## Monday, June 16, 2014

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

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 1024.

#### 3 comments:

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