Geometry Problem. Post your solution in the comments box below.

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

This entry contributed by Ajit Athle.

Click the figure below to see the complete problem 922.

## Tuesday, September 10, 2013

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

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http://img59.imageshack.us/img59/9554/7fts.png

ReplyDeleteDraw 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)

Anonymous Erina NJ

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

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

Hoi,

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

Anonymus Erina Nj

ReplyDeleteEven 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

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

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

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

Erina-NJ

Erina: Your problem has been published at

Deletewww.gogeometry.com/school-college/p923-infographic-quadrilateral-obtuse-inequality-diagonal.htm

Thanks

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

ReplyDeleteThen 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