2015 AIME II Problems/Problem 15

Problem

Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

$[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]$

Hint

$[ABC] = \frac{1}{2}ab \text{sin} C$ 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 $M$ be the intersection of $\overline{BC}$ and the common internal tangent of $\mathcal P$ and $\mathcal Q.$ We claim that $M$ is the circumcenter of right $\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\overline{BC},$ implying that $\angle{BAC} = 90.$ Now draw $\overline{PA}, \overline{PB}, \overline{PM},$ where $P$ is the center of circle $\mathcal P.$ Quadrilateral $PAMB$ is cyclic, and by Pythagorean Theorem $PM = \sqrt{5},$ so by Ptolemy on $PAMB$ we have $AB \sqrt{5} = 2 \cdot 1 + 2 \cdot 1 = 4 \iff AB = \dfrac{4 \sqrt{5}}{5}.$ Do the same thing on cyclic quadrilateral $QAMC$ (where $Q$ is the center of circle $\mathcal Q$ and get $AC = \frac{8 \sqrt{5}}{5}.$

Let $\angle A = \angle{DAB}.$ By Law of Sines, $BD = 2R \sin A = 2 \sin A.$ Note that $\angle{D} = \angle{ABC}$ from inscribed angles, so $\begin{align*} [ABD] &= \dfrac{1}{2} BD \cdot AB \cdot \sin{\angle B} \\ &= \dfrac{1}{2} \cdot \dfrac{4 \sqrt{5}}{5} \cdot 2 \sin A \sin{\left(180 - \angle A - \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \sin{\left(\angle A + \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos D + \cos A \sin D\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos{\angle{ABC}} + \cos A \sin{\angle{ABC}}\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\dfrac{\sqrt{5} \sin A}{5} + \dfrac{2 \sqrt{5} \cos A}{5}\right) \\ &= \dfrac{4}{5} \cdot \sin A \left(\sin A + 2 \cos A\right) \end{align*}$ after angle addition identity.

Similarly, $\angle{EAC} = 90 - \angle A,$ and by Law of Sines $CE = 8 \sin{\angle{EAC}} = 8 \cos A.$ Note that $\angle{E} = \angle{ACB}$ from inscribed angles, so $\begin{align*} [ACE] &= \dfrac{1}{2} AC \cdot CE \sin{\angle C} \\ &= \dfrac{1}{2} \cdot \dfrac{8 \sqrt{5}}{5} \cdot 8 \cos A \sin{\left[180 - \left(90 - \angle A\right) - \angle E\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \sin{\left[\left(90 - \angle A\right) + \angle{ACB}\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \left(\dfrac{2 \sqrt{5} \cos A}{5} + \dfrac{\sqrt{5} \sin A}{5}\right) \\ &= \dfrac{32}{5} \cdot \cos A \left(\sin A + 2 \cos A\right) \end{align*}$ after angle addition identity. Setting the two areas equal, we get $\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \frac{1}{\sqrt{65}}$ after Pythagorean Identity. Now plug back in and the common area is $\frac{64}{65} \iff \boxed{129}.$

Solution 2

$[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label("$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}$",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]$

Call $O_1$ and $O_2$ the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$ . Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$ , respectively. Using the fact that $\triangle{O_1BN} \sim \triangle{O_2CN}$ and setting $NO_1 = k$ , we have that $\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}$ . We can do some more length chasing using triangles similar to $O_1BN$ to get that $AK = AL = \frac{24}{15}$ , $BK = \frac{12}{15}$ , and $CL = \frac{48}{15}$ . Now, consider the circles $\mathcal{P}$ and $\mathcal{Q}$ on the coordinate plane, where $A$ is the origin. If the line $\ell$ through $A$ intersects $\mathcal{P}$ at $D$ and $\mathcal{Q}$ at $E$ then $4 \cdot DA = AE$ . To verify this, notice that $\triangle{AO_1D} \sim \triangle{EO_2A}$ from the fact that both triangles are isosceles with $\angle{O_1AD} \cong \angle{O_2AE}$ , which are corresponding angles. Since $O_2A = 4\cdot O_1A$ , we can conclude that $4 \cdot DA = AE$ .

