## Monday, July 2, 2012

### Problem 778: Triangle, Distance from the Orthocenter to a Vertex, Circle, Circumradius, Side, Square

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

Click the figure below to see the complete problem 778.

1. http://img213.imageshack.us/img213/5918/problem778.png

Draw lines per attached sketch
We have LC//BH and LB//CE
So BLCH is a parallelogram and
BH=LC
In right triangle ACL we have AC^2+CL^2=AL^2
Or BH^2+AC^2=4.R^2

2. http://img232.imageshack.us/img232/3764/p778resuelto.png
(look the construction of the picture).

As AP is diameter, <PCA = 90° and <ABP=90°. Note that PC and HC are paralell to BH and BP respectively. Then BH = PC (because BPCH is paralelogram) and by pythagorean theorem we have that PC^2 + AC^2=4R^2. Hence BH^2+AC^2=4R^2

3. BH = 2R cos(ABC)...by Prob 777
AC = 2R sin(ABC)...by Sine rule

BH^2 + AC^2 = 4R^2

4. By using the result in problem 777, the statement would be reduce to AC = 2R sinABC
,which is obvious under sine law.

5. Let AO meet the circle at X.

< AXB = < B so < BCE = 90. - B = < CAX = CBX hence BX//HC and since BF//XC BXCF is a parellelogram and it thus follows that BH = XC

Now applying Pythagoras to right Tr. ACX the result follows

Sumith Peiris
Moratuwa
Sri Lanka