151. Miscellaneous problems for the admission at math.
by Virgil Nicula, Oct 9, 2010, 4:21 PM
PP1. Let 
be a triangle with 
, the centroid 
and the incircle 
. Prove that 
.
Proof 1 (metric). From the well-known relations
a.s.o. and 
obtain that
.
![$9a(c-b)=\left[2\left(a^2+c^2\right)-b^2\right]-\left[2\left(a^2+b^2\right)-c^2\right]$](//latex.artofproblemsolving.com/3/0/b/30b9428cc35d6456118295bd2a48a53c6234d3ac.png)
.
Proof 2 (synthetic).
![$\blacktriangleright\ \left\{\begin{array}{ccc}
r(s-a)=2\cdot [AIE]=r\cdot IA\cdot\sin\left(\widehat{BIE}\right)=r\cdot IA\cdot\cos\frac A2 & \implies & r\cdot IA\cdot\cos\frac A2=r(s-a)\\\
ra=2\cdot [BIC]=IB\cdot IC\cdot\sin\left(\widehat{BIC}\right)=IB\cdot IC\cdot\cos\frac A2 & \implies & IB\cdot IC\cdot\cos\frac A2=ra\end{array}\right\|$](//latex.artofproblemsolving.com/e/d/8/ed88674dc0653fb078511a59406f8a5d8070db2e.png)
, where 
.
Therefore,
.
Denote the midpoint 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, 
, 
, 
for which 
. Is well-known that 
, 
and
. Therefore, 
.
PP2. Prove that
. Particular case. 
.
Proof. Denote
. Thus ![$E(x)=1+4\cos (x+y)\cdot\left[\cos (x+y)+\cos (x-y)\right]=$](//latex.artofproblemsolving.com/9/0/6/906a2d4ca75be8121666ca6c0f14a006d6294799.png)
![$\left[2\cdot\underline{\cos (x+y)}+\cos(x-y)\right]^2+\sin^2 (x-y)$](//latex.artofproblemsolving.com/5/8/5/5852ead74a60ce2ccfcb09c0cd38f1295d2d0ec7.png)
. In conclusion, 
and
and 
.
PP3. Prove that the exponential equation
.
Proof. Given equation is equivalently with the equation
. Prove easily that 
(strict increasing) on ![$(-\infty , 0]$](//latex.artofproblemsolving.com/7/9/f/79f23d26894398114d1117f8e7c99b58dbf4c057.png)
and
(strict decreasing) on 
. Thus, 
and 
are solutions. Suppose that 
. In this case observe that :
for any 
for any .
for any 
for any 
.
Therefore, for any
the our equation doesn't solutions. In conclusion, the given equation admits only the solutions 
and 
.
PP4. Prove that for any
exists the chain of implications : 
.
Proof. Observe that
and 
. Thus 
.
Otherwise,
exists 
, 
so that 
.
If
, 
are the nonreal roots of this equation with real coeficients, then 
and 
, i.e. 
.
Observe that 
.
PP5. Consider the real numbers
and 
. Prove that 
and 
.
Proof.
.
I used the notation
, i.e. the real numbers 
, 
have same sign.
Suppose
. Thus 
. Observe that 
, absurd because 
.
Thus,
and 
.
PP6. Ascertain the least natural number
for which its first digit is 
and if move this digit on the last place obtain a number 
so that 
.
Proof. If denote
, then 
. From the relation 
obtain
, i.e. 
.
By trials, the least number
is 
and the required number is 
. Verification : 
.
PP7. Let
with orthocenter 
and circumcircle 
. Show 
.
Proof.
and 
. Observe that
![$-1-2\cos A[\cos (B+C)+\cos (B-C)]=$](//latex.artofproblemsolving.com/3/a/e/3ae66bb1de65099c2382c298bec45f5b08f38a0a.png)
, i.e. 
. In conclusion, the relation 
becomes 
,
i.e.
. But ![$1-8\cdot\prod\cos A=1-4\cos A[\cos (B+C)+\cos (B-C)]=$](//latex.artofproblemsolving.com/1/b/2/1b2ff12735dde8cc3fd1f4fcfc1853ab6a167bc3.png)
![$\left[2\cos A-\cos (B-C)\right]^2+\sin^2(B-C)$](//latex.artofproblemsolving.com/2/1/b/21bf35ee6b34d89e2405cbd62bb7d7fb376403c4.png)
. Thus, 
, i.e. 
with the equality iff 
, i.e. 
