Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: Add three times the first equation to the second equation and factor to get $(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375$ . Taking the cube root yields $xy(x+y)=15$ . Noting that the first equation is $(xy)^3\cdot(xy(x+y))=810$ , we find that $(xy)^3=\frac{810}{15}=54$ . Plugging this into the second equation and dividing yields $x^3+y^3 = \frac{945}{54} = \frac{35}{2}$ . Thus the sum required, as noted in Solution 1, is $54+\frac{35}{2}\cdot2 = \boxed{089}$

Answer:  089
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: As with the other solutions, factor. But this time, let $a=xy$ and $b=x+y$ . Then $a^4b=810$ . Notice that $x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)$ , so we are looking for $2b(b^2-3a)+a^3$ . Now, if we divide the second equation by the first one, we get $7/6 = \frac{b^2-3a}{a}$ ; then $\frac{b^2}{a}=\frac{25}{6}$ . Therefore, $a = \frac{6}{25}b^2$ . Substituting $\frac{6}{25}b^2$ for $a$ in equation 1, simplifying, and then taking the cube root gives us $b^3 = \frac{5^3}{2}.$ Finding $a^3$ by cubing $a = \frac{6}{25}b^2$ on both sides and simplifying using our previous substitution, we get $a^3 = 54$ . Substituting this into the first equation and then dividing by $27$ , we get $2b(b^2 - 3a) = 35$ . Our final answer is $35+54=\boxed{089}$

Answer:  089
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: Factor the given equations as: \[x^4y^4(x+y)=810\] \[x^3y^3(x^3+y^3)=x^3y^3(x+y)(x^2-xy+y^2)=945\] We note that these expressions (as well as the desired expression) can be written exclusively in terms of $x+y$ and $xy$ . We make the substitution $s=x+y$ and $p=xy$ (for sum and product, respectively).
\[x^4y^4(x+y)=p^4s=810\] \[x^3y^3(x+y)(x^2-xy+y^2)=p^3(s)(s^2-3p)=s^3p^3-3p^4s=945\]
We see that $p^4s$ shows up in both equations, so we can eliminate it and find $sp$ , after which we can get $p^3$ from the first equation. If you rewrite the desired expression using $s$ and $p$ , it becomes clear that you don't need to actually find the values of $s$ and $p$ , but I will do so for the sake of completion.
\[s^3p^3=945+3p^4s\] \[s^3p^3=945+3(810)=3375\] \[sp=15\]
\[p^3=\frac{810}{sp}=54\] \[p=3\cdot2^{1/3}\] \[s=\frac{15}{p}=5\cdot2^{-1/3}\]
The desired expression can be written as: \[2(x^3+y^3)+(xy)^3=2(x+y)(x^2-xy+y^2)+(xy)^3\] \[2(s)(s^2-3p)+p^3=2s^3-6sp+p^3\]
Plugging in $s$ and $p$ , we get: \[2(5\cdot2^{-1/3})^3-6(15)+54=125-90+54=\boxed{089}\]

Answer:  089
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: Factor the first and second equations as $(xy)^4(x+y)=810$ and $(xy)^3(x+y)(x^2-xy+y^2)=945$ , respectively. Dividing them (allowed, since neither are $0$ ), we have \[\frac{xy}{x^2-xy+y^2}=\frac67\] or \[x^2-\frac{13}{6}xy+y^2=0.\] Plugging into the quadratic formula and solving for $x$ in terms of $y,$ we have \[x=\frac{\frac{13y}{6}\pm \sqrt{\frac{169y^2}{36}-4y^2}}{2}=\frac{2y}3 , \frac{3y}2 .\] WLOG, let $x=\frac{3y}2 .$ Plugging into our first equation, we have \[\left(\frac{3}{2}y\right)^4\left(\frac52 y\right)=810 \implies y^3 = 4 .\] Plugging this result (and the one for $x$ in terms of $y$ ) into our desired expression, we have
\begin{align*} 2x^3+(xy)^3+2y^3 &= \frac{27}{4}y^3 + \left(\frac{3}{2} y^2\right)^3 +2y^3 \\ &= \frac{35}{4}y^3 +\frac{27}{8}(y^3)^2 \\ &= 35+54 \\ &= \boxed{089} ~ASAB

