## Saturday, July 11, 2015

### Geometry Problem 1132: Triangle, Excircle, Tangency Point, Parallel, Midpoint

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

Click the diagram below to enlarge it.

#### 4 comments:

1. http://s11.postimg.org/mjwfzd603/pro_1132.png

Let 2p= perimeter of triangle ABC
Let AB2 and AC2 cut BC at G and H
We have AB1=AC1= p and triangles A1CB1, ACG, BA1C1 and ABH are isosceles
So GA1=AB1= p and HA1=AC1= p
So A is the midpoint of HG
Since A1C2//AB2 and A1B2//AH => C2 and B2 are midpoints of AH and AG
C3B3 =1/2 BC

2. Let <BC1A1=a and CB1A1=b.
<BA1C1=<B2A1C=a and <A1C1B1=b.
Since AC2 parallel to A1C1, <AC2B1=<B2A1B1=<AC1B1=a+b so AC2C1B1 is a circle
Similarly, <AB2C1=<C2A1C1=<AB1C1=a+b so AB2B1C1 is a circle.
AC1EB1 is circle because <AC1E=<AB1E=90
All six points A,C2,C1,E,B1,B2 are cyclic
Triangles AC3C2 and A1BC1 are isosceles so <AC3B3=2a=<ABC and C3B3 parallel to BC.
Triangles A1BC1 and A1C2C1 are isosceles so EBC2 is a line with <AC2E=<AC1E=90
Because AC2B=90 and triangle AC3C2 isosceles, C3A=C3C2=C3B and B3C3=BC/2

3. < B1AB2 = <AB1A 1 = < A1C1B1 = C/2, so AC1B1B2 is con-cyclic

Similarly AB1C1C2 is con-cyclic

Hence AB2B1C1C2 are all con-cyclic

So B/2 = < AC1B2 = < AC2B2 = < C2B2C1 = B2A1C, so BC // B3C3

If the diagonals of parallelogram AC2A1B2 cut at X, AX = XA1

Hence applying the mid point theorem to Tr. ABC, XC3 = A1B/2 and XB3 = A1C/2

Therefore B3C3 =BC/2

Sumith Peiris
Moratuwa
Sri Lanka

4. Considering usual triangle notations
we know AC1EB1 are concyclic (since m(AC1E)=m(AB1E)=90) ------(1)
Since AC2//B2C1 => m(C2AB)=m(BC1B2)=B/2 -----(2)
Also in the triangle C1A1B1, m(C1B1A1)=B/2-----------(3)
From (2) & (3), C2 lies on the circle (1)
Similarly it can be shown that B2 also lie on the same circle (1)
Hence m(AC2B2)=m(AC1B2)=B/2 => m(AC3B3)=B and m(AB3C3)=C
=> C3B3//BC -------(4)
Connect C2E and m(C2EA)=m(C2B2A)=C/2
We also know that m(BEA)=C/2 => C2,B & E are collinear
Hereafter,it can be easily found that C3 is the circum-center of AC2B -----(5)
From (4) & (5) it can be deduced that C3 and B3 are the mid-points of AB & AC respectively and the result follows