{32+32\cos2\theta}</math>. From Pythagorean theorem, we have <cmath>AB^2 + AC^2 = BC^2 \implies \cos 2\theta = -\frac{3}{5} \implies \cos \theta = \frac{1}{\sqrt5}, \sin \theta = \frac{2}{\sqrt 5}</cmath> | ||
+ | In other words, <math>AB = \frac{4}{\sqrt5}, AC = \frac{8}{\sqrt5}</math>. | ||
+ | |||
+ | Using the area condition, we have: | ||
+ | <cmath>\begin{align*} | ||
+ | \frac12 AD*AB \sin \angle DAB &= \frac 12 AE*AC \sin(90-\angle DAB) \ | ||
+ | AD*\frac{4}{\sqrt5} \sin \angle DAB &= 4AD*\frac{8}{\sqrt5} \cos \angle DAB \ | ||
+ | \sin \angle DAB &= 8 \cos \angle DAB \ | ||
+ | \implies \sin \angle DAB &= \frac{8}{\sqrt{65}} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Now, for brevity, let <math>\angle D = \angle ADB</math> and <math>\angle A = \angle DAB</math>. | ||
+ | |||
+ | From Law of Sines on <math>\triangle ABD</math>, we have | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{AB}{\sin \angle D} &= \frac{AD}{\sin (180-\angle A - \angle D)} \ | ||
+ | \frac{\frac{4}{\sqrt5}}{\frac{2}{\sqrt5}} &= \frac{AD}{\sin \angle A \cos\angle D + \sin \angle D\cos\angle A} \ | ||
+ | 2 &= \frac{AD}{\frac{2}{\sqrt{13}}} \ | ||
+ | AD &= \frac{4}{\sqrt{13}} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | It remains to find the area of <math>\triangle ABD</math>. This is just <cmath>\frac12 AD*AB*\sin \angle A = \frac12 * \frac{4}{\sqrt{13}}*\frac{4}{\sqrt5}*\frac{8}{\sqrt{65}} = \frac{64}{65}</cmath> for an answer of <math>\boxed{129}.</math> | ||
+ | |||
+ | This solution was brought to you by Leonard_my_dude. | ||
+ | |||
+ | ==Solution 8 (Synthetic-Trigonometry)== | ||
+ | Add in the line <math>k</math> as the internal tangent between the two circles. Let <math>M</math> be the midpoint of <math>BC</math>; It is well-known that <math>M</math> is on <math>k</math> and because <math>k</math> is the radical axis of the two circles, <math>AM=BM=CM</math>. Therefore because <math>M</math> is the circumcenter of <math>\triangle{BAC}</math>, <math>\angle{BAC}=90^{\circ}</math>. Let <math>O_P</math> be the center of circle <math>\mathcal{P}</math> and likewise let <math>O_Q</math> be the center of circle <math>\mathcal{Q}</math>. It is well known that by homothety <math>O_P, A,</math> and <math>O_Q</math> are collinear. It is well-known that <math>\angle{ABC}=\angle{ADB}=b</math>, and likewise <math>\angle{ACB}=\angle{AEC}=c</math>. By homothety, <math>AD=4AE</math>, therefore since the two triangles mentioned in the problem, the length of the altitude from <math>B</math> to <math>AD</math> is four times the length of the altitude from <math>C</math> to <math>AE</math>. Using the Pythagorean Theorem, <math>BC = 4</math>. By angle-chasing, <math>O_{P}AMB</math> is cyclic, and likewise <math>O_{Q}CMA</math> is cyclic. Use the Pythagorean Theorem for <math>\triangle{O_{P}BM}</math> to get <math>O_{P}M=\sqrt{5}</math>. Then by Ptolemy's Theorem <math>AB=\frac{4\sqrt{5}}{5} \implies AC=\frac{8\sqrt{5}}{5}</math>. Now to compute the area, using what we know about the length of the altitude from <math>B</math> to <math>AD</math> is four times the length of the