## Tuesday, December 5, 2017

### Geometry Problem 1350: Triangle with three Intersecting Circles, Incircle, 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. 1. https://photos.app.goo.gl/Yubn9z3lzvSmV5w03

Denote A’, B’, C’ as points of contact of incircle I to triangle ABC
a, b,c and 2p as 3 sides and perimeter of triangle ABC
Denote k= B’C/B’A , s= C’A/C’B and q=A’C/A’B
Define point B5 on line AC such that B4B’=B4B5 ( see sketch)
Since B4 is on the radical line of circles I and B1 so
So B4A.B4C=B4B’^2=B4B5^2 => ( ACB’B5)= -1
And k= (B’C/B’A)=(B5C/B5A) => B’C/k=B’A/1=b/(1+k) => B’A= b/(1+k)
Similarly B5A=b/(1-k)
We have AB4= ½(AB’+AB5)= b/(1-k^2) and CB4=AB4- b= b.k^2/(1-k^2)
And B4C/B4A= k^2
Similarly we also have C4A/C4B= s^2 and A4B/A4C= 1/q^2
And (B4C/B4A) x(C4A/C4B)x(A4B/A4C)= k^2 . s^2/q^2
Replace k=(p-c)/(p-a) and s=(p-a)/(p-b) , q=(p-c)/(p-b) in above expression
We will have (B4C/B4A) x(C4A/C4B)x(A4B/A4C)= 1
And A4, B4 ,C4 are collinear per Menelaus’s theorem

2. Let T1,T2,T3 be the points of contact of the in-circle
with the sides BC,CA,AB resp.
Let A4T1=x,B4T2=y,C4T3=z
By Tangent-Secant property:
x(x+a)=A4C.A4B=A4A3.A4A2 referred to circle A1 while
A4A3.A4A2=A4T1^2 referred to in-circle.
Note A4T1=A4C+CT1=x+(s-c)
So x(x+a)=[x+(s-c)]^2,
xa=(s-c)^2 + 2x(s-c),
x(a+2c-2s)=(s-c)^2,
x(c-b)=(s-c)^2,
x=(s-c)^2/(c-b) and
x+a=(s-b)^2/(c-b)on further simplification.
Hence A4C/A4B=(s-c)^2/(s-b)^2
By cyclic symmetry B4C/B4A=(s-a)^2/(s-c)^2
and C4A/C4B=(s-b)^2/(s-a)^2
Product of the above 3 ratios being 1 (numerically),
by Menealau C4,A4,B4 are collinear.

3. Problem 1350
Denote the points of contact of the incircle with the sides BC, CA, AB as P, Q, R respectively. We first evaluate the ratio BA4:A4C in terms of a, b, c, and s (respectively the lengths of the sides BC, CA, AB and the semi perimeter of the triangle). Cyclically permuting we get the values of the similar ratios CB4:B4A and AC4:C4B. We next show that the product of these 3 ratios is -1, from which fact follows the collinearity of the points C4, A4, and B4 by the converse of Menelau’s Theorem.
First consider the in-circle I for which A4P is a tangent and A4 A3 A2 is a secant. By the Tangent – Secant Property we get A4P^2 = A4A3. A4A2
Next consider the circle A1 for which both A4 A3 A2 and A4CB are secants. By the Secants Property we get A4A3 . A4A2 = A4C . A4 B.
Hence it follows A4C . A4B = A4B3^2. It is well known that CP = s - c
At this stage we let A4C = x . So A4P = A4C + CP = x + s -c
So x(x + a) = (x + s - c)2,
Cancelling x^2 on either side we get xa = (s – c)^2 + 2(s-c)x,
x(a + 2c – 2s) = (s – c)^2,
x( c – b) = (s – c)^2,
x = (s – c)^2 / (c – b) ......(i)
Also 4(x + a)(c – b) = 4x(c – b) + 4a(c – b)
= 4(s – c)^2 + 4a(c – b) = (2s – 2c)^2 + 4ac – 4ab
= (a + b –c)^2 + 4ac – 4ab
= a^2 + b^2 + c^2 + 2ab -2bc – 2ca + 4ac – 4ab
= a^2 + b^2 + c^2 – 2ab – 2bc + 2ac = (c + a – b)^2
= (2s – 2b)^2= 4(s – b)^2, So
x + a = (s – b)^2 / (c – b) .......(ii)
Hence from (i) and (ii)
BA4 / A4C = (x + a)/x = (s – b)^2 /(s – c)^2 in magnitude.
Similarly by Cyclic Symmetry:
CB4 / B4A = (s - c)^2 / (s – a)^2 ,
AC4 / C4B = (s – a)^2 / (s – b)^2
With due regard to sign we obtain on multiplication:

[BA4 / A4C] [CB4 / B4A] [AC4 / C4B] = -1
Hence by Menelau’s Theorem
C4, A4, and B4 are collinear.

4. Pravin: The above post is only a detailed version of my earlier post. Thank you.

5. Referring to Problem 1350 and its figure.
Let J be the center of the circle through B, C, and I
Do the the following results hold good ?
(i) R' = R / 2 cos A
(ii)<OJI = C - A
(iii) OI = R sin [(C - A)/2]/cos A

6. All solutions posted here seems to be a bit to complex.
Its easy to see A_4,B_4,C_4 all lie on power axis of incircle (i) and circum crircle of ABC (o)
for example A_4 lies on power axis of i and A_1 (line A_2A_3)
as well as on power axis of o and A_1 (line BC)
So A_4 lies on power axis of o and i.
Same for B_4 and C_4.

1. Zakrzew

You got an excellent solution !

Peter tRan