Just find a few parallelograms to finish

Let the midpoints of $PQ$ and $RS$ be $M_1$ and $M_2$ It is obvious that $O_1M_1 \parallel \ell \parallel O_2M_2$ (I had to write so much in the exam ) and also notice that $O_1M_1=O_2M_2=\frac{BC}{2}$ which means that $O_1O_2 \cap M_1M_2$ is the center of homothety taking $O_1M_1 \longrightarrow O_2M_2$ with factor $-1$ which means that this point is actually the midpoint which finishes the proof $\blacksquare$ NOOOO I MISREAD THE QUESTION

BVKRB- wrote: Let the midpoints of $PQ$ and $RS$ be $M_1$ and $M_2$ It is obvious that $O_1M_1 \parallel \ell \parallel O_2M_2$ (I had to write so much in the exam ) and also notice that $O_1M_1=O_2M_2=\frac{BC}{2}$ which means that $O_1O_2 \cap M_1M_2$ is the center of homothety taking $O_1M_1 \longrightarrow O_2M_2$ with factor $-1$ which means that this point is actually the midpoint which finishes the proof $\blacksquare$ True bruh Just because I was afraid of grader checking my copy and giving 0 i had to write 6 pages of s**thouse.

Let the circles intersect at T and U. Keep axis $TU$ as Y axis and $O_{1}O_{2}$ as X axis Assume T to be $(0,1)$ and U to be $(0,-1)$ The circles can be written as $O_{1} : x^2 + y^2 + ax -1 = 0$ and $O_{2} : x^2 + y^2 + bx -1 = 0 \implies O_{1} = (\frac{-a}{2},0)$ and $O_{2} = (\frac{-b}{2},0) $ Assume line ABCD to be $L : y=mx+c$ A and C are $O_1 \cap L$ . Let $A=(x_{11}, y_{11})$ and $C=(x_{12}, y_{12})$ Similarly B and D are $O_2 \cap L$ . Let $B=(x_{21}, y_{21})$ and $D=(x_{22}, y_{22})$ $PQ: y+\frac{x}{m} = \frac{y_{11}+ y_{21}}{2} + \frac{x_{11}+ x_{21}}{2m} $ $RS: y+\frac{x}{m} = \frac{y_{12}+ y_{22}}{2} + \frac{x_{12}+ x_{22}}{2m} $ Let E and F be the midpoints of PR and QS Because PQ || RS, $EF: y+\frac{x}{m} = \frac{y_{12}+ y_{22}+y_{11}+ y_{21}}{4} + \frac{x_{11}+ x_{21}+x_{12}+ x_{22}}{4m} $ $\implies EF \cap O_{1}O_{2} = (0, m( \frac{y_{12}+ y_{22}+y_{11}+ y_{21}}{4} + \frac{x_{11}+ x_{21}+x_{12}+ x_{22}}{4m})) = (0, \frac{mS_{y}}{4} + \frac{S_{x}}{4})$ Let $S=\frac{mS_{y}}{4} + \frac{S_{x}}{4}$ Observe that $S_{y} = mS_{x} + 4c \implies S = S_{x}(\frac{1+m^2}{4})+mc$ But we know that $x_{11} + x_{12} = - \frac{2mc+a}{1+m^2}$ and $x_{21} + x_{22} = - \frac{2mc+b}{1+m^2}$ Using $S_{x} = x_{11}+ x_{21}+x_{12}+ x_{22} $ and $S = S_{x}(\frac{1+m^2}{4})+mc$ we get $S = \frac{-a-b}{4}$ So EF passes through $ ( \frac{-a-b}{4}, 0) $ which is the midpoint of $O_{1}O_{2}$