altitude from <math>C</math> to <math>AE</math>, letting <math>x</math> be the length of the altitude from <math>C</math> to <math>AE</math>, <math>x=\frac{8}{5\sqrt{13}}</math>. From the Law of Sines, <math>\frac{BD}{\sin{A}}=2 \implies \sin{A}=\frac{4x}{AB} \implies BD=\frac{8x}{AB}=\frac{16\sqrt{65}}{65}</math>. Then use the Pythagorean Theorem twice and add up the lengths to get <math>AD=\frac{4}{\sqrt{13}}</math>. Use the formula <math>\frac{b \times h}{2}</math> to get <math>\frac{64}{65} = \boxed{129}</math> as the answer. | ||
+ | |||
+ | ~First | ||
+ | |||
+ | ==Solution 9 (Visual)== | ||
+ | [[File:2015 AIME II 15.png|400px|right]] | ||
+ | [[File:2015 AIME II 15a.png|400px|right]] | ||
+ | Let <math>P</math> and <math>Q</math> be the centers of circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> , respectively. | ||
+ | |||
+ | Let <math>M</math> be midpoint <math>BC, \beta = \angle ACB.</math> | ||
+ | |||
+ | Upper diagram shows that | ||
+ | |||
+ | <math>\sin 2\beta = \frac {4}{5}</math> and <math>AC = 2 AB.</math> Therefore <math>\cos 2\beta = \frac {3}{5}.</math> | ||
+ | |||
+ | Let <math>CH\perp l, BH'\perp l.</math> Lower diagram shows that | ||
+ | |||
+ | <math>\angle CAE = \angle ABH' = \alpha</math> (perpendicular sides) | ||
+ | |||
+ | and <math>\angle CQE = 2\alpha</math> (the same intersept <math>\overset{\Large\frown} {CE}).</math> | ||
+ | <cmath>\tan\alpha = \frac {1}{8}, | ||
+ | \sin2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac {16}{65}, | ||
+ | \cos2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac {63}{65}.</cmath> | ||
+ | The area | ||
+ | <cmath>[ACE] = [AQC]+[CQE]– [AQE].</cmath> | ||
+ | Hence <cmath>[ACE] =\frac{AQ^2}{2} \left(\sin 2\alpha + \sin 2\beta - \sin(2\alpha + 2\beta)\right),</cmath> | ||
+ | <cmath>[ACE] = | ||
+ | 8\left( \frac{16}{65}+\frac{4}{5} - \frac{4}{5}\cdot \frac{63}{65} - \frac{3}{5}\cdot \frac{16}{65}\right) = \frac{64}{65}\implies \boxed {129}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 10 (Similar Triangles, Angle Chasing, and Ptolemy's)== | ||
+ | We begin by extending <math>\overline{AB}</math> upwards until it intersects Circle <math>\mathcal{Q}</math>. We can call this point of intersection <math>F</math>. Connect <math>F</math> with <math>E</math>, <math>C</math>, and <math>A</math> for future use. | ||
+ | |||
+ | Create a trapezoid with points <math>B</math>, <math>C</math>, and the origins of Circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math>. After quick inspection, we can conclude that the distance between the origins is 5 and that <math>\overline{BC}</math> is 4. | ||
+ | |||
+ | (Note: It is important to understand that we could simply use the formula for the distance between the tangent points on a line and two different circles, which is <math>2\cdot \sqrt{a\cdot b}</math>, where <math>a</math> and <math>b</math> are the two respective radii of the circles. In our case, we get <math>2\cdot \sqrt{4} = 4</math>. | ||
+ | |||
