## Tuesday, June 7, 2011

### Problem 615: Right Triangle with a Square, Diagonals, Center, Area of Quadrilateral

Geometry Problem
Level: Mathematics Education, High School, College.

Click the figure below to see the complete problem 615. 1. By Ptolemy's Theorem:
AB.OC+BC.OA=OB.AC
(c+a)(b/√2)=6b, c+a=6√2
Required area
=(ABC)+(COA)
=(1/2)ca + (b²/4)
=(1/4)(2ca+c²+a²)by Pythagoras
=(1/4)(c + a)²
=18 sq units

2. 3. http://s30.postimg.org/qiktgwn6p/pro_615.png

Solution of this problem depend on value b , length of side of square ACDE.
Case 1: b< 6, we have no solution
Case 2: 6< b< 6.sqrt(2) we have 2 solutions. Area of ABCO is 18
Case 3: b> 6.sqrt(2) , we have 2 solutions. Point B is inside the square ACDE
And the area of ACBO is undefined.

4. Area • ac/2 + b^2/4 = (a+c)^2 / 4 using Pythagoras

Applying Ptolemy to cyclic quad ABCO

ab/sqrt2 + cb/sqrt2 = 6b

Hence a+c = 6aqrt2 and so area = 36 X 2 /4 = 18

Sumith Peiris
Moratuwa
Sri Lanka

5. Append triangles congruent to ABC to the other sides of the square, thus creating a new square called BFGH. Connect BO up through G. Since 0 is the center of both squares, then BO is 12. The area of the BFGH is 72 = 12^2 / 2. Since quadrilateral AOCB is 1/4 of the area of the square BFGH, then the area of AOCB is 18.