Tuesday, December 8, 2020

Geometry Problem 1491: Cyclic Quadrilateral, Diagonal, Incircle, Angle, Measurement

Geometry Problem 1491. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

1. I think there is a typo and O1 is the Incentre of Tr. ABD not ABC

Let <ABD = <ACD = 2p.

Its not difficult to infer (considering Tr. ABD) that <AO1D = 90+p = <AO2D (considering Tr. ACD)

Hence AO1O2D is concylic

Again considering Tr. AO1D, <ADO1 = 180 - (90+p) - 40 = 50 - p = <AO2O1

So <O1O2D = (50-p) + (90+p) = 140

Sumith Peiris
Moratuwa
Sri Lanka

1. If we are given the lengths of the 4 sides of ABCD and the length of either diagonal we can easily calculate the length of O1O2 (irrespective of the angle A)

2. Join A to O1. Quadrilateral AO1O2D cyclic
=> Ang O1O2D=140
See sketch for details

3. https://photos.app.goo.gl/BezctJeJzUHUpM9y8

4. https://photos.app.goo.gl/ZYD4fjg9k2jNDHAF9
Let BO1 and CO2 meet at I
I will be the midpoint of arc(AD)
We have isosceles triangles AID, AIO1, and DIO2
In circle ABCD, ∡ (BID)= ∡ (BAD)= ∡ (DIO1)= 80 deg.
In circle I, arc(O1AD)= 360-∡ (DIO1)=280 deg.
So ∡ (O1O2D)= ½ x 280=140 deg.

1. Hi peter ;
AID is an isosceles triangle , Why AIO1 and DIO2 are isosceles triangles ?
How could you prove IO1 , IO2 are radius of Circle I ?

2. https://photos.app.goo.gl/mRovGUixhbgm9TYH6

See attached sketch for details
CI meet arc AB at the midpoint
AI meet arc BC at the midpoint
^(DCM)= u+v
^(DIC)= ext. angle of triangle AIC= u+v
so triangle DIC is isosceles

3. Thanks Peter, It's clear now.