## Friday, January 27, 2017

### Geometry Problem 1310 Square, Center, Right Triangle, Area, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Let AO = a and OF = b so that FC = a-b and AC = sqrt2.a

ab = 30......(1)

From areas of similar triangles

(a^2+b^2)/(a-b)^2 = 15/7 which simplifies to
a^2+b^2 = 225/2 ...,(2) using (1)

Now use Ptolemy in cyclic quadrilateral AOBC, noting that BC = (a-b)a/sqrt(a^2+b^2) using Pythagoras and we get after much simplification

x = 2ab/sqrt2(a^2+b^2) = 60/sqrt225 (from (2)) = 4

Sumith Peiris
Moratuwa
Sri Lanka

2. To Sumith
On what relation is included x

1. Note that AF = 15/sqrt2

Using Ptolemy in AOBC

sqrt2.a.x + a.BC = AB.a

Now note that BC = sqrt2. a(a-b)/15

and AB = 15/sqrt2 + b.(a-b)/sqrt2/15

Now substitute these in the Ptolemy equation

3. Full Ptolemy isn't really necc. once you derive AF = 15/sqrt(2). Then since its cyclic OBF is similar to ACF and x / AC = OF / AF or x = AC * OF / AF.

AC = sqrt(2) * a since its a 45-45-90 and OF = 30 / a using the AFO's area.

So AC * OF = 30 * sqrt(2) and 30 * sqrt(2) / (15 / sqrt(2)) = 4.

1. 2. 4. AC = a \/2 (a sqrt2)