Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, July 11, 2015

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

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ReplyDeleteLet 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

Let <BC1A1=a and CB1A1=b.

ReplyDelete<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

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

ReplyDeleteSimilarly 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

Considering usual triangle notations

ReplyDeletewe 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