Let ABCD be a convex quadrilateral. If M,N,K are the midpoints of the segments AB,BC, and CD, respectively, and there is also a point P inside the quadrilateral ABCD such that, ∠BPN=∠PAD and ∠CPN=∠PDA. Show that AB⋅CD=4PM⋅PK.
Problem
Source: Mexico 2023/3
Tags: geometry, 2023, P3, Mexico, Hi
08.11.2023 12:09
If Q∈(PBC) such that PBQC is harmonic, then by angle chasing △QBC∼△PAD, so PBPC=PAPD⟹PAPB=PDPC. Also, ∠BPC+∠APD=∠APB+∠CPD=180∘, so if R∈(PCD) such that PCRD is harmonic, △PAB∼△RDC. Thus, 4⋅PM⋅PK=4⋅PK⋅KR⋅ABCD. By Power of a Point from K to (PCRD), PK⋅KR=(CD2)2, so 4⋅PM⋅PK=CD2⋅ABCD=AB⋅CD as desired. ◻
09.11.2023 12:29
28.12.2023 03:39
Take R and S such that PARB and PCSD are parallelograms, in which P,M, R collinear, P,K,S collinear and P,N,Q collinear. [asy][asy] import geometry; size(200); pair A = (2, 5), B = (8.62, 7.62), C = (7.78, 0.71), D = (3.95, 1.56), P=(5, 4), R=(5.62,8.62), S=(6.73,-1.74), Q=(11.4,4.33); draw (P--Q, grey+dashed); pair M = midpoint(A--B); pair N = midpoint(B--C); pair K = midpoint(C--D); draw (P--N); filldraw ( A--P--B--R--cycle , opacity (1) + lightcyan , blue ) ; filldraw ( D--P--C--S--cycle , opacity (1) + lightcyan , blue ) ; draw (B--Q, green+dashed); draw (P--R, blue+dashed); draw (P--S, blue+dashed); draw (A--B--C--D--cycle); draw(arc(P, 0.5, degrees(B-P), degrees(A-P)), red); draw(arc(P, 0.5, degrees(C-P), degrees(D-P)), blue); draw(arc(B, 0.5, degrees(P-B), degrees(R-B)), blue); draw(arc(C, 0.5, degrees(P-C), degrees(S-C)), red); filldraw ( P--Q--C--cycle , opacity (0) + lightgreen , black) ; filldraw ( A--P--D--cycle , opacity (0) + lightgreen , black ) ; draw (B--C,black); dot("A", A, dir(180)); dot("B", B, dir(0)); dot("C", C, dir(350)); dot("D", D, dir(200)); dot("P", P, dir(110)); dot("M", M, dir(330)); dot("N", N, dir(0)); dot("K", K, dir(30)); dot("R", R, dir(90)); dot("S", S, dir(350)); dot("Q", Q, dir(30)); [/asy][/asy] Now, ∠CPB+∠APD=180∘ so ∠BPA+∠DPC=180∘ and as △PAD∼△CQP we obtain PAPD=PBPCthen PARB and PCSD are similar therefore PRAB=CDPS⇒AB⋅CD=4PM⋅PKas desired
28.12.2023 08:50
Very nice, guys! Thanks for all!
03.02.2024 23:20
Construct parallelograms PDXC and PAYB. Then ∠DPA=180∘−∠CPN−∠BPN, so ∠DPC+∠APB=180∘. It follows that ∠DXC+∠AYB=180∘ so PDXC and PAYB are similar. From this, we find that CDPY=PXAB⟹AB⋅CD=4PM⋅PK.
