Problem:
Find a four digit number which is a perfect square, knowing that the first two digit number exceeds by $1$ the last two digit number.
Solution:
Let $x$ be the first two digit number and let $y^2$ be the four digit number. Then, $31 < y < 100$ and
$$100(x+1)+x=y^2 \iff 101x=(y-10)(y+10).$$ So, $101$ divides one between $y-10$ and $y+10$. But $y-10<101$, so $101|(y+10)$. Since $41<y+10<110$, $y+10=101$, which gives $y=91$. Therefore, $y^2=8281$.
Friday, December 21, 2012
Thursday, December 20, 2012
KöMaL (Metresis) 1894, Problem 10
Problem:
Find all integers such that their fifth power minus three times their square is equal to $216$.
Solution:
Suppose that $n$ is an integer such that $n^5-3n^2=216=2^3\cdot3^3$. Then $$n^2(n^3-3)=216.$$ From this equality we get $n>1$. Let $d=\gcd(n^2,n^3-3)$. Then $d=1,3$. If $d=1$, then both $n^2$ and $(n^3-3)$ are perfect cubes, impossible since $$(n-1)^3 < n^3-3 < n^3.$$ If $d=3$, then $3^2|n^2$, so $n^2=9a, n^3-3=3b$ and $\gcd(a,b)=1$. Therefore, $ab=8$ and $b+1=3^{3k-1}$ for some $k \in \mathbb{N}$, which gives $a=1,b=8$, i.e. $n=3$.
Note. It was faster to note that $n^2$ and $(n^3-3)$ have different parity, so one of the two is odd. If $n^2$ is odd, $n^2=9$ which yields $n=3$, if $(n^3-3)$ is odd, then $n^3-3=1,3,9,27$, i.e. no solution.
Find all integers such that their fifth power minus three times their square is equal to $216$.
Solution:
Suppose that $n$ is an integer such that $n^5-3n^2=216=2^3\cdot3^3$. Then $$n^2(n^3-3)=216.$$ From this equality we get $n>1$. Let $d=\gcd(n^2,n^3-3)$. Then $d=1,3$. If $d=1$, then both $n^2$ and $(n^3-3)$ are perfect cubes, impossible since $$(n-1)^3 < n^3-3 < n^3.$$ If $d=3$, then $3^2|n^2$, so $n^2=9a, n^3-3=3b$ and $\gcd(a,b)=1$. Therefore, $ab=8$ and $b+1=3^{3k-1}$ for some $k \in \mathbb{N}$, which gives $a=1,b=8$, i.e. $n=3$.
Note. It was faster to note that $n^2$ and $(n^3-3)$ have different parity, so one of the two is odd. If $n^2$ is odd, $n^2=9$ which yields $n=3$, if $(n^3-3)$ is odd, then $n^3-3=1,3,9,27$, i.e. no solution.
Tuesday, November 27, 2012
Mathematical Reflections 2012, Issue 5 - Problem U243
Problem:
Let $f:(a,b) \longrightarrow \mathbb{R}$ be a differentiable function such that $f'(a)=f'(b)=0$ and with the
property that there is a real valued function $g$ for which $g(f'(x))=f(x)$ for all $x$ in $\mathbb{R}$.
Prove that $f$ is constant.
Solution:
Since $f$ is differentiable in $(a,b)$ and $f'(a)=f'(b)=0$, $f$ is continuous in $[a,b]$. By Weierstrass's Theorem there exist maximum $M$ and minimum $m$ in $[a,b]$. If these are both attained at $a$ and $b$, $f$ is constant. If one of these is not attained at $a$ or $b$, then there exists $c \in (a,b)$ such that $f(c)=M$ or $f(c)=m$. Since $f$ is differentiable $f'(c)=0$, so $g(0)=f(c)=f(a)=f(b)$, which means that $f$ is constant in $(a,b)$.
Let $f:(a,b) \longrightarrow \mathbb{R}$ be a differentiable function such that $f'(a)=f'(b)=0$ and with the
property that there is a real valued function $g$ for which $g(f'(x))=f(x)$ for all $x$ in $\mathbb{R}$.
Prove that $f$ is constant.
Proposed by Mihai Piticari and Sorin Radulescu.