Answer:  089
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: Take $w=x+y$ and $z=xy$ . Remark that \begin{align*} ~&2x^{3}+(xy)^{3}+2y^{3} \\ =~&2(x^{3}+y^{3})+(xy)^{3} \\ =~&2\left[(x+y)^{3}-3xy(x+y)\right]+(xy)^{3} \\ =~&2(w^{3}-3wz)+z^{3} \\ =~& 2w^{3}-6wz+z^{3}.\end{align*} The given equations imply that \[wz^{4}=810~~~\text{and}~~~(wz)^{3}-3wz^{4}=945.\] Substituting the first equation into the second, we have that $(wz)^{3}=945+3\cdot 810=3375$ , thus $wz=\sqrt[3]{3375}=15$ . Now \[z^{3}=\frac{wz^{4}}{wz}=\frac{810}{15}=54\] and \[w^{3}=\frac{3375}{54}=\frac{125}{2}.\] Thus \begin{align*}2w^{3}-6wz+z^{3}&=2\left(\frac{125}{2}\right)-6(15)+54 \\ &=125-90+54 \\ &=\boxed{89}

Answer:  89
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: $x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \frac{x^2-xy+y^2}{xy}=\frac{7}{6}, \frac{x^2+y^2}{xy}=\frac{13}{6}$
Let $x^2+y^2=13k; xy=6k$ , then we can see $(x+y)^2-12k=13k, x+y=5\sqrt{k}$ , now, we see $x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}$
The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\cdot 35k^{\frac{3}{2}}=\boxed{89}$

Answer:  89
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: Let's first put the left hand sides of the equations in factored forms. Doing this we obtain $(xy)^4 \cdot (x + y) = 810$ and $(xy)(x + y)(x^2 - xy + y^2) = 945$ . Now, we will subtract and add the equations to gather information on $x$ and $y$ . When we subtract the equations and clean it up via factoring, we yield $(xy)^3 \cdot (x + y) \cdot (x - y)^2 = 115$ , and when we add them, we yield $(xy)^3 \cdot (x + y) \cdot (x^2 + y^2) = 1755$ . Now with some intuition, you should divide the equations to obtain $\frac{(x^2 + y^2)}{(x - y)^2} = 13$ . Now, we clean this up to obtain the following factoring of $0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)$ . This implies that $x = \frac{3y}{2}$ . We plug that into the target expression to reduce it down to one variable, and get that target expression is $2x^3 + (xy)^3 + 2y^3 = \frac{27}{4} \cdot y^3 + \frac{27}{8} \cdot y^6 + 2y^3$ . This means that if we can find a way to get $y^3$ , then the rest is trivial. We get $y^3$ by plugging in $x = \frac{3y}{2}$ into $x^3 \cdot y^6 + y^3 \cdot x^6 = 945$ . However, this time we only factor as $(xy)^3 \cdot (x^3 + y^3)$ because we particularly want a cubic degree on $y$ . Plugging in $x = \frac{3y}{2}$ we get $y^3 = 4$ . Now lets plug this into our target expression to get $\frac{27}{4} \cdot 4 + \frac{27}{8} \cdot 4^2 + 2 \cdot 4 = \boxed{89}$

Answer:  89
Problem: Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution: Let $x, y$ be roots of some polynomial $P(\alpha)$ . Then we have that $P(\alpha) = (\alpha - x)(\alpha - y) \implies P(\alpha) = \alpha^2 - (x + y) \alpha + xy$ . Now from the first equation we have that $x + y = 810/(xy)^4$ . For convenience, denote $z = xy$
Define $P_k = x^k + y^k$ , where $x, y$ are roots of the above polynomial. Then we have that by Newton's formulas $P_k = (x + y)P_{k - 1} - xyP_{k - 2} = 810/z^4P_{k - 1} - zP_{k - 2}$ , where $P_0 = 2$ and $P_1 = x + y = 810/z^4$ . We desire $2P_3 + z^3$ . Now building up this recurrence;
\[P_2 = \frac{810}{z^4}P_1 - zP_0 = \frac{810^2}{z^8} - 2z.\] Then we have that \[P_3 = \frac{810}{z^4} \left(\frac{810^2}{z^8} - 3z \right).\] Plugging this value of $P_3$ into the second equation $z^3P_3 = 810$ , yields $z^9 = 164025$ . Now $z^3 = 54$ . To compute $P_3$ is now trivial, as returning this new value of $z$ into the second equation yields $P_3 = 35/2$ . Hence $z^3 + 2P_3 = 54 + 35 = \boxed{089}$

Answer:  089
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]

Solution: 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 $\boxed{129}.$

Answer:  129
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]

