## Saturday, November 24, 2018

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. #### 6 comments:

1. Triunghiurile GMH și CMA sunt similare ,
de 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.

1. ACGH is not a paralelogram ?

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

3. Mark Δ(delta) the line that goes through the points D and E.
Δ ∩ 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

4. 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.

5. ACHG 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.

Erina,NJ