Solution:
Since $f$ is differentiable in $(a,b)$ and $f'(a)=f'(b)=0$, $f$ is continuous in $[a,b]$. By Weierstrass's Theorem there exist maximum $M$ and minimum $m$ in $[a,b]$. If these are both attained at $a$ and $b$, $f$ is constant. If one of these is not attained at $a$ or $b$, then there exists $c \in (a,b)$ such that $f(c)=M$ or $f(c)=m$. Since $f$ is differentiable $f'(c)=0$, so $g(0)=f(c)=f(a)=f(b)$, which means that $f$ is constant in $(a,b)$.
Mathematical Reflections 2012, Issue 5 - Problem U241
Problem:
Let $a>b$ be positive real numbers. Prove that $$c_n=\dfrac{\sqrt[n+1]{a^{n+1}-b^{n+1}}}{\sqrt[n]{a^n-b^n}}$$
is a decreasing sequence and find its limit.
Solution:
Let $c_n=\dfrac{\sqrt[n+1]{1-x^{n+1}}}{\sqrt[n]{1-x^n}}$, where $x=b/a < 1$. We first prove that $c_n > 1$. Let $b_n=c^{n(n+1)}_n$. Then $$b_n(x)=\dfrac{(1-x^{n+1})^n}{(1-x^n)^{n+1}}$$ and $$b'_n(x)=\dfrac{n(n+1)x^{n-1}(1-x^{n+1})^{n-1}(1-x^n)^n(1-x)}{(1-x^n)^{2(n+1)}}.$$ So, $b'_n(x) > 0$ for all $x \in (0,1)$, so $b_n(x)$ is increasing in $(0,1)$ and $b_n(x) > b_n(0)=1$, i.e. $c^{n(n+1)}_n > 1$, which yields $c_n > 1$. Now, we prove that $c_n$ is a decreasing sequence. We have $$\dfrac{b_{n-1}}{b_n}=\left(\dfrac{(1-x^n)^2}{(1-x^{n-1})(1-x^{n+1})}\right)^n > 1,$$ so $c_{n-1}^{(n-1)n}>c_n^{n(n+1)}=c_n^{(n-1)n}c_n^{2n}>c_n^{(n-1)n}$, i.e. $c_{n-1}>c_n$. Moreover,
$$\lim_{n \to \infty} \log c_n=\lim_{n \to \infty} \left[\dfrac{1}{n+1}\log (1-x^{n+1})-\dfrac{1}{n}\log(1-x^n)\right]=0,$$ so $$\lim_{n \to \infty} c_n= \lim_{n \to \infty} e^{\log c_n}=1.$$
Let $a>b$ be positive real numbers. Prove that $$c_n=\dfrac{\sqrt[n+1]{a^{n+1}-b^{n+1}}}{\sqrt[n]{a^n-b^n}}$$
is a decreasing sequence and find its limit.
Proposed by Ivan Borsenco.
Solution:
Let $c_n=\dfrac{\sqrt[n+1]{1-x^{n+1}}}{\sqrt[n]{1-x^n}}$, where $x=b/a < 1$. We first prove that $c_n > 1$. Let $b_n=c^{n(n+1)}_n$. Then $$b_n(x)=\dfrac{(1-x^{n+1})^n}{(1-x^n)^{n+1}}$$ and $$b'_n(x)=\dfrac{n(n+1)x^{n-1}(1-x^{n+1})^{n-1}(1-x^n)^n(1-x)}{(1-x^n)^{2(n+1)}}.$$ So, $b'_n(x) > 0$ for all $x \in (0,1)$, so $b_n(x)$ is increasing in $(0,1)$ and $b_n(x) > b_n(0)=1$, i.e. $c^{n(n+1)}_n > 1$, which yields $c_n > 1$. Now, we prove that $c_n$ is a decreasing sequence. We have $$\dfrac{b_{n-1}}{b_n}=\left(\dfrac{(1-x^n)^2}{(1-x^{n-1})(1-x^{n+1})}\right)^n > 1,$$ so $c_{n-1}^{(n-1)n}>c_n^{n(n+1)}=c_n^{(n-1)n}c_n^{2n}>c_n^{(n-1)n}$, i.e. $c_{n-1}>c_n$. Moreover,
$$\lim_{n \to \infty} \log c_n=\lim_{n \to \infty} \left[\dfrac{1}{n+1}\log (1-x^{n+1})-\dfrac{1}{n}\log(1-x^n)\right]=0,$$ so $$\lim_{n \to \infty} c_n= \lim_{n \to \infty} e^{\log c_n}=1.$$
Mathematical Reflections 2012, Issue 5 - Problem S243
Problem:
A group of boys and girls went to a dance party. It is known that for every pair of boys, there are exactly two girls who danced with both of them; and for every pair of girls there are exactly two boys who danced with both of them. Prove that the numbers of girls and boys are equal.