.
PP8. Find all
,where 
which verifies 
.
Proof. It's immediate to see that
and that knowledge of 
gives a unique possible value for 
(the quadratic has two
roots, but only one is positive). So there is at most one such sequence. And since
is obviously a solution, this is the unique one.
PP9. Prove that for any
, 
exists the identity 
.
Proof. For
consider the roots 
, 
of the equation 
obtain for 
that
, where 
. In conclusion, .
PP10. Prove that
and in any acute there is the inequality 
.
Proof. Show easily that
. Thus, 
![$\sum a^2\left[\left(a^2+c^2-b^2\right)-\left(b^2+c^2-a^2\right)\right]\ge 0\iff$](//latex.artofproblemsolving.com/e/f/a/efada0e9f2a49f0e06247d8a12f7b435c4fdec8f.png)
, what is truly.
Another application of the identity
is Weitzenbock's inequality (<== click) 
. Indeed,
.
If is acute, then 
.
PP11. Prove that in exists the inequality 
(Schur for 
).
Proof. Can use the remarkable identity
(see here). Therefore, 
because
. Indeed, 
a.s.o.
In conclusion,
.
Remark.
and 
.
PP12. Eliminate the parameter
between the relations 
.
Proof.
. Therefore, 
and 
.
Thus,
. Substitute in here the expressions
for
and 
. If I made no mistake, that should yield 
, where 
and 
.
Remark. Prove easily that
a.s.o.
PP13. Prove that
.
Proof 1 (complex numbers). Let
which satisfy 
, i.e.
. Note the fact 
for 
(symmetry w.r.t. 
-axis in the complex plane).
In conclusion,
, i.e. .
Proof 2 (trigonometric). Denote
, i.e. .
PP14. Show that for any
, exists the inequality ![$\left[\frac{1+(n+1)^{n+1}}{n+2}\right]^{n-1} > \left(\frac{1+n^{n}}{n+1}\right)^{n}$](//latex.artofproblemsolving.com/8/1/e/81e6b8a4cc0875c5376de8ff9af893cf56e43f12.png)
.
Proof. Observe that
. Let 
.
Then
. The only factor that is not immediately obviously positive for 
is
. It must be shown that 
for 
, which is
obvious. Since
for 
obtain that 
is increasing which implies for 
such that 
, it is true that 
.
PP15. Prove that the roots of any polynomial
with 
have the modulus lower than 
.
Proof. For a root we have 
, hence 
.
Assume
. Tthen 
, absurd.
PP16. Solve the trigonometrical equartion
.
Proof. Observe that
. Therefore, 
and 
.
Remark. Prove easily that
Particularly,
.
PP17. Prove that for any
the equation 
has at least a real zero ![$x_0\in[0,2\pi ]$](//latex.artofproblemsolving.com/b/1/e/b1ee89ba6877d0c129c99503eefcee3d0d3e40a8.png)
.
Proof. Consider the function
. Observe that 
and 
.
Apply the Rolle's theorem to the function
and obtain that exists at least so that 
, i.e. 
.
PP18. Let![$f,g\in\mathbb{R}[X]$](//latex.artofproblemsolving.com/0/c/f/0cf50b02432fcb53f18b1635a3fc35942474e33c.png)
two polynomials that have only real roots such that 
,
where
and 
are the roots of 
, respectively and 
their degrees. Prove that and have the same degree and the same roots.
Proof. By condition for all
we have 
, i.e. 
thus 
thus 
(constant).
PP19. Let
, 
be two positive numbers for which 
. Find the maximum value of the sum .
Proof.
, where 
, i.e. 
.
In conclusion,
. Obtain the maximum value for 
and 
, i.e. 
and 
.
PP20. Prove that
.
Proof 1. Prove easily that
.
Proof 2.
. Therefore, 
. I used and the identity 
, i.e. 
.
Remark. The equation
has the roots 
, 
and 
. The equation 
has the roots
, 
and 
. For example, 
.
Prove easily that
and 
.
PP21.Find
and 
without l'Hospital rule.
Proof.
.
I used the remarkable limits
, where 
and 
.
Define
, 
and 
. Observe that 
, where :
. I used the remarkable limit 
. I used the remarkable limit .
.
I used the remarkable limits and 
.
In conclusion,
.
