See complete Problem 244 at:

gogeometry.com/problem/p244_square_centers_diagonal_vertex.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, February 4, 2009

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

Labels:
center,
congruence,
diagonal,
distance,
similarity,
square

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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.

ReplyDeleteNow, 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.

draw DH // a & DH = a

ReplyDeletejoin 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

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

ReplyDeleteH is a point in a line // to OM and in a distance a

ReplyDeletefrom 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

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

ReplyDeleteHence Triangles COM and DCE are similar and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

See the

ReplyDeletedrawingDefine 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