Join $OP$, extend it to $O'$ such that $OP=PO'$. Bisect $O'P$, let this point be $M$. Construct $MX$ and $MD$, with $X$ at $A$ such that $M, X, D$ are collinear and $MX=MD$. Bisect $OX$, this point is $C$, join $CD$.

You did write RMO rd1452002 ?

Mr.Techworm wrote: You did write RMO rd1452002 ? When was RMO2017?

Mr.Techworm wrote: You did write RMO rd1452002 ? Yep misinnyo wrote: Mr.Techworm wrote: You did write RMO rd1452002 ? When was RMO2017? Ended 50 minutes ago, RMO stands for Regional MO ( Indian) not Romanian MO

the paper was easier than last years

Just extend $OP$ to $X$ such that $OP:PX=2:1$. Draw a line through $X$ parallet to $OB$ which meets $OA$ at $C$. Extend $CP$ to meet $OB$ at $D$. $CD$ is the required line.

Take any point X on OA join XP and take a point K on line XP such that it divides XP externally in the ratio 3:2. Now draw a line KD parallel to OA which intersects OB at D.Extend DP to meet OA at C . Hence CD is the required segment and the proof follows by similar triangles.

Let $D \in AO$ such that $PD \perp AO$. Extend $PE$, and let $Q$ be on the otherside of $Q$ such that $2PE = PQ$. Let the parallel of $AO$ at $Q$ intersect $BO$ at $D$. extend $DP$ to meet $AO$ at $C$

This was the easiest RMO2017 problem

Extend OP to X(interior to angle AOB) such that OP:PX=2:1.Now extend OX to Y such that OX=OY. Draw lines through Y parallel to OA and OB to cut OB and OA at D and K respectively.The midpoint of OK is the point C.(ODYK form a parallelogram with X as midpoint of DK.So point P is the centroid of triangle ODK.Hence DP:CP=2:1.

I am from 12th and i have a merit certificate can i write inmo 2018.

hemant1729 wrote: I am from 12th and i have a merit certificate can i write inmo 2018. Only till class 11, sorry. Certificate from where, though?

hemant1729 wrote: I am from 12th and i have a merit certificate can i write inmo 2018. No.you can't.

hemant1729 wrote: Extend OP to X(interior to angle AOB) such that OP:PX=2:1.Now extend OX to Y such that OX=OY. Draw lines through Y parallel to OA and OB to cut OB and OA at D and K respectively.The midpoint of OK is the point C.(ODYK form a parallelogram with X as midpoint of DK.So point P is the centroid of triangle ODK.Hence DP:CP=2:1. You don't need to do all this. A line from vertex to opposite side being divided in the ratio 2:1 itself proves that the point is nothing but the centroid.

THISIS ANSWER

I think it is in class 10 ncert math book.

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PeterLM10 wrote: hemant1729 wrote: Extend OP to X(interior to angle AOB) such that OP:PX=2:1.Now extend OX to Y such that OX=OY. Draw lines through Y parallel to OA and OB to cut OB and OA at D and K respectively.The midpoint of OK is the point C.(ODYK form a parallelogram with X as midpoint of DK.So point P is the centroid of triangle ODK.Hence DP:CP=2:1. You don't need to do all this. A line from vertex to opposite side being divided in the ratio 2:1 itself proves that the point is nothing but the centroid. Yes but how would you indirectly construct a triangle whose centroid is the given pt (one of my friends fakesolved this method so I'm interested in seeing this)

Can someone check if my solution is correct? I am not too sure about it, as this is the first geometric(ish) problem of RMO that I have solved myself.

I think it is wrong somewhere, but I can't find the error. I would greatly appreciate if someone can help

Math-wiz wrote: Can someone check if my solution is correct? I am not too sure about it, as this is the first geometric(ish) problem of RMO that I have solved myself.

I think it is wrong somewhere, but I can't find the error. I would greatly appreciate if someone can help I think there is a notation problem in your solution. $D$ should be on $OB$ and not $OA$. Please make your solution a bit more clearer

Jupiter_is_BIG wrote: Math-wiz wrote: Can someone check if my solution is correct? I am not too sure about it, as this is the first geometric(ish) problem of RMO that I have solved myself.

