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# Solutions

91.
A square and a regular pentagon are inscribed in a circle. The nine vertices are all distinct and divide the circumference into nine arcs. Prove that at least one of them does not exceed 1/40 of the circumference of the circle.
Solution. Let the four points of the square be at $A$, $B$, $C$, $D$. We can partition the circumference into eight arcs as follows: (1) four closed arcs (containing endpoints), each of length 1/20 of the circumference centered at $A$, $B$, $C$ and $D$; call these the shorter arcs: (2) four open arcs (not containing endpoints), each of length 1/5 of the circumference and interpolated between two of the arcs centred at the vertices of the square; call these the longer arcs.
Suppose a regular pentagon is inscribed in a circle. Since any pair of adjacent vertices terminate a closed arc whose length is 1/5 of the circumference, no two vertices of the pentagon can belong to the same longer arc. Since there are only four longer arcs, at least one vertex of the pentagon must lie inside a shorter arc and so be no more distant that $\frac{1}{2}×\frac{1}{20}=\frac{1}{40}$ from a vertex of the square.

92.
Consider the sequence $200125$, $2000125$, $20000125$, $\dots$, $200\dots 00125$, $\dots$ (in which the $n$th number has $n+1$ digits equal to zero). Prove that none of these numbers is the square, cube or fifth power of an integer.
Solution. Each number has the form $2×{10}^{n+1}+125$ where $n\ge 4$. Since both ${10}^{n}$ and $120$ are multiples of 8, each number in the sequence is congruent to 5, modulo 8, and so cannot be square.
Observe that ${10}^{n+1}={10}^{n-2}×8×125$, so that

$2×{10}^{n+1}+125=125\left(16×{10}^{n-2}+1\right)={5}^{3}\left(16×{10}^{n-2}+1\right) .$

If such a number is a $k$th power, it must be divisible by ${5}^{k}$. Since $16×{10}^{n-2}+1$ is not a multiple of 5, no number in the sequence can be a $k$th power for $k\ge 4$.
Let ${u}_{n}=16×{10}^{n-2}+1$, for $n\ge 5$. Suppose that $\left(10{x}_{n}+1\right){}^{3}={u}_{n}$. Then

$1000{x}_{n}^{3}+300{x}_{n}^{2}+30{x}_{n}=16×{10}^{n-2}$

so that ${x}_{n}={10}^{n-3}{y}_{n}$ for some ${y}_{n}$ and we find that

$16={10}^{3n-6}{y}_{n}^{3}+3×{10}^{2n-4}{y}_{n}^{2}+3{y}_{n}\ge {y}_{n}^{2}+3{y}_{n}$

so that ${y}_{n}=1$ or 2. It is straightfoward to check that neither works, so that ${u}_{n}$ can never be a cube. Hence no number in the given sequence can be a cube.

93.
For any natural number $n$, prove the following inequalities:

${2}^{\left(n-1\right)/\left({2}^{n-2}\right)}\le \sqrt{2}\sqrt[4]{4}\sqrt[8]{8}\dots \sqrt[{2}^{n}]{{2}^{n}}<4 .$

Solution. The middle member of the inequality is 2 raised to the power $s\equiv \frac{1}{2}+\frac{2}{4}+\dots +\frac{n}{{2}^{n}}$. Note that

$s-\frac{1}{2}s=\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{{2}^{n}}-\frac{n}{{2}^{n+1}}=1-\frac{1}{{2}^{n}}-\frac{n}{{2}^{n+1}}=1-\frac{n+2}{{2}^{n+1}}$

whence $s=2-\left(n+2\right){2}^{-n}$. Clearly, $s<2$.
On the other hand,

$s-\frac{n-1}{{2}^{n-2}}=\frac{{2}^{n+1}-\left(n+2\right)-4\left(n-1\right)}{{2}^{n}}=\frac{{2}^{n+1}-\left(5n-2\right)}{{2}^{n}} .$

When $n=1$, ${2}^{n+1}-\left(5n-2\right)=1>0$; when $n=2$, ${2}^{n+1}-\left(5n-2\right)=0$, and when $n=3$, ${2}^{n+1}-\left(5n-2\right)=3>0$. Suppose, as an induction hypothesis, that ${2}^{k+1}>5k-2$ for some $k\ge 3$. Then

${2}^{k+2}>10k-4=5\left(k+1\right)-2+\left(5k-7\right)>5\left(k+1\right)-2 ,$

so that, for each positive integer $n$, ${2}^{n+1}\ge 5n-2$, with equality if and only if $n=2$. The desired result follows.

