Prove angle relation given parallelogram

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Prove angle relation given parallelogram

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Source: CSEMO 2018 Grade 11 Q3

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#1 h Jul 30, 2018, 5:28 AM • 1 Y

Y by Adventure10

Let $O$ be the circumcenter of $\triangle ABC$ , where $\angle ABC> 90^{\circ}$ and $M$ is the midpoint of $BC$ . Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$ . $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$ . If quadrilateral $ADOE$ is a parallelogram, prove that $\angle OPE = \angle AMB.$

This post has been edited 1 time. Last edited by mofumofu, Jul 30, 2018, 5:43 AM

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#2 h Jul 31, 2018, 11:39 AM • 1 Y

Y by Adventure10

Nice problem

Let $N$ be the midpoint of $DE$ (which coincides with the midpoint of $AO$ ), $X, Y$ be the projection of $B, C$ on line $DE$ .

Easy length-chasing with $XP=XR, YP=YR$ leads : $XP=YN$ . Hence, $MP=MN$ ; or $P, N$ lies on the circle $(M, MB)$ .

Consider the radical axis theorem on circles $(O), (ANP), (BCNP)$ imply that $PN, BC$ and the tangent from point $A$ at $(O)$ are concurrent.
Say that they concur at $R$ .

Then $\angle RAO=\angle OMR \implies (R, A, M, O) \text{ are concylic } \implies \angle RMA=\angle ROA$ , and also $NO^2=NA^2=NP \times NR \implies \angle RON=\angle NPO$ .

Hence, $\angle RMA=\angle NPO$ , as desired. $\blacksquare$

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#3 h Mar 26, 2021, 9:39 AM • 3 Y

Y by GorgonMathDota, Mango247, Mango247

Extension of Grade 10 P3. Nevertheless, still a beautiful problem.

CSEMO 2018 Grade 11 Q3 wrote:

Let $O$ be the circumcenter of $\triangle ABC$ , where $\angle ABC> 90^{\circ}$ and $M$ is the midpoint of $BC$ . Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$ . $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$ . If quadrilateral $ADOE$ is a parallelogram, prove that $\angle OPE = \angle AMB.$

Let $F$ be the midpoint of $DE$ , $I$ be the intersection of $BD$ and $CE$ , and $J$ be the intersection of $BC$ and $DE$ . Since $ADOE$ is a parallelogram, $AO$ bisects $DE$ .
Claim 01. $BPFC$ is cyclic.
Proof. Notice that $DB \perp CE$ . Since $F$ is the midpoint of $DE$ , we have
$\measuredangle FIB \equiv \measuredangle FID = \measuredangle IDF \equiv \measuredangle BDP = \measuredangle DPB = \measuredangle FPB$ Therefore $BPFI$ is cyclic. Similarly, we get $BFCI$ being cyclic. Since $\measuredangle BIC = \measuredangle BPC = 90^{\circ}$ as well, we have $BPCI$ is cyclic. Therefore, we conclude that $BPFC$ is cyclic.

Claim 02. $AJ$ tangent to $(ABC)$ .
Proof. By radical axis theorem on $(ABC), (APF), (BPFC)$ . We have $PF \cap BC = J$ lie on the radical axis of both $(ABC)$ and $(APF)$ . However, $(APF)$ has diameter $AF \equiv AO$ , and therefore these circles are tangent to each other at $A$ . Hence, $AJ$ is the radical axis of these circle, proving that $AJ$ is tangent to $(ABC)$ .

Claim 03. $OA$ tangent to $(JPO)$ .
Proof. First, we claim that $OA$ is tangent to $(JAP)$ . This is obvious as $(JAP)$ has diameter $AJ$ , which is perpendicular to $AO$ . Since $DE$ bisects $AO$ , $F$ is the midpoint of $AO$ . Now, we have
$FJ \cdot FP = \text{Pow}_F (JAP) = FA^2 = FO^2$ Therefore, $AO$ is tangent to $(JPO)$ .

Claim 04. $OAJM$ is cyclic.
Proof. This follows from $\measuredangle OAJ = 90^{\circ} = \measuredangle OMJ$ .

To conclude, notice that
$\measuredangle OPE \equiv \measuredangle OPF = \measuredangle FOJ = \measuredangle AOJ = \measuredangle AMJ= \measuredangle AMB$

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