I think it is wrong somewhere, but I can't find the error. I would greatly appreciate if someone can help I think there is a notation problem in your solution. $D$ should be on $OB$ and not $OA$. Please make your solution a bit more clearer Ohh sorry I got confused with the letters. Fixed it

Math-wiz wrote: Can someone check if my solution is correct? I am not too sure about it, as this is the first geometric(ish) problem of RMO that I have solved myself.

I think it is wrong somewhere, but I can't find the error. I would greatly appreciate if someone can help Why do you think it's wrong? Seems correct

Maybe cuz it's different than any other solution that I have ever seen to this problem. Moreover, it's the first RMO geo problem that I have solved myselves. Otherwise I can't even solve the RMO 1990 problems of Geo.

@above Your solution is correct!! Bravo!!

Jupiter_is_BIG wrote: @above Your solution is correct!! Bravo!! Thanks for checking

Math-wiz wrote: Maybe cuz it's different than any other solution that I have ever seen to this problem. Moreover, it's the first RMO geo problem that I have solved myselves. Otherwise I can't even solve the RMO 1990 problems of Geo. There is a first problem fro everyone. Even tho I think you are just messing around

A ruler with length marking rit? So I can measure altitudes?

No, without markings

It's easy to draw paralall line to $OA$ which pass through $P$ . For construction join $O,P $ then draw an angle at$ P$ wose magnitude is equal to $\angle AOP$. So we get paralall line to $AO$ . suppose this line met $OB$ at $K$ now draw a point $D$ on $OB$ such that $OK=KD$ .Now join $D,P$ and suppose it cut $OA$ at $C$ . Using $\textbf{Thale's} $theorem $CP:CD =1:2$.

Math-wiz wrote: Can someone check if my solution is correct? I am not too sure about it, as this is the first geometric(ish) problem of RMO that I have solved myself.

I think it is wrong somewhere, but I can't find the error. I would greatly appreciate if someone can help This solution is correct but please explain why CD is our required segment just by writing one or two lines of explanation (I know that it is trivial, but still you need to write that in order to complete your proof, otherwise you might loose some marks )

This was a little bit trivial I could not guess it at first

Vrangr wrote: Join $OP$, extend it to $O'$ such that $OP=PO'$. Bisect $O'P$, let this point be $M$. Construct $MX$ and $MD$, with $X$ at $A$ such that $M, X, D$ are collinear and $MX=MD$. Bisect $OX$, this point is $C$, join $CD$. Bruh this is asking too much. I think you need to specify how to draw $M, X, D$ such that they are collinear and $MX=MD$.

My solution: Draw a perpendicular line $l_1$ on $\overrightarrow{OB}$ passing through $P$. Now draw a perpendicular line $l_2$ on $l_1$ passing through $P$. Call $\overrightarrow{OA} \cap l_2 = A'$. Measure the length $PA'$ using compass. Then, start cutting $\overrightarrow{OB}$ from $O$ with the compass, preserving the measurement of line $PA'$. Cut it $3$ times in this manner. And then call the $3^{rd}$ cut $D$. And then extend $DP$ and you get $\overrightarrow{OA} \cap \overrightarrow{DP} = C$

Hmm I love constructions Join $OP$ , extend it to a point $P'$ such that $OP:PP' = 1 : 2$ (can be done by duplicating $OP$ twice) then draw a line parallel to $OB$ , let it intersect $OA$ at $C'$ then join $CP$ extend it to meet $OB$ at $D'$, notice that $C'PP'$ similar to $D'PO$ (AAA criteria and parallel line), now the ratio must be the same so $C'P:PD' = 2:1$ which was to be constructed I noticed that I proved for 2:1 but the the idea for 1:2 is the same just ratio at the start diff sry bout that

Draw a line $l$ parallel to $OB$ through $P$, and let it meet ray $OA$ at $X$. Reflect $O$ about $X$ to get $X'$ and then reflect $X$ about $X'$ to get $C$. Then make line $CP$ intersect $OB$ at $D$. Then $CD$ is the required segment. $\square$