06.03.2024 02:51
The main claim is that the same angle conditions also hold "on the other side", i.e. Claim: ∠APM=∠PDC and symmetric variants. Let ∠BPN=θ1 and ∠CPN=θ2. Then, by ratio lemma we have PCPB=sinθ1sinθ2=PDPA. Now, let ∠APM=θ3 and ∠BPM=θ4. Note that ∠APD+∠PBC=180, so ∠APB+∠DPC=180, which means that ∠PDC+∠PCD=θ3+θ4.Furthermore, sin∠PDCsin∠PCD=PCPD=PBPA=sinθ3sinθ4,and since ∠PDC+∠PCD=θ3+θ4, we must have ∠PDC=θ3 and ∠PCD=θ4, as desired. The condition we want to show can be rearranged as BMPM×CKPK=1sin∠MPBsin∠KPCsin∠ABPsin∠DCP.However, as shown in our previous claim, ∠MPB=∠DCP and ∠KPC=∠ABP, so we are done. Remark: There is a simpler way to finish after using the ratio lemma, which I saw from other solutions on the AoPS thread. By erecting parallelograms outside of the quadrilateral, we can directly take advantage of ∠APB+∠DPC=180 and the equal ratios from the ratio lemma to induce similar triangles which proves Claim 7.1 directly. In fact, if we construct two parallelograms, one on either side, we don't even have to do the ratio lemma! Noting that the two parallelograms are similar suffices, as @dolphinday explained above. The lesson here is that when there are many midpoint conditions involved, constructing parallelograms is often useful, especially if we have length or angle conditions that "do not touch" as it lets us "move" angles and lengths.
07.03.2024 18:43
Solved with MathLuis. Very fun problem. The great god carried - as always. Note that the given condition rewrites to, AB⋅CD=4PM⋅PKPM⋅PK=AB2⋅CD2PM⋅PK=BM⋅KCPMBM=KCKPwhich upon applying the Sine Rule on triangles △BMP and △PKC, suffices to show sin∠PABsin∠APM=sin∠DPKsin∠PDC(1). We now show the following key claim. Claim : PAPB=PDPC Proof : By Ratio Lemma on △BPC we have that CPBP=sin∠NPBsin∠CPN=sin∠PADsin∠PDA=PDPAwhich upon rearrangement yields the desired ratio condition. Then, note that by applying the Ratio Lemma on △PDC and Law of Sines on △PCK we have PDCP=sin∠CPKsin∠DPK=CKsin∠PCKPKsin∠DPKSimilarly we obtain that, PAPB=sin∠BPMsin∠APM=BMsin∠PBAPMsin∠APMThus, CKsin∠PCKPKsin∠DPK=BMsin∠PBAPMsin∠APMSo, comparing this with the desired relation (1), it suffices to show △BMP∼△CKP. This is our final claim. Claim : The triangles △BMP and △CPK are similar. Proof : Let Q the Miquel point of ABCD. We consider the spiral similarity taking △APB→△DRC. Using this spiral similarity centered at Q we get that CPPD=BPPA=CRRDThis along with ∠APD=180−∠PAD−∠PDA=180−∠BPN−∠CPN=180−∠BPC gives that CRDP is cyclic and also an harmonic quadrilateral. Now to check all of the following chain of similarities use force overlay and the spiral similarity at Q, we basically have △BMP∼△CKR∼△RDP∼△PKCwhich was the desired claim. As per the previous reduction, this means that we are done.
01.04.2024 09:44
We first note ∠APB+∠CPD=180. Ratio lemma tells us CPBP=sin∠BPNsin∠CPN=sin∠PADsin∠PDA=DPAP. Thus we can perform a spiral similarity at P sending A↦D which makes B′PC collinear. Since this transformation preserves the value of AB⋅CDPM⋅PK, we are finished by noting that PK and PM′ are midlines in △B′CD. ◼
16.04.2024 12:16
quad2
11.08.2024 04:13
@v4913 Why does the similarity between QBC and PAD implies the first equality?
23.11.2024 21:50
@dolphinday You wrote: ∠DPC+∠APB=180∘. wouldn't that imply ∠APD+∠BPC=180∘? and hence triangle DPA isoceles? ( btw, I had the same approach as you but I couldn't carry the thought further than this T-T ) Thanks in advance!