![\[\mathrm {END}\]](//latex.artofproblemsolving.com/d/6/6/d66ea1f1cc355a322f4ee314034c12e0a3a40326.png)
be a triangle with 
, the centroid 
and the incircle 
. Prove that 
.Proof 1 (metric). From the well-known relations
a.s.o. and 
obtain that
.
![$9a(c-b)=\left[2\left(a^2+c^2\right)-b^2\right]-\left[2\left(a^2+b^2\right)-c^2\right]$](http://latex.artofproblemsolving.com/3/0/b/30b9428cc35d6456118295bd2a48a53c6234d3ac.png)
.Proof 2 (synthetic).
![$\blacktriangleright\ \left\{\begin{array}{ccc}
r(s-a)=2\cdot [AIE]=r\cdot IA\cdot\sin\left(\widehat{BIE}\right)=r\cdot IA\cdot\cos\frac A2 & \implies & r\cdot IA\cdot\cos\frac A2=r(s-a)\\\
ra=2\cdot [BIC]=IB\cdot IC\cdot\sin\left(\widehat{BIC}\right)=IB\cdot IC\cdot\cos\frac A2 & \implies & IB\cdot IC\cdot\cos\frac A2=ra\end{array}\right\|$](http://latex.artofproblemsolving.com/e/d/8/ed88674dc0653fb078511a59406f8a5d8070db2e.png)
, where 
.Therefore,
. Denote the midpoint 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, 
, 
, 
for which 
. Is well-known that 
, 
and
. Therefore, 
.PP2. Prove that
. Particular case. 
.Proof. Denote
. Thus ![$E(x)=1+4\cos (x+y)\cdot\left[\cos (x+y)+\cos (x-y)\right]=$](http://latex.artofproblemsolving.com/9/0/6/906a2d4ca75be8121666ca6c0f14a006d6294799.png)
![$\left[2\cdot\underline{\cos (x+y)}+\cos(x-y)\right]^2+\sin^2 (x-y)$](http://latex.artofproblemsolving.com/5/8/5/5852ead74a60ce2ccfcb09c0cd38f1295d2d0ec7.png)
. In conclusion, 
and
and 
.PP3. Prove that the exponential equation
.Proof. Given equation is equivalently with the equation
. Prove easily that 
(strict increasing) on ![$(-\infty , 0]$](http://latex.artofproblemsolving.com/7/9/f/79f23d26894398114d1117f8e7c99b58dbf4c057.png)
and
(strict decreasing) on 
. Thus, 
and 
are solutions. Suppose that 
. In this case observe that :
for any 
for any .
for any 
for any 
.Therefore, for any
the our equation doesn't solutions. In conclusion, the given equation admits only the solutions 
and 
. PP4. Prove that for any
exists the chain of implications : 
.Proof. Observe that
and 
. Thus 
.Otherwise,
exists 
, 
so that 
.If
, 
are the nonreal roots of this equation with real coeficients, then 
and 
, i.e. 
.Observe that
.PP5. Consider the real numbers
and 
. Prove that 
and 
.Proof.
.I used the notation
, i.e. the real numbers 
, 
have same sign.Suppose
. Thus 
. Observe that 
, absurd because 
.Thus,
and 
.Quote:
Proposed problem. Study the nature of the interdependent sequencies which are defined so : 
and 
.
and 
.PP6. Ascertain the least natural number
for which its first digit is 
and if move this digit on the last place obtain a number 
so that 
.Proof. If denote
, then 
. From the relation 
obtain
, i.e. 
.By trials, the least number
is 
and the required number is 
. Verification : 
.PP7. Let
with orthocenter 
and circumcircle 
. Show 
.Proof.
and 
. Observe that
![$-1-2\cos A[\cos (B+C)+\cos (B-C)]=$](http://latex.artofproblemsolving.com/3/a/e/3ae66bb1de65099c2382c298bec45f5b08f38a0a.png)
, i.e. 
. In conclusion, the relation 
becomes 
,i.e.
. But ![$1-8\cdot\prod\cos A=1-4\cos A[\cos (B+C)+\cos (B-C)]=$](http://latex.artofproblemsolving.com/1/b/2/1b2ff12735dde8cc3fd1f4fcfc1853ab6a167bc3.png)
![$\left[2\cos A-\cos (B-C)\right]^2+\sin^2(B-C)$](http://latex.artofproblemsolving.com/2/1/b/21bf35ee6b34d89e2405cbd62bb7d7fb376403c4.png)
. Thus, 
, i.e. 
with the equality iff 
, i.e. 
