## Wednesday, February 4, 2009

### Elearn Geometry Problem 244: Squares, Diagonals, Centers, Distance

See complete Problem 244 at:
gogeometry.com/problem/p244_square_centers_diagonal_vertex.htm

Level: High School, SAT Prep, College geometry

1. Imagine initially that the smaller square had GC on BC, then, that it made a counter-clockwise rotation of θ , around C. Every point on the smaller square, including M and E, will describe an arc of circle of θ, with center in C.

Now, look at triangles OMC and DEC. They are similar, because angle BCG = angle OCM = angle DCE = θ, and corresponds to the amount of the rotation done by the smaller square relatively to the bigger one. By the same token, angle COM = angle CDE = φ, and corresponds to the angle described by OM and DE, relatively to their initial positions OC and DC, respectively.

Therefore, CD/CO = CE/CM = DE/OM = √2

Then, DE = (√2)*OM, or b = (√2)*a q.e.d.

2. draw DH // a & DH = a
join H to M and to E

have to prove HE = a & ang DHE = 90

tr COM and tr HME have
1) HM = OC ( HM = OD )
2) ang OCM = ang HME ( 90 - MCK & 90-FMH, CK perpe CO)
3) CM = ME ( CM = ME, 1/2 diagonals)

=> an COM = MHE & HE = a
=> an DHE = DHM + MHE or DOP + COM = 90

3. Hy, I don't understand where is point H. Help please.

4. H is a point in a line // to OM and in a distance a
from D
We had to achieve in conclusion
DEH is a right isoceles triangle
in it from pythagore theorem a² + a² = b² or b = a√2

5. <OCM = < DCE and CD/CO = CE/CM = sqrt2

Hence Triangles COM and DCE are similar and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

6. See the drawing

Define H such as ΔHOB is the rotation of ΔMOC of -Π
=> OM ⊥ OH, OM=OH=a, MC ⊥ HB
MC ⊥ ME, MC ⊥ HB => ME // HB
Hence MEBH is a parallelogram
=> HM=BE=b
=> b is the hypothenuse of a rectangle triangle with side=a
Therefore 2a^2=b^2