Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Saturday, November 24, 2018
Geometry Problem 1402: Triangle, Three Tangent Circles, Parallel Diameter, Collinear Points
Subscribe to:
Post Comments (Atom)
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.