Let $M_1, M_2, N_1, N_2, M'$ and $M''$ be the midpoints of $AB, CD, AC, BD, QS$ and $PR$ respectively. Let $M'M''$ cut $K_1K_2$ and $O_1O_2$ at $X$ and $Y$ respectively. Then $$PQ \parallel O_1K_1 \parallel M'M'' \parallel O_2K_2 \parallel RS$$and $X$ is the midpoint of $M_1M_2.$ Now we need to prove that $Y$ is the midpoint of $O_1O_2,$ for which we need to prove that $X$ is the midpoint of $K_1K_2,$ i.e. \begin{align*} & K_1X = K_2X \\ \text{or}\; & M_1X - M_1K_1 = M_2X - M_2K_2 \\ \text{or}\; & M_1K_1 =M_2K_2 \; [\because M_1X=M_2X] \end{align*}But \begin{align*} M_1K_1 &= K_1A-M_1A \\ &= \frac{AC-AB}{2} \\ &= \frac{BC}{2} \end{align*}Similarly, we can show that $M_2K_2=\frac{BC}{2}.$ $QED.$

Sg48 wrote: Let the circles intersect at T and U. Keep axis $TU$ as Y axis and $O_{1}O_{2}$ as X axis Assume T to be $(0,1)$ and U to be $(0,-1)$ The circles can be written as $O_{1} : x^2 + y^2 + ax -1 = 0$ and $O_{2} : x^2 + y^2 + bx -1 = 0 \implies O_{1} = (\frac{-a}{2},0)$ and $O_{2} = (\frac{-b}{2},0) $ Assume line ABCD to be $L : y=mx+c$ A and C are $O_1 \cap L$ . Let $A=(x_{11}, y_{11})$ and $C=(x_{12}, y_{12})$ Similarly B and D are $O_2 \cap L$ . Let $B=(x_{21}, y_{21})$ and $D=(x_{22}, y_{22})$ $PQ: y+\frac{x}{m} = \frac{y_{11}+ y_{21}}{2} + \frac{x_{11}+ x_{21}}{2m} $ $RS: y+\frac{x}{m} = \frac{y_{12}+ y_{22}}{2} + \frac{x_{12}+ x_{22}}{2m} $ Let E and F be the midpoints of PR and QS Because PQ || RS, $EF: y+\frac{x}{m} = \frac{y_{12}+ y_{22}+y_{11}+ y_{21}}{4} + \frac{x_{11}+ x_{21}+x_{12}+ x_{22}}{4m} $ $\implies EF \cap O_{1}O_{2} = (0, m( \frac{y_{12}+ y_{22}+y_{11}+ y_{21}}{4} + \frac{x_{11}+ x_{21}+x_{12}+ x_{22}}{4m})) = (0, \frac{mS_{y}}{4} + \frac{S_{x}}{4})$ Let $S=\frac{mS_{y}}{4} + \frac{S_{x}}{4}$ Observe that $S_{y} = mS_{x} + 4c \implies S = S_{x}(\frac{1+m^2}{4})+mc$ But we know that $x_{11} + x_{12} = - \frac{2mc+a}{1+m^2}$ and $x_{21} + x_{22} = - \frac{2mc+b}{1+m^2}$ Using $S_{x} = x_{11}+ x_{21}+x_{12}+ x_{22} $ and $S = S_{x}(\frac{1+m^2}{4})+mc$ we get $S = \frac{-a-b}{4}$ So EF passes through $ ( \frac{-a-b}{4}, 0) $ which is the midpoint of $O_{1}O_{2}$ Sir may I ask how much time did u take to finish this sol??

Quote: Sir may I ask how much time did u take to finish this sol?? ~10 minutes if you have experience with Coordinate Geometry, the main idea is to not solve the quadratic but use sum of roots 15 more minutes if you include writing in fair

Let the midpoints of $PR$ and $QS$ be $O_1$ and $O_2$. Note that $O_1$ and $O_2$ are equidistant from $PQ$ and $RS$. We will prove the midpoint of $O_1$ and $O_2$ are also equidistant from $PQ$ and $RS$. Note that $PQ$ is parallel to $RS$, so this is equivalent to $d(O_1, PQ)=d(O_2, RS)$, which is obvious as both are $\frac{BC}{2}$.