Hence, we need to find the slope $m$ of line $\ell$ such that the perpendicular distance $n$ from $B$ to $AD$ is four times the perpendicular distance $p$ from $C$ to $AE$ . This will mean that the product of the bases and heights of triangles $ACE$ and $DBA$ will be equal, which in turn means that their areas will be equal. Let the line $\ell$ have the equation $y = -mx \implies mx + y = 0$ , and let $m$ be a positive real number so that the negative slope of $\ell$ is preserved. Setting $A = (0,0)$ , the coordinates of $B$ are $(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)$ , and the coordinates of $C$ are $(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)$ . Using the point-to-line distance formula and the condition $n = 4p$ , we have $\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}$ $\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.$ If $m > 2$ , then clearly $B$ and $C$ would not lie on the same side of $\ell$ . Thus since $m > 0$ , we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have $\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.$ Thus, the equation of $\ell$ is $y = -\frac{3}{2}x$ .

Then we can find the coordinates of $D$ by finding the point $(x,y)$ other than $A = (0,0)$ where the circle $\mathcal{P}$ intersects $\ell$ . $\mathcal{P}$ can be represented with the equation $(x + 1)^2 + y^2 = 1$ , and substituting $y = -\frac{3}{2}x$ into this equation yields $x = 0, -\frac{8}{13}$ as solutions. Discarding $x = 0$ , the $y$ -coordinate of $D$ is $-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}$ . The distance from $D$ to $A$ is then $\frac{4}{\sqrt{13}}.$ The perpendicular distance from $B$ to $AD$ or the height of $\triangle{DBA}$ is $\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.$ Finally, the common area is $\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}$ , and $m + n = 64 + 65 = \boxed{129}$ .

Solution 3

By homothety, we deduce that $AE = 4 AD$ . (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of $P$ and $Q$ to $l$ .) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from $B$ to $l$ is four times that from $C$ to $l$ . Let the distance from $C$ be $x$ and the distance from $B$ be $4x$ .

Let $P$ and $Q$ be the centers of their respective circles. Then dropping a perpendicular from $P$ to $Q$ creates a $3-4-5$ right triangle, from which $BC = 4$ and, if $\alpha = \angle{AQC}$ , that $\cos \alpha = \dfrac{3}{5}$ . Then $\angle{BPA} = 180^\circ - \alpha$ , and the Law of Cosines on triangles $APB$ and $AQC$ gives $AB = \dfrac{4}{\sqrt{5}}$ and $AC = \dfrac{8}{\sqrt{5}}.$

Now, using the Pythagorean Theorem to express the length of the projection of $BC$ onto line $l$ gives $\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.$ Squaring and simplifying gives $\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,$ and squaring and solving gives $x = \dfrac{8}{5\sqrt{13}}.$

By the Law of Sines on triangle $ABD$ , we have $\frac{BD}{\sin A} = 2.$ But we know $\sin A = \dfrac{4x}{AB}$ , and so a small computation gives $BD = \dfrac{16}{\sqrt{65}}.$ The Pythagorean Theorem now gives $AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},$ and so the common area is $\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.$ The answer is $\boxed{129}.$

Alternate Path to x

Call the intersection of lines $l$ and $BC$ $E$ .You can use similar triangles to find that the distance from $B$ to $E$ is four times the distance from $C$ to $E$ . Then draw a perpendicular from $A$ to $BC$ and call the point $F$ . $AF = \frac{8}{5}$ and $FE = FC + CE = \frac{16}{5} + \frac{4}{3} = \frac{68}{15}$ , so by the Pythagorean Theorem, $AE = \dfrac{20\sqrt{13}}{5}$ . You can now use similar triangles to find that $x = \dfrac{8}{5\sqrt{13}}$ and continue on like in solution 2.

Solution 4

$DE$ goes through $A$ , the point of tangency of both circles. So $DE$ intercepts equal arcs in circle $P$ and $Q$ : homothety. Hence, $AE=4AD$ . We will use such similarity later.

The diagonal distance between the centers of the circles is $4+1=5$ . The difference in heights is $4-1=3$ . So $BC=\sqrt{5^2-3^2}=4$ .

The triangle connecting the centers with a side parallel to $BC$ is a $3-4-5$ right triangle. Since $O_PA=1$ , the height of $A$ is $1+3/5=8/5$ . Drop an altitude from $A$ to $BC$ and call it $I$ : $IB=4/5$ and $IC=4-4/5=32/5$ . Since right $\triangle AIB\sim\triangle CIB$ , $ABC$ is a right triangle also; $IB:IA:IC$ form a geometric progression $\times 2$ .