+ | Using similar triangles or homotheties, <math>AE=4\cdot AD</math> and <math>AF=4\cdot BA</math>. <math>BC^2 = BA\cdot BF</math>. <cmath>16 = BA\cdot (5\cdot BA)</cmath> <cmath>BA = \frac{4}{\sqrt{5}}</cmath> <cmath>AF = \frac{16}{\sqrt{5}}</cmath> <cmath>BF = 4\cdot \sqrt{5}</cmath> | ||
+ | |||
+ | Inspecting <math>\triangle{BFC}</math>, we recognize that it is a right triangle (<math>\angle{BCF} = 90</math>) as the final length (<math>\overline{FC}</math>) being 8 would allow for an <math>x-2x-x\sqrt{5}</math> triangle. Hence, the diameter of circle <math>\mathcal{Q}</math> = <math>CF</math>. This also means that <math>\angle{CAF} = \angle{CEF} = 90</math>. | ||
+ | |||
+ | From the fact that <math>\triangle{ABC}</math> is a right triangle: <cmath>AB^2 + AC^2 = BC^2</cmath> <cmath>AC = \frac{8}{\sqrt{5}}</cmath> | ||
+ | (Note: We could have also used <math>\triangle{FAC}</math>.) | ||
+ | |||
+ | Our next step is to start angle chasing to find any other similar triangles or shared angles. Label the intersection between <math>\overline{FC}</math> and <math>\overline{DE}</math> as <math>M</math>. Label <math>\angle{CAE} = \angle{CFE} = \theta</math>. Since <math>\angle{BAC} = 90</math>, <math>\angle{EAF} = \angle{DAB} = 90-\theta</math>. Now, use the sine formula and the fact that the areas of <math>\triangle{DBA}</math> and <math>\triangle{ACE}</math> are equal to get: | ||
+ | <cmath>\frac{1}{2}\cdot AC\cdot AE\cdot \sin{\theta} = \frac{1}{2}\cdot AD\cdot AB\cdot \sin{(90-\theta)}</cmath> | ||
+ | Since <math>\sin{(90-\theta)} = \cos{\theta}</math>: | ||
+ | <cmath>\tan{\theta} = \frac{AD}{2\cdot AE}</cmath> | ||
+ | |||
+ | Using right <math>\triangle{FCE}</math>, since <math>\angle{ACM} = \angle{CFE}</math>, <math>\tan{\theta} = \frac{CE}{FE}</math>. Hence, plugging into the previous equation: | ||
+ | <cmath>\frac{CE}{FE} = \frac{AD}{2\cdot AE}</cmath> | ||
+ | Using the Pythagorean theorem on <math>\triangle{FCE}</math>, <math>FE = \sqrt{(64 - CE^2)}</math>. We also know that <math>AE=4\cdot AD</math> | ||
+ | Plugging back in: | ||
+ | <cmath>\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{AD}{2\cdot (4\cdot AD)}</cmath> | ||
+ | <cmath>\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{1}{8}</cmath>. | ||
+ | From here, we can square both sides and bring everything to one side to get: | ||
+ | <cmath>EF^2 + 64EF - 64 = 0</cmath>. | ||
+ | <cmath>EF = \frac{64}{\sqrt{65}}</cmath> | ||
+ | <cmath>CE = \frac{8}{\sqrt{65}}</cmath> | ||
+ | We should also return to the fact that <math>\sin{\theta} = \frac{CE}{CF}</math> from <math>\triangle{FCE}</math>, so <cmath>\sin{\theta} = \frac{1}{\sqrt{65}}</cmath> | ||
+ | |||
+ | From the fact that <math>\angle{CAF} = \angle{CEF} = 90</math>, we can use Ptolemy's Theorem on quadrilateral <math>ACEF</math>. <math>AC\cdot EF + CE\cdot FA = CF\cdot AE</math>. Plugging in and solving, we get that <math>AE = \frac{16}{\sqrt{13}}</math>. | ||
+ | |||
+ | We now have all of our pieces to use the Sine Formula on <math>\triangle{ACE}</math>. <cmath>\frac{1}{2}\cdot AC\cdot AF\cdot \sin{\theta}</cmath> | ||
+ | <cmath>\frac{1}{2}\cdot \frac{8}{\sqrt{5}}\cdot \frac{16}{\sqrt{13}}\cdot \frac{1}{\sqrt{65}} = \frac{64}{65} = \boxed {129}</cmath> | ||