Note that for a trapezium , if you join the midpoints of two slant sides the segment is parallel to bases .Extend $O_1O_2$ let they intersect $PQ$ and $RS$ at $X,Y$ now note that midpoint of$PR,XY$ and $QS$ are collinear .Now prove that $d(O_1,PQ)=d(O_2,RS) = BC/2$ and by congruence prove that $O_1X=O_2Y$ so midpoint of $XY$=midpoint of $O_1O_2$

chakrabortyahan wrote: Note that for a trapezium , if you join the midpoints of two slant sides the segment is parallel to bases .Extend $O_1O_2$ let they intersect $PQ$ and $RS$ at $X,Y$ now note that midpoint of$PR,XY$ and $QS$ are collinear .Now prove that $d(O_1,PQ)=d(O_2,RS) = BC/2$ and by congruence prove that $O_1X=O_2Y$ so midpoint of $XY$=midpoint of $O_1O_2$ I think we need to show(atleast state) that O1,O2 lie inside the plane bounded by PQ and RS cuz graders might be really harsh this year

@2above same solution

chakrabortyahan wrote: Note that for a trapezium , if you join the midpoints of two slant sides the segment is parallel to bases .Extend $O_1O_2$ let they intersect $PQ$ and $RS$ at $X,Y$ now note that midpoint of$PR,XY$ and $QS$ are collinear .Now prove that $d(O_1,PQ)=d(O_2,RS) = BC/2$ and by congruence prove that $O_1X=O_2Y$ so midpoint of $XY$=midpoint of $O_1O_2$ ayy lessgo, same solution, altho mine one was a bit more convoluted sanyalarnab wrote: I think we need to show(atleast state) that O1,O2 lie inside the plane bounded by PQ and RS cuz graders might be really harsh this year no komments

sanyalarnab wrote: chakrabortyahan wrote: Note that for a trapezium , if you join the midpoints of two slant sides the segment is parallel to bases .Extend $O_1O_2$ let they intersect $PQ$ and $RS$ at $X,Y$ now note that midpoint of$PR,XY$ and $QS$ are collinear .Now prove that $d(O_1,PQ)=d(O_2,RS) = BC/2$ and by congruence prove that $O_1X=O_2Y$ so midpoint of $XY$=midpoint of $O_1O_2$ I think we need to show(atleast state) that O1,O2 lie inside the plane bounded by PQ and RS cuz graders might be really harsh this year You should b e postbanned for spreading panic.The result is simply not true if that doesn't hold

Consider the following diagram By homothety at $A$ with ratio $\frac{1}{2}$ and by the homothety at $D$ with ratio $\frac{1}{2}$, we obtain $M_1N_1=\frac{BC}{2}=M_2N_2$. Since $M_1M_2 \perp PQ,RS$ ,the perpendicular bisector of $M_1M_2$ is the midline of trapezoid $PQSR$. Clearly perpendicular bisector of $M_1M_2$ pass through midpoint of $O_1O_2$ since $O_1M_1 \parallel O_2M_2$.

sanyalarnab wrote: chakrabortyahan wrote: Note that for a trapezium , if you join the midpoints of two slant sides the segment is parallel to bases .Extend $O_1O_2$ let they intersect $PQ$ and $RS$ at $X,Y$ now note that midpoint of$PR,XY$ and $QS$ are collinear .Now prove that $d(O_1,PQ)=d(O_2,RS) = BC/2$ and by congruence prove that $O_1X=O_2Y$ so midpoint of $XY$=midpoint of $O_1O_2$ I think we need to show(atleast state) that O1,O2 lie inside the plane bounded by PQ and RS cuz graders might be really harsh this year bhai ye baat to obvious h na iske marks to nhi ktne chahiye

