## Thursday, January 26, 2012

### Problem 721: Triangle, Orthocenter, Circumcircle, Altitude, Collinear

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

Click the figure below to see the problem 721 details.

1. http://img688.imageshack.us/img688/225/problem721.png
Let AA” , BB”, CC” cut BC, AC , AB at F,G,E respect.
We have G, F, E are midpoints of HB’, HA’, HC’ ( properties of orthocenter)
Triangles HB’B”, HC’C”, HA’A” are isosceles ( see picture)
1. Note that quadrilateral BFHE is cyclic >> angle(B) supplement to angle (EHF)
But angle (B)= angle (PC’C)+ angle (AA’P)= angle (y+z)>> angle ( EHF) supplement to angle (y+z) >> C”, H,A”” are collinear

2. Quadrilateral HGCA” is cyclic >> angle (GHA”) supplement to angle (C)
But angle (C )= angle (BB’P) – angle (AA’P)= angle (x-z) >> angle (GHA”) supplement to angle (x-z) >> B”, H, A” are collinear
So 4 points B”, C”, H, A” are collinear
Peter Tran

2. http://img688.imageshack.us/img688/225/problem721.png
minor correction due to typo errors or confusion in my previous solution
Let AA” , BB”, CC” cut BC, AC , AB at F,G,E respect.
We have G, F, E are midpoints of HB’, HA’, HC’ ( properties of orthocenter)
Triangles HB’B”, HC’C”, HA’A” are isosceles ( see picture)
1. Note that quadrilateral BFHE is cyclic >> angle(B) supplement to angle (EHF)
But angle (B)= angle (PC’C)+ angle (AA’P)= angle (y+z)>> angle ( EHF) supplement to angle (y+z) >> C”, H,A”” are collinear

2. Quadrilateral HGCF is cyclic => angle (GHF) supplement to angle (C)
But angle (C )= angle (BB’P) – angle (AA’P)= angle (x-z)
angle(B"HB')+angle(B'HA")=x+angle(GHF)-z =angle(GHF)+ angle(C)=180=> B", H, A" are collinear

So 4 points B”, C”, H, A” are collinear
Peter Tran