Solution:
Let $m$ be the number of boys and $n$ be the number of girls. There are $\displaystyle{m \choose 2}$ possibile pairs of boys and $\displaystyle{n \choose 2}$ possible pairs of girls. Suppose that $\displaystyle{m \choose 2} > \displaystyle{n \choose 2}$. Then, there are two distinct pairs of boys which dance with a pair of girls, but this means that there are at least three boys which dance with two girls, contradiction. With the same reasoning we can prove that it can't be $\displaystyle{m \choose 2} < \displaystyle{n \choose 2}$, so $\displaystyle{m \choose 2} = \displaystyle{n \choose 2}$, i.e. $m=n$.
A group of boys and girls went to a dance party. It is known that for every pair of boys, there are exactly two girls who danced with both of them; and for every pair of girls there are exactly two boys who danced with both of them. Prove that the numbers of girls and boys are equal.
Proposed by Iurie Boreico.
Solution:
Let $m$ be the number of boys and $n$ be the number of girls. There are $\displaystyle{m \choose 2}$ possibile pairs of boys and $\displaystyle{n \choose 2}$ possible pairs of girls. Suppose that $\displaystyle{m \choose 2} > \displaystyle{n \choose 2}$. Then, there are two distinct pairs of boys which dance with a pair of girls, but this means that there are at least three boys which dance with two girls, contradiction. With the same reasoning we can prove that it can't be $\displaystyle{m \choose 2} < \displaystyle{n \choose 2}$, so $\displaystyle{m \choose 2} = \displaystyle{n \choose 2}$, i.e. $m=n$.
Mathematical Reflections 2012, Issue 5 - Problem S241
Problem:
Let $p$ and $q$ be odd primes such that $\dfrac{p^3-q^3}{3} \geq 2pq+3.$ Prove that
$$\dfrac{p^3-q^3}{4} \geq 3pq+16.$$
Solution:
Clearly, $p>q$, so $p-2 \geq q$. It must be $p-2 \neq q$, otherwise we would have $$p^3-q^3=p^3-(p-2)^3=6p^2-12p+8=6pq+8,$$ which contradicts the given inequality. So $p-4 \geq q$ and $$p^3-q^3 \geq p^3-(p-4)^3=12p^2-48p+64=12p(p-4)+64 \geq 12pq+64,$$ which gives the conclusion.
Let $p$ and $q$ be odd primes such that $\dfrac{p^3-q^3}{3} \geq 2pq+3.$ Prove that
$$\dfrac{p^3-q^3}{4} \geq 3pq+16.$$
Proposed by Titu Andreescu.
Solution:
Clearly, $p>q$, so $p-2 \geq q$. It must be $p-2 \neq q$, otherwise we would have $$p^3-q^3=p^3-(p-2)^3=6p^2-12p+8=6pq+8,$$ which contradicts the given inequality. So $p-4 \geq q$ and $$p^3-q^3 \geq p^3-(p-4)^3=12p^2-48p+64=12p(p-4)+64 \geq 12pq+64,$$ which gives the conclusion.
Mathematical Reflections 2012, Issue 5 - Problem J245
Problem:
Find all triples $(x,y,z)$ of positive real numbers satisfying simultaneously the inequalities $x+y+z-2xyz \leq 1$ and $$xy+yz+zx+\dfrac{1}{xyz} \leq 4.$$
Solution:
Using HM-AM Inequality, from the first inequality we have
$$\dfrac{9xyz}{xy+yz+zx} \leq x+y+z \leq 1+2xyz,$$ which gives $\dfrac{9xyz}{2xyz+1} \leq xy+yz+zx$. From the second inequality, we have $$\dfrac{9xyz}{2xyz+1}+\dfrac{1}{xyz} \leq 4.$$ Putting $t=xyz>0$ and clearing the denominators, we obtain $$9t^2+2t+1 \leq 4t(2t+1),$$ i.e. $(t-1)^2 \leq 0$, which gives $t=1$. So, the first inequality is $x+y+z \leq 3xyz$, but by AM-GM Inequality $3xyz \leq x+y+z$, which implies $x+y+z=3xyz$. The equality holds if and only if $x=y=z=1$, so the only triple which satisfies the two inequalities is $(1,1,1)$.