94.
$\mathrm{ABC}$ is a right triangle with arms $a$ and $b$ and hypotenuse $c=‖\mathrm{AB}‖$; the area of the triangle is $s$ square units and its perimeter is $2p$ units. The numbers $a$, $b$ and $c$ are positive integers. Prove that $s$ and $p$ are also positive integers and that $s$ is a multiple of $p$.
Solution. Since ${a}^{2}+{b}^{2}={c}^{2}$ and squares are congruent to 0 or 1, modulo 4, it is not possible for $a$ and $b$ to be both odd. Since, at least one of them is even, $s=\frac{1}{2}\mathrm{ab}$ is an integer. Also, since either two or none of $a$, $b$, $c$ are odd, $p=\frac{1}{2}\left(a+b+c\right)$ is an integer. Let $r$ be the inradius of the triangle, so that $s=\mathrm{rp}$. It remains to find $r$ and show that it is an integer. In any triangle with sides $a$, $b$, $c$, the length of the tangents from vector to incircle are $\frac{1}{2}\left(a+b-c\right)$, $\frac{1}{2}\left(b+c-a\right)$ and $\frac{1}{2}\left(b+c-a\right)$. For a right triangle with hypotenuse $c$, one of the lengths is $r$, namely $r=\frac{1}{2}\left(a+b-c\right)$. Since one of the legs of the triangle is even, the other leg and the hypotenuse must have the same parity, so that $a+b-c$ is even and $r$ is an integer.

95.
The triangle $\mathrm{ABC}$ is isosceles is isosceles with equal sides $\mathrm{AC}$ and $\mathrm{BC}$. Two of its angles measure ${40}^{ˆ}$. The interior point $M$ is such that $\angle \mathrm{MAB}={10}^{ˆ}$ and $\angle \mathrm{MBA}={20}^{ˆ}$. Determine the measure of $\angle \mathrm{CMB}$.
Solution. Let $\mathrm{BM}$ be produced to meet $\mathrm{AC}$ at $N$. Since $\angle \mathrm{AMB}={150}^{ˆ}$, we have that $\angle \mathrm{NMA}={30}^{ˆ}=\angle \mathrm{NAM}$ so that $\mathrm{NA}=\mathrm{NM}$. Let $a=‖\mathrm{BC}‖=‖\mathrm{AC}‖$, $c=‖\mathrm{AB}‖$, $u=‖\mathrm{CN}‖$ and $v=‖\mathrm{NA}‖=‖\mathrm{NM}‖$. Since $\mathrm{BN}$ bisects angle $\mathrm{CBA}$, we have that $v/u=c/a$.
By the Law of Sines,

$\frac{c}{a}=\frac{\mathrm{sin}{100}^{ˆ}}{\mathrm{sin}{40}^{ˆ}}=\frac{\mathrm{sin}{80}^{ˆ}}{\mathrm{sin}{40}^{ˆ}}=2\mathrm{cos}{40}^{ˆ}$

and

$\frac{v}{u}=\frac{\mathrm{sin}\left(\alpha -{60}^{ˆ}\right)}{\mathrm{sin}\alpha }$

where $\alpha =\angle \mathrm{CMB}$. Hence

$\mathrm{sin}\left(\alpha -{60}^{ˆ}\right)=2\mathrm{sin}\alpha \mathrm{cos}{40}^{ˆ}=\mathrm{sin}\left(\alpha +{40}^{ˆ}\right)+\mathrm{sin}\left(\alpha -{40}^{ˆ}\right)$

whence

$2\mathrm{cos}\left(\alpha -{50}^{ˆ}\right)\mathrm{sin}{10}^{ˆ}=\mathrm{sin}\left(\alpha -{60}^{ˆ}\right)-\mathrm{sin}\left(\alpha -{40}^{ˆ}\right)=\mathrm{sin}\left(\alpha +{40}^{{}^{c}\mathrm{irc}}\right) .$

Since $\alpha +{40}^{ˆ}=\left(\alpha -{50}^{ˆ}\right)+{90}^{ˆ}$, $\mathrm{sin}\left(\alpha +{40}^{ˆ}\right)=\mathrm{cos}\left(\alpha -{50}^{ˆ}\right)$. Therefore, we find that $2\mathrm{cos}\left(\alpha -{50}^{ˆ}\right)\mathrm{sin}{10}^{ˆ}=\mathrm{cos}\left(\alpha -{50}^{ˆ}\right)$. Since $\mathrm{sin}{10}^{ˆ}\ne \frac{1}{2}$, we have that $\mathrm{cos}\left(\alpha -{50}^{ˆ}\right)=0$ and $\alpha ={140}^{ˆ}$.

96.
Find all prime numbers $p$ for which all three of the numbers ${p}^{2}-2$, $2{p}^{2}-1$ and $3{p}^{2}+4$ are also prime.
Solution. Modulo 7, we find that ${p}^{2}-2\equiv 0$ when $p\equiv 3,4$, $2{p}^{2}-1\equiv 0$ when $p\equiv 2,5$ and $3{p}^{2}+4\equiv 0$ when $p\equiv 1,6$. Thus, when $p>7$, at least one of the four numbers is a proper multiple of 7. The only primes $p$ for which all numbers are prime at 3 and 7, and we get the quadruples $\left(3,7,17,31\right)$ and $\left(7,47,97,151\right)$.
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