+ | |||
+ | ~Solution by: armang32324 | ||
+ | |||
+ | ==Solution 11 (Trig)== | ||
+ | Let the center of the larger circle be <math>O,</math> and the center of the smaller circle be <math>P.</math> It is not hard to find the areas of <math>ACO</math> and <math>ABP</math> using pythagorean theorem, which are <math>\frac{32}{5}</math> and <math>\frac{2}{5}</math> respectively. Assign <math>\angle AOC=a,\angle COE=c,\angle DPB=d,\angle BPA=b.</math> We can figure out that <math>\angle AOE=\angle DOA=\theta</math> using vertical angles and isosceles triangles. Now, using <math>[ABC]=\frac{1}{2}ab\sin C</math> | ||
+ | <cmath>[ACE]=[AOC]+[COE]-[AOC]=\dfrac{32}{5}-8\sin c-8\sin \theta,</cmath> | ||
+ | <cmath>[DAB]=[APD]+[APB]+[BPD]=\dfrac{2}{5}+\dfrac{1}{2}\sin \theta+\dfrac{1}{2}\sin d.</cmath> | ||
+ | We can also figure out that <math>\sin a=\dfrac{4}{5},\cos a=\dfrac{3}{5},\sin b=\dfrac{4}{5}, \cos b=-\dfrac{3}{5}.</math> Also, <math>c=\theta-a</math> and <math>d=360-\theta-b.</math> Using sum and difference identities: | ||
+ | <cmath>\sin c=\dfrac{3}{5}\sin \theta-\dfrac{4}{5}\cos \theta,</cmath> | ||
+ | <cmath>\sin d=\dfrac{3}{5}\sin\theta-\dfrac{4}{5}\cos\theta.</cmath> | ||
+ | (We can also notice that <math>c+d=360-\theta-b+\theta-a=360-(a+b)=180</math> which means that <math>\sin c=\sin d.</math>) | ||
+ | Substituting in the equations for <math>\sin c</math> and <math>\sin d</math> into the equations for <math>[ACE]</math> and <math>[DAB],</math> setting them equal, and simplifying: | ||
+ | <cmath>3=2\sin\theta+3\cos\theta.</cmath> | ||
+ | Solving this equation we get that <math>\sin\theta=\frac{12}{13}</math> and <math>\cos\theta=\frac{5}{13}.</math> Doing a lot of substitution gives us <cmath>[ACE]=[DAB]=\dfrac{64}{65},</cmath> which means the answer is <math>64+65=\boxed{129}.</math> | ||
+ | |||
+ | ~[[BS2012]] | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/LBCmQc2awMk | ||
− | + | ~MathProblemSolvingSkills.com | |
− | |||
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:15, 20 December 2023
Contents
[hide]- 1 Problem
- 2 Hint
- 3 Solution 1 (guys trig is fast)
- 4 Solution 2
- 5 Solution 3
- 6 Alternate Path to x
- 7 Solution 4
- 8 Solution 5 (HARD computation)
- 9 Solution 6 (Simple computation)
- 10 Solution 7
- 11 Solution 8 (Synthetic-Trigonometry)
- 12 Solution 9 (Visual)
- 13 Solution 10 (Similar Triangles, Angle Chasing, and Ptolemy's)
- 14 Solution 11 (Trig)
- 15 Video Solution
- 16 See also
Problem
Circles and have radii and , respectively, and are externally tangent at point . Point is on and point is on so that line is a common external tangent of the two circles. A line through intersects again at and intersects again at . Points and lie on the same side of , and the areas of and are equal. This common area is , where and are relatively prime positive integers. Find .
Hint
is your friend for a quick solve. If you know about homotheties, go ahead, but you'll still need to do quite a bit of computation. If you're completely lost and you have a lot of time left in your mocking of this AIME, go ahead and use analytic geometry.
