Geometry Problem

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

Click the figure below to see the problem 733 details.

## Monday, February 27, 2012

### Problem 733: Triangle, Orthocenter, Altitude, Reflection in a line, Circumcircle, Concurrency

Labels:
altitude,
circumcircle,
concurrent,
orthocenter,
reflection,
triangle

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http://img718.imageshack.us/img718/9332/problem733.png

ReplyDeletebelow is my comment with some typo correction and clarification.

Let BH, AH and CH cut the circle at B’, A’ and C’ and A, B, C are angles of triangle ABC

Let angles x, y, z denotes angles shown on attached sketch.

A’, B’, C’ are points of symmetric of H over BC, AC and AB ( property of orthocenter)

So ∠(C’AB’)=2(∠BAH+∠HAC)= 2∠ A

1. Let P is the intersection of C’C1 to B’B1

In triangle C1PB1 ∠ ( C1PB1)= 2(y+z)-180= 180-2∠A

So quadrilateral AC’PB’ is cyclic => point P will be on the circle.

2. Let P1 is the intersection of A1A’ to C1C’

∠ (C’BA’)= 2∠ (B)

In triangle BA1C1 external angle B=x+z

In Triangle A1C1P1, angle(C1P1A1)= 180-2x-2z=180-2B

So quadrilateral C’BA’P1 is cyclic => point P1 will be on the circle

Both points P and P1 are the intersection of line C2 to the circle => P coincide to P1

So A1A2, B1B2 , C1C2 are concurrent at P on circle

i cannot open the attached sketch..plz open this for...i just need this for my presentation.

ReplyDeletehttp://s15.postimg.org/lsh45cs6j/Problem_733.png

ReplyDeleteSee link above for the sketch

Peter Tran