Find all triples $(x,y,z)$ of positive real numbers satisfying simultaneously the inequalities $x+y+z-2xyz \leq 1$ and $$xy+yz+zx+\dfrac{1}{xyz} \leq 4.$$
Proposed by Titu Andreescu.
Solution:
Using HM-AM Inequality, from the first inequality we have
$$\dfrac{9xyz}{xy+yz+zx} \leq x+y+z \leq 1+2xyz,$$ which gives $\dfrac{9xyz}{2xyz+1} \leq xy+yz+zx$. From the second inequality, we have $$\dfrac{9xyz}{2xyz+1}+\dfrac{1}{xyz} \leq 4.$$ Putting $t=xyz>0$ and clearing the denominators, we obtain $$9t^2+2t+1 \leq 4t(2t+1),$$ i.e. $(t-1)^2 \leq 0$, which gives $t=1$. So, the first inequality is $x+y+z \leq 3xyz$, but by AM-GM Inequality $3xyz \leq x+y+z$, which implies $x+y+z=3xyz$. The equality holds if and only if $x=y=z=1$, so the only triple which satisfies the two inequalities is $(1,1,1)$.
Mathematical Reflections 2012, Issue 5 - Problem J244
Problem:
Let $a$ and $b$ positive real numbers. Prove that
$$1 \leq \dfrac{\sqrt[n]{a^n+b^n}}{\sqrt[n+1]{a^{n+1}+b^{n+1}}} \leq \sqrt[n(n+1)]{2}.$$
Let $a$ and $b$ positive real numbers. Prove that
$$1 \leq \dfrac{\sqrt[n]{a^n+b^n}}{\sqrt[n+1]{a^{n+1}+b^{n+1}}} \leq \sqrt[n(n+1)]{2}.$$
Proposed by Ivan Borsenco.
Solution:
Since we can collect $a$ or $b$ in the given expression, it is sufficient to prove that
$$1 \leq \dfrac{\sqrt[n]{1+x^n}}{\sqrt[n+1]{1+x^{n+1}}} \leq \sqrt[n(n+1)]{2} \qquad \forall x \in \mathbb{R}^+,$$ or equivalently
$$1 \leq \dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n} \leq 2, \qquad \forall x \in \mathbb{R}^+.$$ Put $f(x)=\dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n}$. Then, $$f'(x)=\dfrac{n(n+1)x^{n-1}(1+x)^n(1+x^{n+1})^{n-1}(1-x)}{[(1+x^{n+1})^n]^2},$$ and $f'(x) > 0$ if and only if $x<1$, i.e. $f(1)$ is a maximum. Moreover, $f(0)=1$ and $\lim_{x \to +\infty} f(x)=1^+$, so $f(0) \leq f(x) \leq f(1)$ for all positive real numbers $x$, which gives $1 \leq f(x) \leq 2$ for all positive real numbers $x$.
$$1 \leq \dfrac{\sqrt[n]{1+x^n}}{\sqrt[n+1]{1+x^{n+1}}} \leq \sqrt[n(n+1)]{2} \qquad \forall x \in \mathbb{R}^+,$$ or equivalently
$$1 \leq \dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n} \leq 2, \qquad \forall x \in \mathbb{R}^+.$$ Put $f(x)=\dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n}$. Then, $$f'(x)=\dfrac{n(n+1)x^{n-1}(1+x)^n(1+x^{n+1})^{n-1}(1-x)}{[(1+x^{n+1})^n]^2},$$ and $f'(x) > 0$ if and only if $x<1$, i.e. $f(1)$ is a maximum. Moreover, $f(0)=1$ and $\lim_{x \to +\infty} f(x)=1^+$, so $f(0) \leq f(x) \leq f(1)$ for all positive real numbers $x$, which gives $1 \leq f(x) \leq 2$ for all positive real numbers $x$.
