Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, November 24, 2018

### Geometry Problem 1402: Triangle, Three Tangent Circles, Parallel Diameter, Collinear Points

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Triunghiurile GMH și CMA sunt similare ,

ReplyDeletede unde GH/AC = GM/MC = MH/MA

și cum unghiurile GMA = CMH rezultă GA // HC

adică ACHG paralelogram, de unde

GH = AC, GM = MC, MH = MA.

Fie BM intersectat cu AC în punctul P.

Cum M este punctul de mijloc al diagonalelor rezultă

BM = MP, BG =BH, AP=PC.

Fie acum punctul N intersecția lui DE cu BP.

Atunci din teorema transversalei în triunghiul BAC rezultă că

NP/NB *AC = (DA/DB)*PC + (EC/EB )*AP = AP*(DA/DB + EC/EB) =

= AP*(DA+EC)/DB = DA + EC = AF + FC =AC . De aici NP/NB =1,

Adică NP = NB și de aici punctele M=N.

Florin Popa, Comănești , România.

ACGH is not a paralelogram ?

DeleteIt is the same thing to be proved M, F, intersections of internal tg are collinear

ReplyDeleteMark Δ(delta) the line that goes through the points D and E.

ReplyDeleteΔ ∩ AC =Q; Δ ∩ GH = P; Δ ∩ AB=D₀

1. ΔAMC ͠ ΔMGH; AM/MH=CM/MG=AC/GH=(R₁+R₃/2R₂)

2. ΔCEQ ͠ ΔBPE; (CQ/BP=CE/BE X/Y+R₂= R₃/R₂)

3. ΔCMQ ͠ ΔMPG; (CQ/PG=CM/MG X/Y= R₁+R₃/2R₂)

4. ΔAQD₀ ͠ ΔPBD₀; (AQ/PB=AD₀/D₀B R₁+R₃+X/R₂+Y= Z/R₁+R₂-Z)

By solving the system:

X= (R₁+R₃)R₃/R₁-R₃

y= 2R₂*R₃/R₁-R₃

Z=R₁

AD₀=R₁=AD D₀ Ξ D

E,M,D are colinear

Erina, NJ

Also F,D,G and F,E,H are collinear. Problem would have been more challenging if FD and FE extended meets circle "b" at G and H. Then we have to prove GH is diameter and also parallel to AC.

ReplyDeleteACHG is a paralelogram if radius of circle B is arithmetic average of radius of the circle A and the circle C. In all the other cases ACGH trapezus.

ReplyDeleteErina,NJ