## Tuesday, September 10, 2013

### Geometry Problem 922: Circle, Tangent Lines, Tangency Point, Chord, Secant, Harmonic Mean

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

This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 922.

1. http://img59.imageshack.us/img59/9554/7fts.png
Draw circle diameter PO
Let M is the midpoint of AB and PO cut DE at N
We have PA.PB=a.b= PE^2 ( power from P to circle O)
PM=(a+b)/2
O,N,C,M are cocyclic => PC.PM=PN.PO=PE^2 ( relation in right triangle)
Or c. (a+b)/2 = PE^2 => a.b= c.(a+b)/2 => c= 2.a.b/(a+b)

2. Anonymous Erina NJ

PD^2 = A*B PD^2 - PC^2 = DC*CE = AC*CB

a*b-c^2 = (c-a)(b-c)-> c = 2ab/(a+b)

1. Hoi,

Why is PD^2 - PC^2 = DC*CE ?

3. Anonymus Erina Nj

Even the previous solution is from Erina NJ

For Pravin

Sinx = 2 sin10 cos (30-x)
sinx = sin (40-x)- sin (20-x)
sin (40-x)- sinx= Sin (20-x)
2sin (20-x) cos 20-sin (20-x) =0
sin(20-x) [2cos20-1] =0
2cos20-1 different from 0 => sin(20-x) = 0
X=20

1. Erina: This solution has been moved to problem 920. Thanks.

4. Sorry about that Antonio(your system seems to have cut out a few details):

In the quandrangle ABCD, angles A, B, C, are obtuse. Prove that BD is greater than AC

Erina-NJ

5. Let PD = p, DC = m and CE = n and let X be the mid point of DE

Then PX^2 = p^2 - (m+n)^2/4 = c^2 - (n-m)^2/4

So p^2 -c^2 = mn = (c-a)(b-c)

But p^2 = ab

Solving c = 2ab/(a+b)

Sumith Peiris
Moratuwa
Sri Lanka