Solution 1 (guys trig is fast)
Let be the intersection of and the common internal tangent of and We claim that is the circumcenter of right Indeed, we have and by equal tangents to circles, and since is the midpoint of implying that Now draw where is the center of circle Quadrilateral is cyclic, and by Pythagorean Theorem so by Ptolemy on we have Do the same thing on cyclic quadrilateral (where is the center of circle and get
Let By Law of Sines, Note that from inscribed angles, so after angle addition identity.
Similarly, and by Law of Sines Note that from inscribed angles, so after angle addition identity. Setting the two areas equal, we get after Pythagorean Identity. Now plug back in and the common area is
Solution 2
Call and the centers of circles and , respectively, and extend and to meet at point . Call and the feet of the altitudes from to and to , respectively. Using the fact that and setting , we have that . We can do some more length chasing using triangles similar to to get that , , and . Now, consider the circles and on the coordinate plane, where is the origin. If the line through intersects at and at then . To verify this, notice that from the fact that both triangles are isosceles with , which are corresponding angles. Since , we can conclude that .
Hence, we need to find the slope of line such that the perpendicular distance from to is four times the perpendicular distance from to . This will mean that the product of the bases and heights of triangles and will be equal, which in turn means that their areas will be equal. Let the line have the equation , and let be a positive real number so that the negative slope of is preserved. Setting , the coordinates of are , and the coordinates of are . Using the point-to-line distance formula and the condition , we have If , then clearly and would not lie on the same side of . Thus since , we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have Thus, the equation of is .
Then we can find the coordinates of by finding the point other than where the circle intersects . can be represented with the equation , and substituting into this equation yields as solutions. Discarding , the -coordinate of is . The distance from to is then The perpendicular distance from to or the height of is Finally, the common area is , and .
Solution 3
By homothety, we deduce that . (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of and to .) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from to is four times that from to . Let the distance from be and the distance from be .
Let and be the centers of their respective circles. Then dropping a perpendicular from to creates a right triangle, from which and, if , that . Then , and the Law of Cosines on triangles and gives and
Now, using the Pythagorean Theorem to express the length of the projection of onto line gives Squaring and simplifying gives and squaring and solving gives
By the Law of Sines on triangle , we have But we know , and so a small computation gives The Pythagorean Theorem now gives and so the common area is The answer is
Alternate Path to x
Call the intersection of lines and .You can use similar triangles to find that the distance from to is four times the distance from to . Then draw a perpendicular from to and call the point . and , so by the Pythagorean Theorem, . You can now use similar triangles to find that and continue on like in solution 2.
Solution 4
goes through , the point of tangency of both circles. So intercepts equal arcs in circle and : homothety. Hence, . We will use such similarity later.
The diagonal distance between the centers of the circles is . The difference in heights is . So .
The triangle connecting the centers with a side parallel to is a right triangle. Since , the height of is . Drop an altitude from to and call it : and . Since right , is a right triangle also; form a geometric progression .
Extend through to a point on the other side of . By homothety, . By angle chasing through right triangle , we deduce that is a right angle. Since is cyclic, is also right. So is a diameter of . Because of this, , the tangent line. is right and .
so and .
Since , the common area is . because the triangles are similar with a ratio of . So we only need to find now.
Extend through to intersect the tangent at . Because , the altitude from to is times the height from to . So and . We look at right triangle . and . is a right triangle. Hypotenuse intersects at a point, we call it . . So .
By Power of a Point, . So . The height from to is .
Thus, . The area of the whole cyclic quadrilateral is . Lastly, the common area is the area of the quadrilateral, or . So .
Solution 5 (HARD computation)
Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because and , is right angle.
First, , so . And, Then, Since , , , we have Since is four times in scale to , their area ratio is 16. Divide the two equations for the two areas, we have With this angle found, everything else just follows.