Project_smne_into_smth wrote: You should b e postbanned for spreading panic.The result is simply not true if that doesn't hold Yeah so going with you, the result is not true if some statement you have proved in test is not true

The_Great_Learner wrote: sanyalarnab wrote: chakrabortyahan wrote: Note that for a trapezium , if you join the midpoints of two slant sides the segment is parallel to bases .Extend $O_1O_2$ let they intersect $PQ$ and $RS$ at $X,Y$ now note that midpoint of$PR,XY$ and $QS$ are collinear .Now prove that $d(O_1,PQ)=d(O_2,RS) = BC/2$ and by congruence prove that $O_1X=O_2Y$ so midpoint of $XY$=midpoint of $O_1O_2$ I think we need to show(atleast state) that O1,O2 lie inside the plane bounded by PQ and RS cuz graders might be really harsh this year bhai ye baat to obvious h na iske marks to nhi ktne chahiye like I don't think it is a serious punishable mistake if you do not prove these types of things...like in the BMO 1 marking scheme a maximum of 1 mark will be deducted if the candidate has claimed something which he could prove very easily ...but did not..such as if you have used this analogy in P4 but forgot to prove that $O_1O_2$ lies inside $PQRS$ which is obvious...(further 1 or 2 line and I don't think they will deduct mark for that as rmo has returned after 4 years so they cannot expect writeups like pros and many ppl even of class 12 are giving their first rmo ...so..)or in $P2$ you have used the intersection point of $AC , BD$ and did not mention why they must intersect...

This was the hardest problem I solved in the test, took me a long time, but the proof was to identify parallelograms by doing length chasung.

Let the midpoints of segment $PR$, $O_1O_2$ and $RS$ be $G,E,F$ respectively. Observe that $PQ \parallel RS \implies PQ \parallel GF \parallel RS$ this is because $PG=GR$ and $QF=FS$ Let $H$ be the midpoint of $PQ$ and $I$ be the midpoint of $RS$, we do this because the intersection of $IH$ with line $GF$ will divide the line $IH $into equal halves because $PQ \parallel RS$ so their midpoints will be collinear if $PQ \parallel GF \parallel RS$ For this to work we prove the following claim: Claim: $HE=IE$ and $\overline{H-E-I}$ Proof: $\triangle O_1HE \cong \triangle O_2IE$, the collinearity part is complete because of the intercept theorem. Note that $B$ is the orthocenter of $\triangle PCQ$ because the reflection that is $A$ lies on the circle, and we get $BC=2O_1H$. Similarly $2O_2I=BC$ $\Rightarrow O_2I=O_2H$. Notice that, $\angle O_1HE=\angle O_2IE$ this is because $O_1H \parallel O_2 I$ and $O_1H=O_2I$ $\Rightarrow \triangle O_1HE \cong \triangle O_2IE \; \text{\textbf{(SAS)}}\implies IE=HE \implies O_1E=O_2E$. Therefore the midpoints of $PR,\; O_1O_2\;\text{and}\;QS$ are collinear.

I have discussed this question in my RMO 2023 video on my channel "little fermat". Here is the Video

I understand why people were saying that the write-up is tedious. Even though solving in contest would take like 20-25 minutes at most, writing it down would literally exhaust all the time, since you need to write a lot of details for every step. My friend will soon post the solution, but we've hardly written justification to any of the steps, like say a why trapeziums are actually trapezium and all. The justification is simple, but this problem is just bad in terms of write-up. Someone save the kids from the checking scam in RMO

