Proposed Problem
See complete Problem 257 at:
gogeometry.com/problem/p257_equilateral_triangle_circumcircle_distance_sides.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Tuesday, February 24, 2009
Problem 257: Equilateral Triangle, Circumcircle, Point, Vertices, Side, Distances, Squares
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By an earlier problem, we know that d = e + f
ReplyDeletehence, LHS =(e+f)^2+e^2+f^2 = 2(e^2+f^2+ef)
Using cosine rule in Tr. BDC, we've: a^2 =e^2+f^2-2efcos(BDC)=e^2+f^2-2ef(-1/2)=e^2+f^2+ef
Hence, LHS = 2(e^2+f^2+ef) = 2a^2
Ajit: ajitathle@gmail.com
We see that ∠ADB=∠ADC =60
ReplyDeleteIf we apply the cosine rule in triangl ABD with cos(60)=1/2:
d^2+e^2-2decos(∠ADB)=a^2
d^2+e^2-2decos(60)=a^2
d^2+e^2-de=a^2
And then we apply the cosine rule in triangle ADC :
d^2+f^2-2dfcos(∠ADC)=a^2
d^2+f^2-2dfcos(60)=a^2
d^2+f^2-df=a^2
Adding these two equations we get
2d^2+e^2+f^2-de-df=2a^2
2d^2+e^2+f^2-d(e+f)=2a^2
In problem 256 we see that d=e+f :
2d^2+e^2+f^2-d^2=2a^2
d^2+e^2+f^2=2a^2
We can use the extension of ptolemy given here :
ReplyDeletehttp://gogeometry.com/equilic/cyclic_ptolemy_ratio_diagonal.htm
Applying the extension of ptolemy gives
d/a=(a^2+ef)/(ae+af)........ (1.)
and from problem 256 d=e+f
First setting d =e+f in equation 1 gives
d/a=(a^2+ef)/(ae+af)
(e+f)/a=(a^2+ef)/(ae+af)
(e+f)^2=a^2+ef
this leads to
a^2=e^2+f^2+ef ......... (2.)
We also have ad = af+ea ,from this follows
d/a=(a^2+ef)/(ae+af)
d/a=(a^2+ef)/(ad)
d^2=a^2+ef
and we get
a^2=d-ef .......(3.)
Adding 2 and 3 gives
2a^2=d^2+e^2+f^2
Here is my solution...
ReplyDeletehttps://www.youtube.com/watch?v=cx23xF65X_Q
https://photos.app.goo.gl/PoLNTVK1efCJcii19
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