Solution 6 (Simple computation)
Let be the intersection of and . Since the radii of the two circles are 1:4, so we have , and the distance from to line and the distance from to line are in a ratio of 4:1, so . We can easily calculate the length of to be 4, so . Let be the foot of perpendicular line from to , we can know that , so , , , and . Since , so , and . , so the distance from to line is . so the area is The final answer is .
--- by Dan Li
Solution 7
Consider the common tangent from to both circles. Let this intersect at point . From equal tangents, we have , which implies that .
Let the center of be , and the center of be . Angle chasing, we find that with a ratio of . Hence .
We can easily deduce that by dropping an altitude from to . Let . By some simple angle chasing, we obtain that and similarly .
Using LoC, we get that and . From Pythagorean theorem, we have In other words, .
Using the area condition, we have:
Now, for brevity, let and .
From Law of Sines on , we have
It remains to find the area of . This is just for an answer of
This solution was brought to you by Leonard_my_dude.
Solution 8 (Synthetic-Trigonometry)
Add in the line as the internal tangent between the two circles. Let be the midpoint of ; It is well-known that is on and because is the radical axis of the two circles, . Therefore because is the circumcenter of , . Let be the center of circle and likewise let be the center of circle . It is well known that by homothety and are collinear. It is well-known that , and likewise . By homothety, , therefore since the two triangles mentioned in the problem, the length of the altitude from to is four times the length of the altitude from to . Using the Pythagorean Theorem, . By angle-chasing, is cyclic, and likewise is cyclic. Use the Pythagorean Theorem for to get . Then by Ptolemy's Theorem . Now to compute the area, using what we know about the length of the altitude from to is four times the length of the altitude from to , letting be the length of the altitude from to , . From the Law of Sines, . Then use the Pythagorean Theorem twice and add up the lengths to get . Use the formula to get as the answer.
~First
Solution 9 (Visual)
Let and be the centers of circles and , respectively.
Let be midpoint
Upper diagram shows that
and Therefore
Let Lower diagram shows that
(perpendicular sides)
and (the same intersept The area Hence vladimir.shelomovskii@gmail.com, vvsss
Solution 10 (Similar Triangles, Angle Chasing, and Ptolemy's)
We begin by extending upwards until it intersects Circle . We can call this point of intersection . Connect with , , and for future use.
Create a trapezoid with points , , and the origins of Circles and . After quick inspection, we can conclude that the distance between the origins is 5 and that is 4.
(Note: It is important to understand that we could simply use the formula for the distance between the tangent points on a line and two different circles, which is , where and are the two respective radii of the circles. In our case, we get .
Using similar triangles or homotheties, and . .
Inspecting , we recognize that it is a right triangle () as the final length () being 8 would allow for an triangle. Hence, the diameter of circle = . This also means that .
From the fact that is a right triangle: (Note: We could have also used .)
Our next step is to start angle chasing to find any other similar triangles or shared angles. Label the intersection between and as . Label . Since , . Now, use the sine formula and the fact that the areas of and are equal to get: Since :
Using right , since , . Hence, plugging into the previous equation: Using the Pythagorean theorem on , . We also know that Plugging back in: . From here, we can square both sides and bring everything to one side to get: . We should also return to the fact that from , so
From the fact that , we can use Ptolemy's Theorem on quadrilateral . . Plugging in and solving, we get that .
We now have all of our pieces to use the Sine Formula on .
~Solution by: armang32324
Solution 11 (Trig)
Let the center of the larger circle be and the center of the smaller circle be It is not hard to find the areas of and using pythagorean theorem, which are and respectively. Assign We can figure out that using vertical angles and isosceles triangles. Now, using We can also figure out that Also, and Using sum and difference identities: (We can also notice that which means that ) Substituting in the equations for and into the equations for and setting them equal, and simplifying: Solving this equation we get that and Doing a lot of substitution gives us which means the answer is
Video Solution
~MathProblemSolvingSkills.com
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.