Solved with SatisfiedMagma. Solution: Let $U, V$ be the midpoints of $\overline{AC}$ and $\overline{CD}$ respectively. Let $X, Y$ be the midpoints of $\overline{AB}$ and $\overline{CD}$. Finally, let $E$, $F$, $G$, $H$ be the midpoints of $\overline{PR}$, $\overline{O_1O_2}$, $\overline{UV}$ and $\overline{QS}$ respectively. The big claim is the following. [asy][asy] import olympiad; import geometry; size(11cm); defaultpen(fontsize(10.5pt)); pair O1 = (6.09, -0.48); pair O2 = (12.95, 0.87); pair A = (3.36, -2.25); pair B = (7.7, -2.21); path p1 = circle(O1, abs(O1-A)); path p2 = circle(O2, abs(O2-B)); pair X = (A+B)/2; pair C = intersectionpoints(line(A,B), p1)[1]; pair D = intersectionpoints(line(A,B), p2)[1]; pair U = (A+C)/2; pair Y = (C+D)/2; pair V = (B+D)/2; pair P = intersectionpoints(line(X, X+O1-U), p1)[0]; //i love this hack, vectors uwu <3 pair Q = intersectionpoints(line(X, X+O1-U), p1)[1]; pair R = intersectionpoints(line(Y, Y+O2-V), p2)[0]; pair S = intersectionpoints(line(Y, Y+O2-V), p2)[1]; pair E = (P+R)/2; pair F = (O1+O2)/2; pair G = (U+V)/2; pair H = (Q+S)/2; draw(p1^^p2, purple); draw(A--D, deepblue); draw(P--R, deepblue); draw(Q--S, deepblue); draw(O1--U, purple); draw(O2--V, purple); draw(P--A--Q--B--P, purple); draw(P--Q, purple); draw(R--D--S--C--R, purple); draw(R--S, purple); draw(O1--O2, deepblue); draw(E--H, dashed); draw(rightanglemark(A,X,P), deepblue); draw(rightanglemark(B,U,O1), deepblue); draw(rightanglemark(G,V,O2), deepblue); draw(rightanglemark(D,Y,R), deepblue); dot("$O_1$", O1, N); dot("$O_2$", O2, N); dot("$A$", A, dir(230)); dot("$B$", B, dir(270)); dot("$C$", C, dir(280)); dot("$D$", D, dir(310)); dot("$U$", U, dir(270)); dot("$V$", V, dir(270)); dot("$Y$", Y, dir(310)); dot("$X$", X, dir(230)); dot("$P$", P, N); dot("$Q$", Q, dir(270)); dot("$R$", R, N); dot("$S$", S, dir(270)); dot("$E$", E, N); dot("$F$", F, dir(50)); dot("$G$", G, dir(360-50)); dot("$H$", H, dir(270)); [/asy][/asy] Claim: $G$ is also the midpoint of $\overline{XY}$. Proof: We will use directed segments for the proof. Note that it is enough to show that $\overrightarrow{XU} = \overrightarrow{VY}$. We will additionally show that $\overrightarrow{XU} = \overrightarrow{VY} = \frac{1}{2}\overrightarrow{BC}$. Now, note that \begin{align*} \overrightarrow{XU} &= \overrightarrow{AU} - \overrightarrow{AX} \\ &= \overrightarrow{UC} - \overrightarrow{XB} \\ &= (\overrightarrow{UB} + \overrightarrow{BC}) - (\overrightarrow{XU}+\overrightarrow{UB}) \end{align*}which finally gives that $\overrightarrow{XU} = \frac{1}{2}\overrightarrow{BC}$. Hence, the claim is proven by symmetry. $\square$ Looking at trapeziums $PQSR$ and $XYSQ$, we can quickly find that $HG$ is the mid-line of both the trapeziums. This gives us that $H-G-E$ are collinear. It suffices to show that $F$ lies on the same line. Since $RS \parallel H-G-E$, it is enough to show that $FG \parallel RS$ since it will show that $F$ lies on line $HGE$. For this, simply look at trapezium $O_1O_2VU$. This time, $FG$ is the mid-line of this trapezium. But clearly, $FG \parallel O_2V \parallel RY$. This completes the solution. $\blacksquare$