314. USAMTS 2009 Round 2 #5.
by Virgil Nicula, Aug 31, 2011, 1:15 AM
USAMTS 2009 Round 2 # 5. Let 
be a triangle with 
, 
and 
. Let ![$P\in [BC]$](//latex.artofproblemsolving.com/0/8/7/087dcfc6e817021079bdd2436e980d096613bc7a.png)
be a point
and let
be the point (other than 
) where the line through and 
intersects the circumcircle of . Prove that 
.
Proof. I"ll prove for any-right triangle . Denote 
. Prove easily that 
, where ![$x\in\left[0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/b/9/8/b98177089d2b2c5991a6a2601746d0bcaedac8dc.png)
. Thus, 
is max. 
is max. 
is min. Denote 
, where 
and 
. Our extremum problem becomes :
Lemma. Are given positive numbers
. Ascertain 
, where 
and 
.
Proof. For
is easily ! Indeed, can suppose w.l.o.g. that 
. Denote 
. Prove easily that ![$t\in \left[1,\sqrt 2\right]$](//latex.artofproblemsolving.com/4/f/b/4fb18a3248ba98daaa154ced66611f15d17a24d5.png)
. Indeed, 
. Therefore, 
is min. 
is max. 
is max. 
is max. Since the
function
is 
(strict increasing) on ![$\left[0,\sqrt 2\right]$](//latex.artofproblemsolving.com/6/0/8/608c2b3b57fffbac4abac8735cb35b51f6429a42.png)
obtain that this function is max. 
, i.e. 
. What happen if 
? This is the actual problem. I"ll use
the Holder's inequality :![$\left(\frac{b}{u}+\frac{c}{v}\right)\left(\frac{b}{u}+\frac{c}{v}\right)(u^2+v^2) \ge (\sqrt[3]{b^2}+\sqrt[3]{c^2})^3$](//latex.artofproblemsolving.com/1/6/f/16f408a9a2be9020d62b085d0501a02fe4990f83.png)
Have equality iff 
, i.e. ![$\tan x=\sqrt[3]{\frac bc}\iff$](//latex.artofproblemsolving.com/a/e/8/ae8a8e72f4388f82ebe21b9094ecc9f10b206cc9.png)
![$\boxed{\frac {PB}{PC}=\frac cb\cdot\tan x=\sqrt[3]{\frac {c^2}{b^2}}}$](//latex.artofproblemsolving.com/2/f/3/2f3062fb1d541e63711c1f620d457bf3c3b4b7d5.png)
.
Otherwise. is max. 
is max. Prove easily that 
![$\sqrt[3]{\frac bc}-\tan x$](//latex.artofproblemsolving.com/7/e/8/7e8272dcd1b52a524ba646832b3ca323d6c90fef.png)
a.s.o. I used the notation 
, i.e. and 
have same sign.
USAMTS 1998 Round 3 # 4 - Trigonometry. Prove that if
, then 
.
Proof.
. Denote 
, where 
because 
.
In conclusion,![$(\forall )\ t\in\left[0,\frac 14\right]$](//latex.artofproblemsolving.com/7/f/2/7f2d95968dd4a30b3533ad28253d8897a79a0788.png)
we have 
.
be a triangle with 
, 
and 
. Let ![$P\in [BC]$](http://latex.artofproblemsolving.com/0/8/7/087dcfc6e817021079bdd2436e980d096613bc7a.png)
be a pointand let
be the point (other than 
) where the line through and 
intersects the circumcircle of . Prove that 
.Proof. I"ll prove for any
-right triangle . Denote 
. Prove easily that 
, where ![$x\in\left[0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/b/9/8/b98177089d2b2c5991a6a2601746d0bcaedac8dc.png)
. Thus, 
is max. 
is max. 
is min. Denote 
, where 
and 
. Our extremum problem becomes :Lemma. Are given positive numbers
. Ascertain 
, where 
and 
.Proof. For
is easily ! Indeed, can suppose w.l.o.g. that 
. Denote 
. Prove easily that ![$t\in \left[1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/4/f/b/4fb18a3248ba98daaa154ced66611f15d17a24d5.png)
. Indeed, 
. Therefore, 
is min. 
is max. 
is max. 
is max. Since thefunction
is 
(strict increasing) on ![$\left[0,\sqrt 2\right]$](http://latex.artofproblemsolving.com/6/0/8/608c2b3b57fffbac4abac8735cb35b51f6429a42.png)
obtain that this function is max. 
, i.e. 
. What happen if 
? This is the actual problem. I"ll usethe Holder's inequality :
![$\left(\frac{b}{u}+\frac{c}{v}\right)\left(\frac{b}{u}+\frac{c}{v}\right)(u^2+v^2) \ge (\sqrt[3]{b^2}+\sqrt[3]{c^2})^3$](http://latex.artofproblemsolving.com/1/6/f/16f408a9a2be9020d62b085d0501a02fe4990f83.png)
Have equality iff 
, i.e. ![$\tan x=\sqrt[3]{\frac bc}\iff$](http://latex.artofproblemsolving.com/a/e/8/ae8a8e72f4388f82ebe21b9094ecc9f10b206cc9.png)
![$\boxed{\frac {PB}{PC}=\frac cb\cdot\tan x=\sqrt[3]{\frac {c^2}{b^2}}}$](http://latex.artofproblemsolving.com/2/f/3/2f3062fb1d541e63711c1f620d457bf3c3b4b7d5.png)
.Otherwise.
is max. 
is max. Prove easily that 
![$\sqrt[3]{\frac bc}-\tan x$](http://latex.artofproblemsolving.com/7/e/8/7e8272dcd1b52a524ba646832b3ca323d6c90fef.png)
a.s.o. I used the notation 
, i.e. and 
have same sign.USAMTS 1998 Round 3 # 4 - Trigonometry. Prove that if
, then 
.Proof.
. Denote 
, where 
because 
.In conclusion,
![$(\forall )\ t\in\left[0,\frac 14\right]$](http://latex.artofproblemsolving.com/7/f/2/7f2d95968dd4a30b3533ad28253d8897a79a0788.png)
we have 
.This post has been edited 46 times. Last edited by Virgil Nicula, Nov 20, 2015, 8:25 AM
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