In square ABCD, AD is 4 centimetres, and M is the midpoint of \( \overline{CD}\). Let O be the intersection of \(\overline{AC}\) and \(\overline{BM}\). What is the ratio of OC to OA? Express your answer as a common fraction.

Guest Feb 25, 2020

#3**+2 **

Let A = (0,4) B = ( 4,4) C =(4,0) D = (0, 0)

The line segment AC lies on a line with an equation of

y = -x + 4

AC = 4sqrt (2)

The segment MB has a slope of [4 -0 ] /[ 4 - 2] = 4/2 = 2

So....this segment lies on a line with an equation of

y= 2 ( x - 2)

y = 2x - 4

The x coordinate of the intersection of these two lines is

-x + 4 = 2x - 4

8 = 3x

x = 8/3

And y = -(8/3) + 4 = 4/3

So OC = sqrt [ (4 - 8/3)^2 + (4/3)^2 ] = sqrt [ ( 4/3)^2 + (4/3)^2 ] = (4/3)sqrt (2)

And OA = AC - OC = (4 - 4/3)sqrt (2) = (8/3)sqrt (2)

So

OC (4/3) sqrt (2) 4 1

___ = ____________ = ___ = _____

OA (8/3) sqrt (2) 8 2

CPhill Feb 25, 2020

#4**+3 **

Melody's method :

Triangle AOB is similar to triangle OCM

This means that

CM / AB = OC / OA

CM = (1/2)AB .....

(1/2) AB OC

_______ = ___

AB OA

1 OC

__ = ____

2 OA

CPhill Feb 25, 2020

#5**+2 **

AD = DC = 4 cm

MC = DC/2 = 2

∠ACD = 45°

AC = AD / sin(45°) = 5.656854249

∠BMC = tan^{-1}(BC/MC) = 63.435°

∠MOC = 180 - 45 - 63.435 = 71.565°

Let F be the foot of altitude from C to MO

CF = sin(∠BMC) * MC = 1.788854382

OC = CF / sin(∠MOC) = 1.885618645

OA = AC - OC = 3.771235604

**The ratio of OC to OA = 1.885618645 / 3.771235604 = 0.5 or 1/2 **

Dragan Feb 25, 2020