Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1282.
Thursday, November 3, 2016
Geometry Problem 1282 Squares, Common Vertex, Midpoint, Perpendicular, 90 Degrees
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https://goo.gl/photos/QQWPcN1k1LiZB8Jt8
ReplyDeleteLet N, Q, E, F, P and R are the projection of B, D, M, B1, B2 and D2 over CA1 ( see sketch)
See sketch for positions of K, G , L and H.
Let u= D2R=A1P
v=B2P=A1R
v1=BN=CQ
u1=CN=DQ
w=FB1=FA1=FC=FD1
1. Since M is the midpoint of BB2=> E is the midpoint of NP
By observation we have EF=MG=NF-1/2.NP= ½(u1-u)
ME=1/2(v-v1)=GF
GB1= w+1/2(v-v1)
D2K=QR= v+2w-v1= 2.GB1
DK=DQ-QK=u1-u=2.MG
So MG/DK=GB1/D2K= ½
2. Triangle MGB1 similar to DKD2 ( case SAS)
So ∠ (MB1G)= ∠ (DD2K)=> D2,L, H, B1 are cocyclic
And ∠ (B1HD2)= ∠ (B1LD2)= 90 degrees
So B1M ⊥DD2
Thanks Peter
DeleteAnother conclusion in your step 2:
DD2 = 2(B1M)
BY CONGRUENCE D1D2=B2B1, BB1=DD1 AND D1D2 ┴ B2B1 , BB1 ┴ DD1
DeleteLet X be the intersection of D1D2 AND B2B1,Y be the intersection of BB1 AND DD1
THEN ( C, Y, B1, A1, X, D1 ):CONCYCLIC
∠(CYD1)=45=∠(XA1D1) => ∠(XD1A1)=∠(YB1C). SO,∠(BB1B2)=90 ISW ∠(DD1D2)=90
Triangle BB1B2 ≡ DD1D2 ( SAS )
M : triangle BB1B2 with the circumcenter
So BB2=DD2=2*MB1
Let W be the intersection of MB1 AND DD2
∠YDD2= ∠B1BB2= ∠MB1B THEN (D.Y.W.B1) CONCYCLIC => ∠DYB1=∠DWB1=90
I wanted to open the sketch but I couldn't so now I don't understand the solution. Could someone please help me. I have to solve this for my mathclass
Deletesee link below for the sketch of the problem 1282
Deletehttps://photos.app.goo.gl/pN56UvkWo7gKq2n23
Peter Tran