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# RMO 2014

Let ABC be a triangle with AB>AC. Let P be a point on line beyond A such that AP+PC=AB. Let M be the mid-point of BC and let Q be a point on the side AB such that CQ intersect AM at right angle. Prove that BQ = 2AP.

Note by Mithil Shah
4 months, 3 weeks ago

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· 4 months, 3 weeks ago

Very nice solution sir .... :) Upvoted ✓ · 4 months, 3 weeks ago

Thanks. I appreciate you. · 4 months, 3 weeks ago

@mithil shah It is not from RMO 2014. It is from INMO. I don't remember the Date.

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$$Constructions:::---$$ Produce $$PC$$ to $$P'$$ such that $$CP'$$ = $$2AP$$. Join $$BP'$$. Then draw $$PD \perp BP'$$. Finally join $$MD$$.

Given that- $$AP$$ + $$PC$$ = $$AB$$

=> $$2AP$$ + $$PC$$ = $$AB + AP$$

=> $$CP'$$ + $$PC$$ = $$PP'$$ ---[Because $$CP'$$ = $$2AP$$.]

=> $$PB = PP'$$

=> $$\angle B = \angle P' =\theta$$ And $$\angle BPD =\angle APD = 90-\theta$$ ---- [$$1$$]

But $$PB = PP'$$ and also $$PD \perp BP'$$. => $$BD = DP'$$

=> $$DM$$ = $$\frac{1}{2}CP'$$ = $$AP$$ ---- [$$2$$]& also, $$DM || P'C$$ [By Midpoint Theorem]

$$DM || P'C$$ => $$\angle PDM = 90 - \angle MDB$$ = $$90 - \angle P' = 90 - \theta$$. ---- [$$3$$]

From the results in $$eq^{n}$$ $$[1 , 2 , 3]$$, we get that $$DM = AP$$ and $$\angle MDP = \angle APD$$ => $$MDPA$$ is isosceles trapezium. And hence, $$AM || PD$$ ,also $$AM \perp PD$$ => $$PD \perp QC$$ but $$PD$$ also bisects $$\angle APP'$$=> $$PQ = PC$$ But also $$PB = PP'$$. => $$BQ = CP' = 2AP$$

$$K.I.P.K.I.G$$ · 4 months, 2 weeks ago

Maybe it is from INMO, i do not remember either. · 4 months, 2 weeks ago

it is from CRMO-2 2014 · 4 months, 2 weeks ago