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.

#Geometry

Note by Mithil Shah
4 years, 2 months ago

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Comments

Ahmad Saad - 4 years, 2 months ago

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Very nice solution sir .... :) Upvoted ✓

Rahil Sehgal - 4 years, 2 months ago

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Thanks. I appreciate you.

Ahmad Saad - 4 years, 2 months ago

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

..

Constructions:::−−−Constructions:::--- Produce PCPC to P′P' such that CP′CP' = 2AP2AP. Join BP′BP'. Then draw PD⊥BP′PD \perp BP'. Finally join MDMD.

Given that- APAP + PCPC = ABAB

=> 2AP2AP + PCPC = AB+APAB + AP

=> CP′CP' + PCPC = PP′PP' ---[Because CP′CP' = 2AP2AP.]

=> PB=PP′PB = PP'

=> ∠B=∠P′=θ\angle B = \angle P' =\theta And ∠BPD=∠APD=90−θ\angle BPD =\angle APD = 90-\theta ---- [11]

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

=> DMDM = 12CP′\frac{1}{2}CP' = APAP ---- [22]& also, DM∣∣P′CDM || P'C [By Midpoint Theorem]

DM∣∣P′CDM || P'C => ∠PDM=90−∠MDB\angle PDM = 90 - \angle MDB = 90−∠P′=90−θ90 - \angle P' = 90 - \theta. ---- [33]

From the results in eqneq^{n} [1,2,3][1 , 2 , 3], we get that DM=APDM = AP and ∠MDP=∠APD\angle MDP = \angle APD => MDPAMDPA is isosceles trapezium. And hence, AM∣∣PDAM || PD ,also AM⊥PDAM \perp PD => PD⊥QCPD \perp QC but PDPD also bisects ∠APP′\angle APP'=> PQ=PCPQ = PC But also PB=PP′PB = PP'. => BQ=CP′=2APBQ = CP' = 2AP

K.I.P.K.I.GK.I.P.K.I.G

Vishwash Kumar ΓΞΩ - 4 years, 2 months ago

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Maybe it is from INMO, i do not remember either.

mithil shah - 4 years, 2 months ago

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it is from CRMO-2 2014

akarsh jain - 4 years, 2 months ago

On which line is P; BA or CA?

Yatin Khanna - 4 years, 2 months ago
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