## Thursday, August 19, 2010

### Problem 506: Triangle with Three Squares, Center, Midpoint, Perpendicular, Congruence

Geometry Problem
Click the figure below to see the complete problem 506 about Triangle with Three Squares, Center, Midpoint, Perpendicular, Congruence.

See also:
Complete Problem 506

Level: High School, SAT Prep, College geometry

#### 3 comments:

1. Let Q is the projection of E on AC ; N is the projection of M on AC
Let K is the projection of B on AC; P is the projection of G on AC
We will prove that N is the midpoint of AC and MN=1/2* AC
Since M is the midpoint of EQ so N is the midpoint of PQ .
We have CP=CQ.cos(90-C)= a.sin(C) = BK
GP=a.cos(C )= CK
QA= c.cos(A) = BK and EQ=c.cos(A)= AK
Since CP=QA so N will be midpoint of AC.
In the trapezoid EQPG we have MN=.5*(EQ+GP)=.5(AK+CK)=.5*AC
So AMC is a right and isosceles triangle and AMCO is a square.
Similarly if we project points E, M, B, G over DH and with the same way as above
We can show that DMH is a right and isosceles triangle .

Peter Tran

2. Aplicamos el teorema de Botema en el segmento AC y obtenemos que AM=MC, que AC=2MX (donde X es proyeccion ortogonal de M en AC) y que <AMC=90.
Como AO=OC y <AOC=90 tenemos que AOCM es cuadrado.
by mathreyes

3. From problem 496, AH and DC meet orthogonally at point P on line EC.
Let M1 be intersection of circumcircle of APC with line EC, which must be proven to coincide with M.
Also from 496, <EPA=<GPC=45.
Using cyclic condition, <M1CA=<M1PA=45 and <M1AC=<GPC=45 making AM1C isosceles right triangle.
AH=CD, AM1=CM1, and <M1AH=<M1CD from cyclic quadrilateral ACPM1, so triangle M1CD is 90 rotation of triangle M1AH about M1.
Triangle EAM1 is reduced rotation of triangle DAC by factor of sqrt(2). Same argument applies for triangles GCM1 and HCA. However, since DC=AH, EM1=GM1=DC/sqrt(2).