Solution: Let $K$ be the intersection of $BC$ and $AE$ . Since the radii of the two circles are 1:4, so we have $AD:AE=1:4$ , and the distance from $B$ to line $l$ and the distance from $C$ to line $l$ are in a ratio of 4:1, so $BK:CK=4:1$ . We can easily calculate the length of $BC$ to be 4, so $CK=\frac{4}{3}$ . Let $J$ be the foot of perpendicular line from $A$ to $BC$ , we can know that $BJ:CJ=1:4$ , so $BJ = 0.8$ $CJ=3.2$ $AJ=1.6$ , and $AK=\sqrt{1.6^2+\left(3.2+\frac{4}{3}\right)^2}=\frac{4}{3}\sqrt{13}$ . Since $CK^2 = EK\cdot AK$ , so $EK=\frac{4}{39}\sqrt{13}$ , and $AE = \frac{4}{3}\sqrt{13} - \frac{4}{39}\sqrt{13}= \frac{16}{13}\sqrt{13}$ $\sin\angle AKB=\frac{AJ}{AK} = \frac{1.6}{\frac{4}{3}\sqrt{13}}=\frac{1.2}{\sqrt{13}}$ , so the distance from $C$ to line $l$ is $d=CK\cdot \sin\angle AKB = \frac{4}{3}\cdot \frac{1.2}{\sqrt{13}}=\frac{1.6}{\sqrt{13}}$ . so the area is \[[ACE] = \frac{1}{2}\cdot AE\cdot d = \frac{1}{2}\cdot\frac{16}{13}\sqrt{13}\frac{1.6}{\sqrt{13}} = \frac{64}{65}\] The final answer is $\boxed{129}$

Answer:  129
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]

Solution: Consider the common tangent from $A$ to both circles. Let this intersect $BC$ at point $K$ . From equal tangents, we have $BK=AK=CK$ , which implies that $\angle BAC = 90^\circ$
Let the center of $\mathcal{P}$ be $O_1$ , and the center of $\mathcal{Q}$ be $O_2$ . Angle chasing, we find that $\triangle O_1DA \sim \triangle O_2EA$ with a ratio of $1:4$ . Hence $4AD = AE$
We can easily deduce that $BC=4$ by dropping an altitude from $O_1$ to $O_2C$ . Let $\angle ABC = \theta$ . By some simple angle chasing, we obtain that $\angle BO_1A = 2\angle BDA = 2\angle ABC = 2\theta,$ and similarly $\angle CO_2A = 180 - 2\theta$
Using LoC, we get that $AB = \sqrt{2-2\cos2\theta}$ and $AC = \sqrt{32+32\cos2\theta}$ . From Pythagorean theorem, we have \[AB^2 + AC^2 = BC^2 \implies \cos 2\theta = -\frac{3}{5} \implies \cos \theta = \frac{1}{\sqrt5}, \sin \theta = \frac{2}{\sqrt 5}\] In other words, $AB = \frac{4}{\sqrt5}, AC = \frac{8}{\sqrt5}$
Using the area condition, we have: \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*}
Now, for brevity, let $\angle D = \angle ADB$ and $\angle A = \angle DAB$
From Law of Sines on $\triangle ABD$ , we have \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*}
It remains to find the area of $\triangle ABD$ . This is just \[\frac12 AD*AB*\sin \angle A = \frac12 * \frac{4}{\sqrt{13}}*\frac{4}{\sqrt5}*\frac{8}{\sqrt{65}} = \frac{64}{65}\] for an answer of $\boxed{129}.$

Answer:  129
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]

Solution: Add in the line $k$ as the internal tangent between the two circles. Let $M$ be the midpoint of $BC$ ; It is well-known that $M$ is on $k$ and because $k$ is the radical axis of the two circles, $AM=BM=CM$ . Therefore because $M$ is the circumcenter of $\triangle{BAC}$ $\angle{BAC}=90^{\circ}$ . Let $O_P$ be the center of circle $\mathcal{P}$ and likewise let $O_Q$ be the center of circle $\mathcal{Q}$ . It is well known that by homothety $O_P, A,$ and $O_Q$ are collinear. It is well-known that $\angle{ABC}=\angle{ADB}=b$ , and likewise $\angle{ACB}=\angle{AEC}=c$ . By homothety, $AD=4AE$ , therefore since the two triangles mentioned in the problem, the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$ . Using the Pythagorean Theorem, $BC = 4$ . By angle-chasing, $O_{P}AMB$ is cyclic, and likewise $O_{Q}CMA$ is cyclic. Use the Pythagorean Theorem for $\triangle{O_{P}BM}$ to get $O_{P}M=\sqrt{5}$ . Then by Ptolemy's Theorem $AB=\frac{4\sqrt{5}}{5} \implies AC=\frac{8\sqrt{5}}{5}$ . Now to compute the area, using what we know about the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$ , letting $x$ be the length of the altitude from $C$ to $AE$ $x=\frac{8}{5\sqrt{13}}$ . From the Law of Sines, $\frac{BD}{\sin{A}}=2 \implies \sin{A}=\frac{4x}{AB} \implies BD=\frac{8x}{AB}=\frac{16\sqrt{65}}{65}$ . Then use the Pythagorean Theorem twice and add up the lengths to get $AD=\frac{4}{\sqrt{13}}$ . Use the formula $\frac{b \times h}{2}$ to get $\frac{64}{65} = \boxed{129}$ as the answer.

