## Tuesday, April 2, 2013

### Problem 868: Cyclic Quadrilateral, Circle, Five Rectangles, Four Centers, Congruence

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

Click the figure below to see the complete problem 868.

1. From problem 866,

∠O1O2O3 = 90°

Similarly, we have
∠O2O3O4 = 90°
∠O3O4O1 = 90°
∠O4O1O2 = 90°

Hence, O1O2O3O4 is a rectangle.

2. ∠O1O2O3 = 90°
Why???

3. thank you
for this problem, i passed my prelim. :-)

4. AO4=CO2, AO1=CO3, <GCK=360-90-90-(180-<BAD)=<BAD, <O1AB=<KCO3, <DAO4=<GCO2, значит треугольники O3CO2 и O1AO4 конгруэнтно друг друга, и O1O4=O2O3. Подобным образом O1O2=O3O4. Поскольку противоположные стороны четырехугольника O1O2O3O4 равны, он параллелограмм. <O1O4O3=<AO4D-<O1O4A-<O3O4D, <O1O2O3=(<BO2C=180-<AO4D)+(<O1O2B=<O3O4D)+(<O3O2C=<O1O4A), значит что <O1O4O3+<O1O2O3=180 и O1O2O3O4 вписанный четырехугольник. Если O1O2O3O4 одновременно параллелограмм и вписанный четырехугольник, то оно прямоугольник.

5. AO4 = CO2, AO1 = CO3, <GCK = 360-90-90- (180- <BAD) = <BAD, <O1AB = <KCO3, <DAO4 = <GCO2, then triangles O3CO2 O1AO4 and congruent to each other, and O1O4 = O2O3. Similarly O1O2 = O3O4. Since the opposite sides of the quadrilateral O1O2O3O4 are equal, it is parallelogram. <O1O4O3 = <AO4D- <O1O4A- <O3O4D, <O1O2O3 = (<BO2C = 180- <AO4D) + (<O1O2B = <O3O4D) + (<O3O2C = <O1O4A), means that <O1O4O3 + <O1O2O3 = 180 and O1O2O3O4 inscribed quadrilateral. If O1O2O3O4 simultaneously parallelogram and inscribed quadrilateral, it is a rectangle.

6. Excellence solution !

7. Triangle AO4D ≡ BO2H hence < ADO4 = HBO2 =á say….(1)

Triangle O1AB ≡ DO3J hence < O1BA = < O3DJ = < 90- O3DC = â say….(2)

Hence < O1BO2 = 360 – B – (90-á) – â = 270 – B + á –â …(3)

< O4DO3 = 180 – B + á + 90 - â = 270 - B + á –â….(4)

From (3) and (4) < O1BO2 = < O1BO2 and so Triangles O4DO3 ≡ O1BO2 (SAS)

Therefore O1O2 = O3O4.
Similarly O2O3 can be shown to be = to O1O4 and so O1O2O3O4 is a //ogram.

Let < BO2O1 = < DO4O3 = q and < CO2O3 = < AO4O1 = n

So in this //ogram < O1O4O3 = 180 - 2á – n- q = < O1O2O3 = q+n+ 180 – 2
(90-á) = 2á +n +q from which

2(2á +n +q) = 180 and so 2á +n +q = 90 and therefore from Triangle AO4D, <
O1O4O3 = 90

So O1O2O3O4 is a //ogram having an angle = to 90 and is hence a rectangle.

Sumith Peiris
Moratuwa
Sri Lanka