Mathematical Reflections 2012, Issue 5 - Problem J243
Problem:
Let $a,b,c$ be real numbers such that $$\left(-\dfrac{a}{2}+\dfrac{b}{3}+\dfrac{c}{6}\right)^3+\left(\dfrac{a}{3}+\dfrac{b}{6}-\dfrac{c}{2}\right)^3+\left(\dfrac{a}{6}-\dfrac{b}{2}+\dfrac{c}{3}\right)^3=\dfrac{1}{8}.$$ Prove that $$(a-3b+2c)(2a+b-3c)(-3a+2b+c)=9.$$
Solution:
From the given equality, we have $$(-3a+2b+c)^3+(2a+b-3c)^3+(a-3b+2c)^3=27.$$
Let $x=-3a+2b+c, y=2a+b-3c, z=a-3b+2c$. Since $x+y+z=0$, using the well known identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx),$$ we get $x^3+y^3+z^3=3xyz$, i.e.
$$3(-3a+2b+c)(2a+b-3c)(a-3b+2c)=27,$$ which gives $(-3a+2b+c)(2a+b-3c)(a-3b+2c)=9$.
Let $a,b,c$ be real numbers such that $$\left(-\dfrac{a}{2}+\dfrac{b}{3}+\dfrac{c}{6}\right)^3+\left(\dfrac{a}{3}+\dfrac{b}{6}-\dfrac{c}{2}\right)^3+\left(\dfrac{a}{6}-\dfrac{b}{2}+\dfrac{c}{3}\right)^3=\dfrac{1}{8}.$$ Prove that $$(a-3b+2c)(2a+b-3c)(-3a+2b+c)=9.$$
Proposed by Titu Andreescu.
Solution:
From the given equality, we have $$(-3a+2b+c)^3+(2a+b-3c)^3+(a-3b+2c)^3=27.$$
Let $x=-3a+2b+c, y=2a+b-3c, z=a-3b+2c$. Since $x+y+z=0$, using the well known identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx),$$ we get $x^3+y^3+z^3=3xyz$, i.e.
$$3(-3a+2b+c)(2a+b-3c)(a-3b+2c)=27,$$ which gives $(-3a+2b+c)(2a+b-3c)(a-3b+2c)=9$.
Mathematical Reflections 2012, Issue 5 - Problem J242
Problem:
Let $ABC$ be a triangle and let $D,E,F$ be the feet of the altitudes from $A,B,C$ to the
sides $BC,CA,AB$, respectively. Let $X,Y,Z$ be the midpoints of segments $EF,FD,DE$ and let $x,y,z$ be the perpendiculars from $X,Y,Z$ to $BC,CA$, and $AB$, respectively. Prove that the lines $x,y,z$ are concurrent.
Let $ABC$ be a triangle and let $D,E,F$ be the feet of the altitudes from $A,B,C$ to the
sides $BC,CA,AB$, respectively. Let $X,Y,Z$ be the midpoints of segments $EF,FD,DE$ and let $x,y,z$ be the perpendiculars from $X,Y,Z$ to $BC,CA$, and $AB$, respectively. Prove that the lines $x,y,z$ are concurrent.
Proposed by Cosmin Pohoata.
Solution:
Obviously, the lines $x,y,z$ are parallel to the altitudes $AD,BE,CF$ respectively. Since the orthocenter of the triangle $ABC$ is the incenter of the orthic triangle $DEF$, the lines $AD,BE,CF$ are bisectrices of the angle $D,E,F$ of the triangle $DEF$ respectively. Moreover, since $X,Y,Z$ are midpoints of the segments $EF,FD,DE$, then $XY,YZ,ZX$ are parallel to $DE,EF,FD$. The angle between the line $XY$ and the line $x$ and the angle $ADE$ are equal (alternate interior angles) and so are the angle between the line $XZ$ and the line $x$ and the angle $ADF$ (alternate exterior angles), so the line $x$ is the bisectrix of the angle $X$ of the triangle $XYZ$. Likewise, the lines $y$ and $z$ are bisectrices of the angles $Y,Z$ respectively. So the lines $x,y,z$ are concurrent since their point of intersection is the incenter of the triangle $XYZ$.
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