## Monday, June 16, 2014

### Geometry Problem 1024: Contact, Gergonne Triangle, Point on an arc of Incircle, Perpendicular, Distances

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

Click the figure below to see the complete problem 1024.

1. Denote inscribed angles of arcs PTa, PTc, and TcTb to be a, b, and c respectively.
a1=PTa*sin(a)=(PTa)^2/(2R)
b1=PTb*sin(b+c)=(PTb)^2/(2R)
c1=PTc*sin(b)=(PTc)^2/(2R)
Plugging these into equation to be proved yields Ptolemy's Theorem in quadrilateral PTaTbTc.

2. Can someone post a solution to this question, I'm having a difficult time proving it

3. https://goo.gl/photos/8vk1AQ8RWkqjdc3Q9

Let PQ is the diameter of circle O and R= radius of circle O
Define a as ∠ (PTaB)= ∠ (PQTa)
. b=∠ (BTcP)= ∠ (PQTc)
. c= ∠ (ATbTc)= ∠ (TbQTc)
See sketch for details
We have
Sin(a)=PTa/2.R
Sin(b)=PTc/2R
Sin(b+c)= PTb/2R
a1=PTa*sin(a)=(PTa)^2/(2R) => PTa= sqrt(2.R.a1)
b1=PTb*sin(b+c)=(PTb)^2/(2R)=> PTb= sqrt(2.r.b1)
c1=PTc*sin(b)=(PTc)^2/(2R) => PTc= sqrt(2.R.c1)
Apply Ptolemy’s theorem in quad. PTaTbTc
we have TaTc. PTb= PTc.TaTb+ PTa.TbTc
Plug these values in above equation we will get the result