Solutions

49.

Find all ordered pairs
$(x,y)$ that are solutions
of the following system of two equations (where
$a$ is a
parameter):
$xy=2$
$(x\frac{2}{a})(y\frac{2}{a})={a}^{2}1\hspace{1em}.$
Find all values of the parameter
$a$ for which the solutions of
the system are two pairs of nonnegative numbers. Find the
minimum value of
$x+y$ for these values of
$a$.
Solution. We need to assume that
$a\ne 0$.
Substitute
$y=x2$ into the second equation to
obtain
${x}^{2}2x\frac{4}{a}(x1)+\frac{4}{{a}^{2}}={a}^{2}1\hspace{1em}.$
This can be manipulated to
$[x(1+\frac{2}{a}){]}^{2}={a}^{2}\hspace{1em},$
from which we find that
$(x,y)=(1+(2/a)+a,1+(2/a)+a),(1+(2/a)a,1+(2/a)a)\hspace{1em}.$
For
$x$ and
$y$ to be nonnegative for both solutions, we require
all of the four inequalities:
$a({a}^{2}+a+2)\ge 0\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}a({a}^{2}a+2)\ge 0\hspace{1em},$
$a(a2)r(a+1)=a({a}^{2}a2)\le 0\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}a(a+2)(a1)=a({a}^{2}+a2)\le 0\hspace{1em}.$
The last two equations entail that
$a\le 2$ or
$0<a\le 1$.
When
$a\le 2$, the first two inequalities fail to hold, while
if
$0<a\le 1$, all four equations hold. Thus, the solutions
are two pairs of nonnegative numbers when
$0<a\le 1$.
When both
$x$ and
$y$ are nonnegative, then
$x+y\ge xy=2$.
When
$a=1$, we obtain the solutions
$(x,y)=(4,2)$ and
$(2,0)$,
and in the latter case,
$x+y=2$. Hence the minimum value of
$x+y$ is 2.
Comment. The minimum of
$x+y$ can be also obtained
with more effort directly from the solutions in terms of
$a$.
Now
$x+y=(4/a)\pm 2a$. Since
$(\frac{4}{a}+2a)6=\frac{2(2a)(1a)}{a}\ge 0$
for
$0<a\le 1$,
$(4/a)+2a$ assumes its minimum
value of 6 when
$a=1$. Since
$(\frac{4}{a}2a)2=\frac{2(2+a)(1a)}{a}\ge 0$
for
$0<a\le 1$,
$(4/a)2a$ assumes its minimum value
of 2 when
$a=1$. Hence the minimum possible value of
$x+y$ is 2.

50.

Let
$n$ be a natural number exceeding 1, and let
${A}_{n}$ be the set of all natural numbers that are
not relatively prime with
$n$ (i.e.,
${A}_{n}=\{x\in N:\hspace{1em}\mathrm{gcd}\hspace{1em}(x,n)\ne 1\}$.
Let us call the number
$n$ magic if for each two numbers
$x,y\in {A}_{n}$, their sum
$x+y$ is also an element of
${A}_{n}$ (i.e.,
$x+y\in {A}_{n}$ for
$x,y\in {A}_{n}$).


(a) Prove that 67 is a magic number.


(b) Prove that 2001 is not a magic number.


(c) Find all magic numbers.
Solution. [O. Bormashenko] (a) 67 is prime, so that
all numbers that are not relatively prime with 67 and are
elements of
${A}_{67}$ are the multiples of 67. The sum
of two such numbers is also a multiple of 67, and hence
belongs to
${A}_{67}$. Therefore 67 is a magic number.
(b)
$2001=3\times 23\times 29$. Now 3 and 23 belong to
${A}_{2001}$, but 26 = 3 + 23 does not, because
gcd (26, 2001) is equal to 1. Thus, 2001 is not a magic number.
(c) First, let us prove that all prime powers are magic numbers.
Suppose that
$n={p}^{k}$ for some prime
$p$ and positive integer
$k$. Then the numbers not relatively prime to
$n$ are
precisely those that are divisible by
$p$ (as this is the
only prime that can divide any divisor of
$n$). The sum of
any two such numbers is also divisible by
$p$, so that
${A}_{n}$ is closed under addition and
$n$ is magic.
Now suppose that
$n$ is not a power of a prime. Then
$n=\mathrm{ab}$,
where
$a$ and
$b$ are two relatively prime numbers exceeding 1.
Clearly,
$a$ and
$b$ belong to
${A}_{n}$. However, gcd
$(a,a+b)$
= gcd
$(a,b)$ = 1, and gcd
$(b,a+b)$ = gcd
$(b,a)$ = 1, so that
$a+b$ is relatively prime to both
$a$ and
$b$, and hence to
$n=\mathrm{ab}$. Thus,
$a+b\backslash not\in {A}_{n}$, so that
$n$ is not magic.
We conclude that the set of prime powers equals the set of
magic numbers.

51.

