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 ?

DeleteI try to translate the solution of Florin to English . see below

DeleteThe triangles GMH and CMA are similar,

whence GH / AC = GM / MC = MH / MA

and how the angles GMA = CMH result in GA // HC

that is, ACHG parallelogram, from where

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

Let BM intersect with AC at point P.

As M is the midpoint of the diagonals it results

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

Let now be the point N the intersection of DE with BP.

Then from the cross-sectional theorem in the triangle BAC it follows that

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

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

That is, NP = NB and hence the points M = N.

Florin Popa, Comănești, Romania.

In my opinion , there are some errors in above solution.

GA is not parallel to HC and ACHG is not a parallelogram

Peter Tran

It 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

https://photos.app.goo.gl/cRCLTb68yREoFXGg7

ReplyDeletenote that triangles DBG and DAF are isosceles => F, D, G are collinear

similarly F, E, H are colinear

Per Pappus theorem applied to colinear points G,B,H and A, F, C .

Pappus line formed by D,M,E will be colinear

Problem solve with Hyperbola (Adriaan van Roomen).

ReplyDeleteBig Circle Radius: R

Small Circle Radius: r

Let the big Circle Midpont A(0,0) => F1 (Hyperbola Foci)

Let the small Circle Midpoint C(R+r,0) => F2 (Hyperbola Foci)

1.)

We know the Hyperbola have the linear eccentricity length:

2*e=R+r

The Hyperbola Center (e,0).

2.)

The Hyperola curve will be construction on the Intersections Point with the big and Small Circle, so:

(x-e)²/a² - y²/b² = 1 (hyp.)

And: e-a=r => a=(R+r)/2 , b=√(R*r) | Because: a²+b²=e²

3.)

On the Hyperbola curve can be construction a other Circle (only touch Circle to the two Circle (F1 , F2)) with the Radius w and Center B(k,l).

-> k²+l² = (R+w)²

(k-e)²/a² - l²/b² = 1 (Point (k,l) on hyp.)

=> Solve (k,l) <=> (R; r; w) or B(k,l) (Point B)

=> Solve Intersect Point's (D and E) from the Circles F1 and F2, to Point B(k,l)

x²+y²=R²

(x-k)²+(y-l)²=w²

-> Solve Point D

(x-(R+r))² + y²=r²

(x-k)²+(y-l)²=w²

-> Solve Point E

4.)

Construction with the Parallel Line to Ox,

so: y=l

And now it can win two other Points, with Intersect of Line and the 3'rd Circle (k,l).

(x-k)²+(y-l)²=w²

y=l

Solve x1 and x2: x1=k+w , x2=k-w, who x1 > x2, and x1=Point H, x2=Point G

5.)

Construction two Lines on Points x1 and F1, x2 and F2.

The Intersect Point give the Point M

6.)

Point D, M and E can construction three Lines, who will Prove the three Lines are the same.