Answer:  129
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]

Solution: Let $P$ and $Q$ be the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively.
Let $M$ be midpoint $BC, \beta = \angle ACB.$
Upper diagram shows that
$\sin 2\beta = \frac {4}{5}$ and $AC = 2 AB.$ Therefore $\cos 2\beta = \frac {3}{5}.$
Let $CH\perp l, BH'\perp l.$ Lower diagram shows that
$\angle CAE = \angle ABH' = \alpha$ (perpendicular sides)
and $\angle CQE = 2\alpha$ (the same intersept $\overset{\Large\frown} {CE}).$ \[\tan\alpha = \frac {1}{8}, \sin2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac {16}{65}, \cos2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac {63}{65}.\] The area \[[ACE] = [AQC]+[CQE]– [AQE].\] Hence \[[ACE] =\frac{AQ^2}{2} \left(\sin 2\alpha + \sin 2\beta - \sin(2\alpha + 2\beta)\right),\] \[[ACE]  = 8\left( \frac{16}{65}+\frac{4}{5} - \frac{4}{5}\cdot \frac{63}{65} - \frac{3}{5}\cdot \frac{16}{65}\right) = \frac{64}{65}\implies \boxed{129}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  129
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]

Solution: Let the center of the larger circle be $O,$ and the center of the smaller circle be $P.$ It is not hard to find the areas of $ACO$ and $ABP$ using pythagorean theorem, which are $\frac{32}{5}$ and $\frac{2}{5}$ respectively. Assign $\angle AOC=a,\angle COE=c,\angle DPB=d,\angle BPA=b.$ We can figure out that $\angle AOE=\angle DOA=\theta$ using vertical angles and isosceles triangles. Now, using $[ABC]=\frac{1}{2}ab\sin C$ \[[ACE]=[AOC]+[COE]-[AOC]=\dfrac{32}{5}-8\sin c-8\sin \theta,\] \[[DAB]=[APD]+[APB]+[BPD]=\dfrac{2}{5}+\dfrac{1}{2}\sin \theta+\dfrac{1}{2}\sin d.\] We can also figure out that $\sin a=\dfrac{4}{5},\cos a=\dfrac{3}{5},\sin b=\dfrac{4}{5}, \cos b=-\dfrac{3}{5}.$ Also, $c=\theta-a$ and $d=360-\theta-b.$ Using sum and difference identities: \[\sin c=\dfrac{3}{5}\sin \theta-\dfrac{4}{5}\cos \theta,\] \[\sin d=\dfrac{3}{5}\sin\theta-\dfrac{4}{5}\cos\theta.\] (We can also notice that $c+d=360-\theta-b+\theta-a=360-(a+b)=180$ which means that $\sin c=\sin d.$ )
Substituting in the equations for $\sin c$ and $\sin d$ into the equations for $[ACE]$ and $[DAB],$ setting them equal, and simplifying: \[3=2\sin\theta+3\cos\theta.\] Solving this equation we get that $\sin\theta=\frac{12}{13}$ and $\cos\theta=\frac{5}{13}.$ Doing a lot of substitution gives us \[[ACE]=[DAB]=\dfrac{64}{65},\] which means the answer is $64+65=\boxed{129}.$

Answer:  129
Problem: The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.
[asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy]

Solution: The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\dfrac{80}{3}$ mm.  According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\sqrt{50^2+\left(\frac{80}{3}\right)^2}=\frac{170}{3}$ mm.  Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side.  Therefore, the minimum of the lace in millimeters is \[6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.\]

Answer:  790