Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Ajit Athle.
Click the figure below to see the problem 767 details.
Monday, June 18, 2012
Problem 767: Equilateral Triangle, Circle tangent to a side, Tangent line, Measurement, Secant
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By theTangent Theorem:
ReplyDeletem^2 = (m+n-d)(m+n-e); n^2=(m+n-g)(m+n-f)
By the Secant Theorem:
de = fg or fg - de = 0
Then
m^2 - n^2 = (m+n)(f+g-d-e)
(m+n)(m-n) = (m+n)(f+g-d-e)
m-n = f+g-d-e
So:
m + d + e = f + g + n
http://www.mathematica.gr/forum/viewtopic.php?f=22&t=27794&p=135568
ReplyDeletehttps://goo.gl/photos/FUFbuawj6LN7feiG6
ReplyDeleteLet N and P are the projection of O over AB and AC
Let M is the midpoint of AC
Note that N and P are the midpoints of ED and FG
Observe that MT= ½(m-n)
BN=1/2(d+e)=BQ
BP=1/2(f+g)
PQ=BP-BQ= ½(f+g-d-e)
Since ABC is equilateral so angle(AC, NO)=angle (AO,BC)= 30 degrees
And MT=PQ => ½(m-n)= ½(f+g-d-e)
So m+d+e=n+f+g