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by math_explorer, Mar 1, 2011, 2:54 AM
Ptolemy's! No.
I really need more Geometry Unbound.
Draw the incenter
. Observe that because 
is isosceles, its altitude from 
is the same as its angle bisector from ; that is, 
. Similarly 
. Therefore 
and 
.
Side-side-side congruence applies.
.
And we can now bash with Law of Cosines... almost.
Let's set
.
Triple-angle formula plus some straightforward offscene term-moving yields
![\[ \frac{MN^2}{BC^2} = 1 + \frac{2BI\times CI (4\sin^2 \alpha \cos \alpha)}{a^2} \]](//latex.artofproblemsolving.com/0/5/8/058c4c8038473a857800fb02de5e0145eddd212a.png)
Now,
; Extended Law of Sines yields
![\[ \frac{MN^2}{BC^2} = 1 - \frac{2BI\times CI \sin \alpha}{aR} \]](//latex.artofproblemsolving.com/7/2/8/728b225d7e87ed098bdfccb87de0219f2d1e078f.png)
Also,
so we've got
![\[ \frac{MN^2}{BC^2} = 1 - \frac{2r}{R} \]](//latex.artofproblemsolving.com/7/6/0/76078a24c200db07e1e1bedaa16c803300b502a9.png)
so
Note to self: stop forgetting the factors of 2.
I really need more Geometry Unbound.
Quote:
APMO 2005.5. In a triangle 
, points 
and 
are on sides 
and 
, respectively, such that 
. Let 
and 
denote the circumradius and the inradius of the triangle , respectively. Express the ratio 
in terms of and .
, points 
and 
are on sides 
and 
, respectively, such that 
. Let 
and 
denote the circumradius and the inradius of the triangle , respectively. Express the ratio 
in terms of and .Draw the incenter
. Observe that because 
is isosceles, its altitude from 
is the same as its angle bisector from ; that is, 
. Similarly 
. Therefore 
and 
.Side-side-side congruence applies.
.And we can now bash with Law of Cosines... almost.
Let's set
.
Triple-angle formula plus some straightforward offscene term-moving yields
![\[ \frac{MN^2}{BC^2} = 1 + \frac{2BI\times CI (4\sin^2 \alpha \cos \alpha)}{a^2} \]](http://latex.artofproblemsolving.com/0/5/8/058c4c8038473a857800fb02de5e0145eddd212a.png)
Now,
; Extended Law of Sines yields![\[ \frac{MN^2}{BC^2} = 1 - \frac{2BI\times CI \sin \alpha}{aR} \]](http://latex.artofproblemsolving.com/7/2/8/728b225d7e87ed098bdfccb87de0219f2d1e078f.png)
Also,
so we've got![\[ \frac{MN^2}{BC^2} = 1 - \frac{2r}{R} \]](http://latex.artofproblemsolving.com/7/6/0/76078a24c200db07e1e1bedaa16c803300b502a9.png)
so
Note to self: stop forgetting the factors of 2.
May 2020
January 2020