## Wednesday, May 22, 2013

### Problem 878: Square, Inscribed Circle, Quadrant, Center, Perpendicular, Metric Relations

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

Click the figure below to see the complete problem 878. 1. since circle O inscribed in the circle then OE = OF = 2 and AE = AF = 4.
AC = 4√2 then AO = 2√2. if HO = x and FH = a then
4^2 = (2√2 + x)^2 + a^2
OH = x = 0.5√2.

1. how did u get from the third line to the last line?

2. Insert Complex Coordinate plane with O(0,0)
The inscribed circle : |z|=2,
The circle BEFD : |z + (2+2i)| = 4
By solving and get the mid point, point H = 1/2 + 1/2i
OH = 1/sqrt2

3. EF is the radical axis of the two circles so the power of circle at H is the same.
x^2 - 2^2 = (OA + x)^2 - 4^2, where OA = sqrt(8)
By solving, x = 1/sqrt2

4. AE=4, AO=2*sqrt2, OE=2
By Law Of Cosines, angle AOE=110°41'17"
Angle EOH=69°17'43"
OH=2*cosEOH=0.707107

5. AE=4, EO=2, and AO=2*sqrt(2). Then 4^2+(2*sqrt(2))^2-2(4)(2*sqrt(2))cos<EAO=2^2, so cos<EAO=5/(4*sqrt(2)), making AH=4*cos<EAO=5/sqrt(2). AH-AO=1/sqrt(2)=x.

6. EH^2 = 4^2 - (2sqrt2+x)^2 = 2^2 - x^2

Solving x = 1/sqrt2

Sumith Peiris
Moratuwa
Sri Lanka