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The 2017 Canadian Open Mathematics Challenge — Nov 2/3

Nicolae Strungaru

## Problem of the Week

by Assoc. Prof. Nicolae Strungaru, Grant MacEwan University

The COMC has three parts. In part A solutions do not require work be shown and may be possible to do in your head. In part B the problems begin to draw on more knowledge and have some more challenging aspects that will need a pencil and paper to solve. By Part C the problems require that work be shown and involve arguments to support the answer.

We have selected a cross section of contest problems from a variety of national and regional contests that we hope will stimulate interest in problem solving and give some experience to get ready to write the Canadian Open Mathematics Competition in November. The problem areas are not tied to particular grade levels, or to the curriculum but cover a number of areas from algebra, through logic and some geometry.

We will post solutions to these problems one week later, but teachers should be aware that determined students may be able to locate solutions elsewhere online before then.

For a more comprehensive set of problems and solutions at each of these levels, please feel welcome to download past official exams and solutions from our archive.

Need more COMC Problems-of-the-Week? Take a look at the set from past years: 2016, 2015, 2014, or 2013!

Week 3

• Problem (posted September 19th)

This week we look at a point inside a square.

A point $E$ inside a square $ABCD$ is such that $AE=5, BE=2\sqrt{2}$ and $CE=3$. Determine the area of $ABCD$.

• Solution (check back on September 26th)

Week 2

• Problem (posted September 12th)

This week we look at an equation with integer roots.

Determine all rational numbers $r$ for which all the solutions of the equation $$rx^2+(r+1)x+r-1=0$$ are integers.

• Solution (posted September 19th)

Problem 5 of the $21^{\mbox{st}}$ Austian Mathematical Olympiad, Final Round, 1990, which appeared in Crux Mathematicorum [1992:100]. We present the similar solutions by Joseph Ling, Pavlos Maragoudakis and Michael Selby which appeared at [1993:138].

If $r=0$ then the equation becomes $x-1=0$, so $x=1$. Therefore $r=0$ is a solution.

If $r \neq 0$, let then $x_1 \leq x_2$ be the roots of the quadratic equation. Then \begin{align} x_1+x_2 &= -\frac{r+1}{r}=-1-\frac{1}{r} \cr x_1x_2 &= \frac{r-1}{r}=1-\frac{1}{r} \,. \end{align}

Then \begin{align} x_1x_2-(x_1+x_2) &= 2 \cr (x_1-1)(x_2-1) &= 3 \,. \end{align}

Therefore, as $x_1 \leq x_2$ are integers, we either have $x_1=2, x_2=4$ or $x_1=-2, x_2=0$.

In the first case we get $r=-\frac{1}{7}$ while in the second we get $r=1$.

Answer: $r \in \{ -\frac{1}{7}, 0, 1 \}$.

Week 1

• Problem (posted September 5th)

We give two entry level problems this week. Give them a try. Look for the source and the solution next week!

Problem A
Prove that the number $$\underbrace{111...111}_{1997} \underbrace{222...222}_{1998} 5$$ is a perfect square.

Problem B
In the diagram, $ABCD$ is a rectangle with $AD=1$, and both $BF$ and $DE$ are perpendicular to the diagonal $AC$. We further have $AE=EF=FC$. Find the length of the side $AB$.

• Solution (posted September 12th)

Problem A

Problem 1 of the 2nd Junior Balkan Math Olympiad, 1998, which appeared in the Skoliad Corner of Crux Mathematicorum at [2002:522]. We present the official solution that appeared at [2003:262].

\begin{align} \underbrace{111...111}_{1997}\underbrace{222...222}_{1998}5 &=\underbrace{111...111}_{1997}\underbrace{000...000}_{1999}+2\cdot \underbrace{111...111}_{1998}0+5 \cr & = \frac{10^{1997}-1}{9}\cdot10^{1999}+2 \cdot \frac{10^{1998}-1}{9}\cdot10+5\cr & =\frac{1}{9} \left(10^{3996}-10^{1999}+2\cdot10^{1999}-20+45 \right) \cr & =\frac{1}{9} \left(10^{3996}+\cdot10^{1999}+25 \right) \cr & =\frac{1}{9} \left(10^{3996}+2 \cdot 5 \cdot10^{1998}+5^2 \right) \cr & =\frac{1}{9} \left(10^{1998}+5 \right)^2 =\left(\frac{10^{1998}+5}{3}\right)^2=\left(\frac{10^{1998}-1}{3}+2\right)^2 \cr &=\left( \underbrace{333...333}_{1998}+2 \right)^2 \end{align}

Problem B

Problem 10 of the British Columbia Secondary School Mathematics Contest, 2008, Junior Final, Part A which appeared in the Skoliad Corner of Crux Mathematicorum at [2008:321-324]. We present the solution by Jixuan Wang that appeared at [2009:269].

Let $x$ be the common length of $AE, EF$ and $FC$. Then, $EC=2x$.

By the Pythagorean Theorem, $AE^2+DE^2=AD^2$ and hence $$DE=\sqrt{1-x^2} \,.$$

Note that $$\angle ECD =90^\circ -\angle EDC =\angle ADE \,,$$ and therefore $$\Delta DEA \sim \Delta CED \,.$$

We therefore get \begin{align} \frac{EA}{DE} &=\frac{ED}{CE} \qquad \Rightarrow \cr \frac{x}{\sqrt{1-x^2}} &= \frac{\sqrt{1-x^2}}{2x} \qquad \Rightarrow \cr 2x^2&=1-x^2 \qquad \Rightarrow \cr x&=\frac{1}{\sqrt{3}} \end{align}

Thus, $AC=3x =\sqrt{3}$ and by the Pythagorean Theorem in $\Delta ABC$ we get $$AB= \sqrt{2} \,.$$