Extend $BA$ through $A$ to a point $G$ on the other side of $\circ Q$ . By homothety, $\triangle DAB\sim\triangle EAG$ . By angle chasing $\triangle DAB$ through right triangle $ABC$ , we deduce that $\angle CEG$ is a right angle. Since $ACEG$ is cyclic, $\angle GAC$ is also right. So $CG$ is a diameter of $\circ G$ . Because of this, $CG \perp BC$ , the tangent line. $\triangle BCG$ is right and $\triangle BCG\sim\triangle ABC\sim\triangle CAG$ .

$AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5$ so $AG=2AC=16\sqrt{5}/5$ and $[\triangle CAG]=64/5$ .

Since $[\triangle DAB]=[\triangle ACE]$ , the common area is $[ACEG]/17$ . $16[\triangle DAB]=[\triangle GAE]$ because the triangles are similar with a ratio of $1:4$ . So we only need to find $[\triangle CEG]$ now.

Extend $DE$ through $E$ to intersect the tangent at $F$ . Because $4DA=AE$ , the altitude from $B$ to $AD$ is $4$ times the height from $C$ to $EA$ . So $BC=3/4BF$ and $BF=16/3$ . We look at right triangle $\triangle AIF$ . $IF=68/15$ and $AI=8/5$ . $\triangle AIF$ is a $17-6-5\sqrt{13}$ right triangle. Hypotenuse $AF$ intersects $CG$ at a point, we call it $H$ . $CH=4/3\div 68/15\cdot 8/5=8/17$ . So $HG=8-8/17=128/17$ .

By Power of a Point, $CH\cdot HG=AH\cdot HE$ . $AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17.$ So $HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})$ . The height from $E$ to $CG$ is $17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65$ .

Thus, $[\triangle CEG]=64/65\cdot 8\div 2=256/65$ . The area of the whole cyclic quadrilateral is $64/5+256/65=(832+256)/65=1088/65$ . Lastly, the common area is $1/17$ the area of the quadrilateral, or $64/65$ . So $64+65=\boxed{129}$ .

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 $AE=XY$ and $AE \parallel XY$ , $XYE$ is right angle.

First, $\frac{NO_1}{NO_1+5} = \frac{1}{4}$ , so $NO_1=\frac{5}{3}$ . And, $\cos{\angle{AO_2C}}=\cos{2\angle{AYC}} = \frac{O_2C}{NO_1+5} = \frac{3}{5}$ $\sin{\angle{AYC}} = \sqrt{\dfrac{1-\cos{2\angle{AYC}}}{2}}=\frac{1}{\sqrt{5}}$ $\cos{\angle{AYC}} = \frac{2}{\sqrt{5}}$ Then, $[AEC] = \frac{1}{2}AE*CE*\sin{\angle{ACE}}=\frac{1}{2}AE*8\sin{\angle{CYE}}*\frac{1}{\sqrt{5}}$ $[CXY] = \frac{1}{2}CX*XY*\sin{\angle{CXY}}=\frac{1}{2}*8\sin{\angle{XYC}}*XY*\sin{\angle{CAY}}$ Since $\angle{CAY} = 90 - \angle{AYC}$ , $\angle{XYC} = 90 - \angle{CYE}$ , $XY = AE$ , we have $[CXY] = \frac{1}{2}*8AE\cos{\angle{CYE}}*\cos{\angle{AYC}}=\frac{1}{2}*8AE\cos{\angle{CYE}}*\frac{2}{\sqrt{5}}$ Since $\triangle{CXY}$ is four times in scale to $\triangle{AEC}$ , their area ratio is 16. Divide the two equations for the two areas, we have $\tan{\angle{CYE}} = \frac{1}{8}$ With this angle found, everything else just follows. $\sin{\angle{CYE}} = \dfrac{\tan{\angle{CYE}}}{\sqrt{1+\tan^2{\angle{CYE}}}}=\dfrac{1}{\sqrt{65}}$ $\cos{\angle{CYE}} = \dfrac{8}{\sqrt{65}}$ $\sin{\angle{AYE}} = \sin({\angle{AYC}+\angle{CYE}} )= \frac{1}{\sqrt{5}}*\frac{8}{\sqrt{65}} + \frac{2}{\sqrt{5}}*\frac{1}{\sqrt{65}} = \frac{2}{\sqrt{13}}$ $AE = 8\sin{\angle{AYE}} = \frac{16}{\sqrt{13}}$ $[AEC] = \frac{1}{2}*8*\frac{16}{\sqrt{13}}*\dfrac{1}{\sqrt{65}}*\frac{1}{\sqrt{5}}=\frac{64}{65}$

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