261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.
by Virgil Nicula, Apr 6, 2011, 11:53 AM
An easy extension. Let 
be a triangle with the circumcircle 
. Condider the points 
, 
and a point
for which 
, 
. Let 
, 
and 
, 
.
Suppose that the line
is tangent to the circumcircle of 
. Prove that 
.
Proof. Observe that
and 
.
Remark. For
obtain the second proposed problem from IMO 2009.
JBMO 2007, PP2. Let
be a quadrilateral with 
, 
and 
. Find 
, where 
.
Proof. Construct the point
in the outside such that 
and 
. This means that 
are collinear and
. It results that 
. And because 
, from the last relation we obtain 
.
Continuing, let the lines
and 
intersect in point 
. Since triangles 
and 
are similar, we obtain 
. Thus the quadrilateral 
is cyclic. This means that
. Also, let 
. Also,
in cyclic quadrilateral
, we have 
. But 
. Now we have calculated all the angles
around and 
, thus from the generalized Ceva trig, we obtain 
.
Lemma (extension). Let
be a convex quadrilateral. Prove that 
.
Remark. The relation
is symmetrically in the variables 
, 
. Thus, I can propose the following problem :
Proof of the proposed problem. For 
the relation from the above lemma becomes 
, i.e.
. Observe that 
and on 
function 
is strict decreasing and 
, i.e. 
.
JBMO 2007, PP1. Let
be a positive real number such that 
. Prove that the equation 
has no real solution.
Proof 1. For
obtain that 
.
Observe that
. In conclusion, the mentioned equation has no real solution.
Proof 2. Show easily that the function
is strict increasing. Therefore, the equation 
, i.e. the equation 
has at most one
positive root. Observe that for the function
there is the relation 
, i.e. 
.
In conclusion, the our equation has a single positive root and 
. The equation 
has no real roots iff 
, what is truly.
Remark. Observe and prove easily that the equation has only one real root, ![$a=\sqrt [3]2+\sqrt [3]4$](//latex.artofproblemsolving.com/9/9/d/99d48a1b3e89349ef0d349e5a82c701450b4e727.png)
.
Generally for any
, , 
, i.e. ![$1+\left[a^2\right]=p^2$](//latex.artofproblemsolving.com/c/7/b/c7b51b47c7818c14e01cc57045e355cbd4e66b03.png)
. Indeed, 
![$a\cdot \left[a^2-\left(p^2-1\right)\right]=$](//latex.artofproblemsolving.com/1/2/7/12715ec50b3a4fc96eb43f692a984ddfcd68b810.png)
![$>0\ \iff\ \sqrt{p^2-1}<a<p\ \iff\ 1+\left[a^2\right]=p^2$](//latex.artofproblemsolving.com/0/1/f/01f2cef1a5eef618648168e7d329ee569669ae29.png)
.
be a triangle with the circumcircle 
. Condider the points 
, 
and a point
for which 
, 
. Let 
, 
and 
, 
.Suppose that the line
is tangent to the circumcircle of 
. Prove that 
.Proof. Observe that
and 
.Remark. For
obtain the second proposed problem from IMO 2009.JBMO 2007, PP2. Let
be a quadrilateral with 
, 
and 
. Find 
, where 
.Proof. Construct the point
in the outside such that 
and 
. This means that 
are collinear and
. It results that 
. And because 
, from the last relation we obtain 
.Continuing, let the lines
and 
intersect in point 
. Since triangles 
and 
are similar, we obtain 
. Thus the quadrilateral 
is cyclic. This means that
. Also, let 
. Also,in cyclic quadrilateral
, we have 
. But 
. Now we have calculated all the anglesaround
and 
, thus from the generalized Ceva trig, we obtain 
.Lemma (extension). Let
be a convex quadrilateral. Prove that 
.
.
![$ \cos x\cdot [\cos (x+2y)-\cos (2z+x-2y)]=\cos y\cdot [\cos (2x+y)-\cos (y+2z-2x)]$](http://latex.artofproblemsolving.com/7/0/6/7060f04642d0c81011987ebbae2f6d39dae1df2a.png)
.Remark. The relation
is symmetrically in the variables 
, 
. Thus, I can propose the following problem :Quote:
Let be a quadrilateral with 
, 
and 
. Find , where .
be a quadrilateral with 
, 
and 
. Find , where .
the relation from the above lemma becomes 
, i.e.
. Observe that 
and on 
function 
is strict decreasing and 
, i.e. 
.JBMO 2007, PP1. Let
be a positive real number such that 
. Prove that the equation 
has no real solution.Proof 1. For
obtain that 
.Observe that
. In conclusion, the mentioned equation has no real solution.Proof 2. Show easily that the function
is strict increasing. Therefore, the equation 
, i.e. the equation 
has at most onepositive root. Observe that for the function
there is the relation 
, i.e. 
.In conclusion, the our equation has a single positive root
and 
. The equation 
has no real roots iff 
, what is truly.Remark. Observe and prove easily that the equation
has only one real root, ![$a=\sqrt [3]2+\sqrt [3]4$](http://latex.artofproblemsolving.com/9/9/d/99d48a1b3e89349ef0d349e5a82c701450b4e727.png)
.Generally for any
, , 
, i.e. ![$1+\left[a^2\right]=p^2$](http://latex.artofproblemsolving.com/c/7/b/c7b51b47c7818c14e01cc57045e355cbd4e66b03.png)
. Indeed, 
![$a\cdot \left[a^2-\left(p^2-1\right)\right]=$](http://latex.artofproblemsolving.com/1/2/7/12715ec50b3a4fc96eb43f692a984ddfcd68b810.png)
![$>0\ \iff\ \sqrt{p^2-1}<a<p\ \iff\ 1+\left[a^2\right]=p^2$](http://latex.artofproblemsolving.com/0/1/f/01f2cef1a5eef618648168e7d329ee569669ae29.png)
.This post has been edited 25 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:58 AM
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