Difference between revisions of "2015 AIME II Problems/Problem 15"
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==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); | ||
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dot(O_1);dot(O_2); | 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); | 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); | ||
+ | |||
</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>. |
− | Hence, we need to find the slope <math>m</math> of line <math>\ell</math> such that the perpendicular distance <math>n</math> from <math>B</math> to <math>AD</math> is four times the perpendicular distance <math>p</math> from <math>C</math> to <math>AE</math>. This will mean that the product of the bases and heights of triangles <math>ACE</math> and <math>DBA</math> will be equal, which in turn means that their areas will be equal. Let the line <math>\ell</math> have the equation <math>y = -mx \implies mx + y = 0</math>, and let <math>m</math> be a positive real number so that the negative slope of <math>\ell</math> is preserved. Setting <math>A = (0,0)</math>, the coordinates of <math>B</math> are <math>(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)</math>, and the coordinates of <math>C</math> are <math>(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)</math>. Using the point-to-line distance formula and the condition <math>n = 4p</math>, we have <cmath>\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}</cmath> <cmath>\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.</cmath> | + | Hence, we need to find the slope <math>m</math> of line <math>\ell</math> such that the perpendicular distance <math>n</math> from <math>B</math> to <math>AD</math> is four times the perpendicular distance <math>p</math> from <math>C</math> to <math>AE</math>. This will mean that the product of the bases and heights of triangles <math>ACE</math> and <math>DBA</math> will be equal, which in turn means that their areas will be equal. Let the line <math>\ell</math> have the equation <math>y = -mx \implies mx + y = 0</math>, and let <math>m</math> be a positive real number so that the negative slope of <math>\ell</math> is preserved. Setting <math>A = (0,0)</math>, the coordinates of <math>B</math> are <math>(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)</math>, and the coordinates of <math>C</math> are <math>(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)</math>. Using the point-to-line distance formula and the condition <math>n = 4p</math>, we have <cmath>\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}</cmath> <cmath>\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.</cmath> If <math>m > 2</math>, then clearly <math>B</math> and <math>C</math> would not lie on the same side of <math>\ell</math>. Thus since <math>m > 0</math>, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have <cmath>\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.</cmath> Thus, the equation of <math>\ell</math> is <math>y = -\frac{3}{2}x</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 < | + | 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>. | ||
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<cmath>AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},</cmath> | <cmath>AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},</cmath> | ||
and so the common area is <math>\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.</math> The answer is <math>\boxed{129}.</math> | and so the common area is <math>\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.</math> The answer is <math>\boxed{129}.</math> | ||
+ | |||
+ | ==Alternate Path to x== | ||
+ | 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 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. | ||
+ | |||
+ | The diagonal distance between the centers of the circles is <math>4+1=5</math>. The difference in heights is <math>4-1=3</math>. So <math>BC=\sqrt{5^2-3^2}=4</math>. | ||
+ | |||
+ | The triangle connecting the centers with a side parallel to <math>BC</math> is a <math>3-4-5</math> right triangle. Since <math>O_PA=1</math>, the height of <math>A</math> is <math>1+3/5=8/5</math>. Drop an altitude from <math>A</math> to <math>BC</math> and call it <math>I</math>: <math>IB=4/5</math> and <math>IC=4-4/5=32/5</math>. Since right <math>\triangle AIB\sim\triangle CIB</math>, <math>ABC</math> is a right triangle also; <math>IB:IA:IC</math> form a geometric progression <math>\times 2</math>. | ||
+ | |||
+ | |||
+ | Extend <math>BA</math> through <math>A</math> to a point <math>G</math> on the other side of <math>\circ Q</math>. By [[homothety]], <math>\triangle DAB\sim\triangle EAG</math>. By angle chasing <math>\triangle DAB</math> through right triangle <math>ABC</math>, we deduce that <math>\angle CEG</math> is a right angle. Since <math>ACEG</math> is cyclic, <math>\angle GAC</math> is also right. So <math>CG</math> is a diameter of <math>\circ G</math>. Because of this, <math>CG \perp BC</math>, the tangent line. <math>\triangle BCG</math> is right and <math>\triangle BCG\sim\triangle ABC\sim\triangle CAG</math>. | ||
+ | |||
+ | |||
+ | <math>AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5</math> so <math>AG=2AC=16\sqrt{5}/5</math> and <math>[\triangle CAG]=64/5</math>. | ||
+ | |||
+ | Since <math>[\triangle DAB]=[\triangle ACE]</math>, the common area is <math>[ACEG]/17</math>. <math>16[\triangle DAB]=[\triangle GAE]</math> because the triangles are similar with a ratio of <math>1:4</math>. So we only need to find <math>[\triangle CEG]</math> now. | ||
+ | |||
+ | |||
+ | Extend <math>DE</math> through <math>E</math> to intersect the tangent at <math>F</math>. Because <math>4DA=AE</math>, the altitude from <math>B</math> to <math>AD</math> is <math>4</math> times the height from <math>C</math> to <math>EA</math>. So <math>BC=3/4BF</math> and <math>BF=16/3</math>. We look at right triangle <math>\triangle AIF</math>. <math>IF=68/15</math> and <math>AI=8/5</math>. <math>\triangle AIF</math> is a <math>17-6-5\sqrt{13}</math> right triangle. Hypotenuse <math>AF</math> intersects <math>CG</math> at a point, we call it <math>H</math>. <math>CH=4/3\div 68/15\cdot 8/5=8/17</math>. So <math>HG=8-8/17=128/17</math>. | ||
+ | |||
+ | |||
+ | By [[Power of a Point Theorem|Power of a Point]], <math>CH\cdot HG=AH\cdot HE</math>. <math>AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17.</math> So <math>HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})</math>. The height from <math>E</math> to <math>CG</math> is <math>17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65</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 5 (HARD computation)== | ||
+ | <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 19:15, 20 December 2023
Contents
- 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
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