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