In the triangle
$\mathrm{ABC}$,
$\mathrm{AB}=15$,
$\mathrm{BC}=13$ and
$\mathrm{AC}=12$. Prove that, for this triangle, the angle bisector
from
$A$, the median from
$B$ and the altitude from
$C$ are
concurrent (i.e., meet in a common point).
Solution. [K. Ho] Denote by
$E$ the intersection
of
$\mathrm{AC}$ with the median from
$B$ (so that
$E$ is the
midpoint of
$\mathrm{AC}$), by
$F$ the intersection of
$\mathrm{AB}$
with the altitude from
$C$ (so that
$\mathrm{CF}\perp \mathrm{AB}$),
by
$D$ the intersection of
$\mathrm{BC}$ and the bisector from
$A$, and by
$x$ the length of
$\mathrm{AF}$. Then
$\mathrm{BF}=\mathrm{AB}\mathrm{AF}=15x$. By the anglebisector theorem,
$\mathrm{CD}:\mathrm{DB}=\mathrm{AC}:\mathrm{AB}=12:15=4:5$. By the pythagorean
theorem for the triangles
$\mathrm{AFC}$ and
$\mathrm{BFC}$,
${\mathrm{AC}}^{2}{\mathrm{AF}}^{2}={\mathrm{CF}}^{2}={\mathrm{BC}}^{2}{\mathrm{BF}}^{2}\&lrArr;{12}^{2}{x}^{2}={13}^{2}(15x){}^{2}=56+30x{x}^{2}$.
Thus,
$x=20/3$ and
$15x=25/3$, so that
$\mathrm{AF}:\mathrm{FB}=4:5$. Since
$(\mathrm{AF}/\mathrm{FB})(\mathrm{BD}/\mathrm{DC})/(\mathrm{CE}/\mathrm{EA})=(4/5)(5/4)1=1$, Ceva's
theorem tells us that
$\mathrm{AD}$,
$\mathrm{BE}$ and
$\mathrm{CF}$ are concurrent.

52.

One solution of the equation
$2{x}^{3}+{\mathrm{ax}}^{2}+\mathrm{bx}+8=0$
is
$1+\sqrt{3}$. Given that
$a$ and
$b$ are rational
numbers, determine its other two solutions.
Solution 1. [R. Barrington Leigh] Since
$1+\sqrt{3}$
satisfies the given equation,
$0=2(1+\sqrt{3}){}^{3}+a(1+\sqrt{3}){}^{2}+b(1+\sqrt{3})+8=(28+4a+b)+(12+2a+b)\sqrt{3}\hspace{1em}.$
Since
$a$ and
$b$ are rational, this is possible only when
$28+4a+b=12+2a+b=0$, or
$(a,b)=(8,4)$.
The equation is thus
$0=2{x}^{3}8{x}^{2}+4x+8$.
By inspection, we find that 2 is a root, and so the three
roots turn out to be
$2$,
$1+\sqrt{3}$ and
$1\sqrt{3}$.
Comment. Once 2 and
$1+\sqrt{3}$ are known to be roots,
we can get the third root by noting that the sum of the roots
is
$(8)/2=4$.
Solution 2. A more structural way of getting the
result is to note that the mapping that takes a
surd
$u+v\sqrt{3}$ (with
$u$ and
$v$ rational) to its
surd conjugate
$uv\sqrt{3}$ preserves addition,
subtraction and multiplication (i.e., the
surd conjugate of the sum (resp. product) or two
surds is equal to the sum (resp. product) of the
surd conjugates. The surd conjugate of a rational
is the rational itself. Tranforming all the elements
of the equation to their surd conjugates, gives the
same equation with
$x$ replaced by its surd conjugate.
Thus, the surd conjugate
$1\sqrt{3}$ of
$1+\sqrt{3}$ also satisfies the equation.
The quadratic with these as roots is
${x}^{2}2x2$, and this quadratic must be a factor
of the given cubic. Since
$2{x}^{3}+{\mathrm{ax}}^{2}+\mathrm{bx}+8=({x}^{2}2x2)(2+[a+4]x)+(2a+b+12)x+(16+2a)$, we must have
$2a+b+12=16+2a=0$, whence
$(a,b)=(8,4)$,
and
$2{x}^{3}+{\mathrm{ax}}^{2}+\mathrm{bx}+8=2({x}^{2}2x2)(x2)$
and the third root is 2.

53.

Prove that among any 17 natural numbers chosen from
the sets
$\{1,2,3,\dots ,24,25\}$, it is always possible
to find two whose product is a perfect square.
Solution. [K. Ho] Consider the following 16 sets:
$\{1,4,9,16,25\}$,
$\{2,8,18\}$,
$\{3,12\}$,
$\{5,20\}$,
$\{6,24\}$,
$\{7\}$,
$\{10\}$,
$\{11\}$,
$\{13\}$,
$\{14\}$,
$\{15\}$,
$\{17\}$,
$\{19\}$,
$\{21\}$,
$\{22\}$,
$\{23\}$. In any of the
sets with more than one element, the product of any two
elements is a perfect square. Any choice of 17 numbers from
among these sets must, by the Pigeonhole Principle,
yield at least two from the same set. The product of these
two must be a square. The desired result follows.

54.

A circle has exactly one common point with each of the
sides of a
$(2n+1)$sided polygon. None of the vertices of the
polygon is a point of the circle. Prove that at least one of the
sides is a tangent of the circle.
Solution. [J.Y. Jin] Assume that none of the sides is
tangent to the circle. Let
${A}_{1},{A}_{2},\dots ,{A}_{2n1}$ be
the consecutive vertices of a
$(2n+1)$sided polygon. Each side
of the polygon has exactly one common point with the circle, none
of the vertices lies on the circle and, by assumption, none of
the sides is tangent to the circle. Therefore, for each side,
one of the endpoints lies inside the circle and the second
endpoint lies outside the circle. Colour the vertices inside
the circle blue and the ones outside the circle red. Apparently,
the colours of the consecutive vertices alternate (
i.e.,
blue, red, blue, red, etc.). Since there are
$2n+1$ vertices,
${A}_{1}$ and
${A}_{2n+1}$ must have the same colour. However,
these two vertices are the endpoints of a side of the polygon
and so they cannot have the same colour. Thus, the assumption
leads to a contradiction, and so at least one of the
sides is tangent to the circle.