## Wednesday, August 11, 2010

### Problem 499: Triangle, Two Squares, Perpendicular

Geometry Problem
Click the figure below to see the complete problem 499 about Triangle, Two Squares, Perpendicular.

Complete Problem 499

Level: High School, SAT Prep, College geometry

1. Let Q is projection of E on AC ; H is the projection of B on AC; P is the projection of F on AC
Let O is the point of intersection of AF to BH and O’ is the point of intersection of CE to BH.
We will prove that HO=HO’.

Tri. AHO similar to tri. APF and we have HO/FP=AH/AP
Replace AH=c.cos(A) ; AP= AC+CP= b+a.cos(90-C)=b+a.sin(C) ; FP=a .cos(C)
We get HO=a.c.cos(A).cos(C)/(b+a.sin(C))

Similarly Tri. CHO’ similar to CQE and we have HO’/EQ=CH/CQ
Replace CH=a. cos(C) ; CQ=b +c.sin(A) ; EQ=c.cos(A)
We get HO’= a.c.cos(A).cos(C)/(b+c.sin(A))
Note that a,sin(C )=c.sin(A)=BH
Replace this value in above formulas for HO and HO’ we will get HO=HO’. Point O coincide with O' and BO will perpendicular to AC at H

Peter Tran

2. Any other, more direct proof?

3. Problem 499 (to Anonymous): Hints:
Construct the external square ACPQ and prove that lines AQ and BO are parallel.

More hints: prove that angles AQB and ABO are congruent, see problem 496, cyclic quadrilateral.

4. Construct square ACPQ
BQ and BP cross EC and AF at N and M respectively.

Now, triangles EAC and BAQ are congruent (SAS), EC is perpendicular to QB. Also, AN bisects the right angle (see problem 496).
In the same way triangles FCA and BCP are congruent; BP is perpendicular to AF and CM bisectes the right angle.

Let angle OBN= a
Quad BNOM is cyclic (opp angles supplimentary)
So angle NMO= a (angles on the cirumference subtended by same chord)
Quad ANMC is cyclic (angles ANC, AMC are both 135)
Angles NMA=NCA= a (angles on the cirumference subtended by same chord)
Angle NCA =NQA= a (from congruency)

Hence angle OBQ= angle BQA = a
BO is parallel to AQ
BO is perpendicular to AC