.PP8. Find all
,where 
which verifies 
.Proof. It's immediate to see that
and that knowledge of 
gives a unique possible value for 
(the quadratic has tworoots, but only one is positive). So there is at most one such sequence. And since
is obviously a solution, this is the unique one.PP9. Prove that for any
, 
exists the identity 
. Proof. For
consider the roots 
, 
of the equation 
obtain for 
that
, where 
. In conclusion, .PP10. Prove that
and in any acute there is the inequality 
. Proof. Show easily that
. Thus, 
![$\sum a^2\left[\left(a^2+c^2-b^2\right)-\left(b^2+c^2-a^2\right)\right]\ge 0\iff$](http://latex.artofproblemsolving.com/e/f/a/efada0e9f2a49f0e06247d8a12f7b435c4fdec8f.png)
, what is truly.Another application of the identity
is Weitzenbock's inequality (<== click) 
. Indeed,
.If
is acute, then 
.PP11. Prove that in
exists the inequality 
(Schur for 
). Proof. Can use the remarkable identity
(see here). Therefore, 
because
. Indeed, 
a.s.o.In conclusion,
.Remark.
and 
.PP12. Eliminate the parameter
between the relations 
.Proof.
. Therefore, 
and 
.Thus,
. Substitute in here the expressionsfor
and 
. If I made no mistake, that should yield 
, where 
and 
.Remark. Prove easily that
a.s.o.PP13. Prove that
.Proof 1 (complex numbers). Let
which satisfy 
, i.e.
. Note the fact 
for 
(symmetry w.r.t. 
-axis in the complex plane).In conclusion,
, i.e. .Proof 2 (trigonometric). Denote
, i.e. .PP14. Show that for any
, exists the inequality ![$\left[\frac{1+(n+1)^{n+1}}{n+2}\right]^{n-1} > \left(\frac{1+n^{n}}{n+1}\right)^{n}$](http://latex.artofproblemsolving.com/8/1/e/81e6b8a4cc0875c5376de8ff9af893cf56e43f12.png)
.Proof. Observe that
. Let 
.Then
. The only factor that is not immediately obviously positive for 
is
. It must be shown that 
for 
, which isobvious. Since
for 
obtain that 
is increasing which implies for 
such that 
, it is true that 
.PP15. Prove that the roots of any polynomial
with 
have the modulus lower than 
.Proof. For a root
we have 
, hence 
.Assume
. Tthen 
, absurd.PP16. Solve the trigonometrical equartion
.Proof. Observe that
. Therefore, 
and 
.Remark. Prove easily that
Particularly,
.PP17. Prove that for any
the equation 
has at least a real zero ![$x_0\in[0,2\pi ]$](http://latex.artofproblemsolving.com/b/1/e/b1ee89ba6877d0c129c99503eefcee3d0d3e40a8.png)
.Proof. Consider the function
. Observe that 
and 
.Apply the Rolle's theorem to the function
and obtain that exists at least so that 
, i.e. 
.PP18. Let
![$f,g\in\mathbb{R}[X]$](http://latex.artofproblemsolving.com/0/c/f/0cf50b02432fcb53f18b1635a3fc35942474e33c.png)
two polynomials that have only real roots such that 
,where
and 
are the roots of 
, respectively and 
their degrees. Prove that and have the same degree and the same roots.Proof. By condition for all
we have 
, i.e. 
thus 
thus 
(constant).PP19. Let
, 
be two positive numbers for which 
. Find the maximum value of the sum .Proof.
, where 
, i.e. 
.In conclusion,
. Obtain the maximum value for 
and 
, i.e. 
and 
.PP20. Prove that
.Proof 1. Prove easily that
.Proof 2.
. Therefore, 
. I used and the identity 
, i.e. 
.Remark. The equation
has the roots 
, 
and 
. The equation 
has the roots
, 
and 
. For example, 
.Prove easily that
and 
.PP21.Find
and 
without l'Hospital rule.Proof.
.I used the remarkable limits
, where 
and 
.Define
, 
and 
. Observe that 
, where :
. I used the remarkable limit 
. I used the remarkable limit .
.I used the remarkable limits
and 
.In conclusion,
.![\[\mathrm {END}\]](http://latex.artofproblemsolving.com/d/6/6/d66ea1f1cc355a322f4ee314034c12e0a3a40326.png)
This post has been edited 191 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:26 PM
August 2018
January 2018