Problem 01

Find all real roots of the equation

$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$ ,

where $p$ is a real parameter.

Solution

Assuming $x \geq 0$ , square the equation, obtaining $4\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2$ . If we have $p + 4 \geq 4x^2$ , we can square again, obtaining $x^2 = \frac {(p - 4)^2}{4(4 - 2p)} \implies x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}$

We must have $4 - 2p > 0 \iff p < 2$ , so we have $x = \frac {4 - p}{2\sqrt {4 - 2p}}$

However, this is only a solution when $p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\leq(p - 4)^2 \iff 0\leq p(3p - 4)$

so we have $p\leq 0$ or $p \geq \frac {4}{3}$

But if $p < 0$ , then $\sqrt {x^2 - p} > x$ contradiction.

So we have $x = \frac {4 - p}{2\sqrt {4 - 2p}}$ for $p = 0, \frac {4}{3}\leq p < 2$ .

Problem 02

Point $A$ and segment $BC$ are given. Determine the locus of points in space which are the vertices of right angles with one side passing through $A$ , and the other side intersecting the segment $BC$ .

Solution

Let $\omega_1$ be the circle with diameter $AB$ , and let $\omega_2$ be the circle with diameter $AC$ . Then the locus is simply the set of points inside either $\omega_1$ or $\omega_2$ , but not both.

To see this, suppose the right angle's ray that does not pass through $A$ intersects segment $BC$ at $X$ . Then the right angle's vertex must lie on the circle with diameter $AX$ . So, for a particular $X$ , the desired locus is a circle with diameter $AX$ . Accounting for all possible $X$ , the total locus is the union of the circumferences of all circles that have a diameter $AX$ , where $X$ is some point on $BC$ .

As $X$ moves from $B$ to $C$ , the motion of the circle with diameter $AX$ is continuous and fluid. Any point $P$ lying within $\omega_1$ but outside $\omega_2$ will eventually be intersected by this moving circle, since it went from inside to outside the circle. This applies similarly to all points inside $\omega_2$ but outside $\omega_1$ . Also, the points inside both $\omega_1$ and $\omega_2$ are never intersected by this moving circle, as it always stays inside.

Problem 03

In an $n$ -gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation

$a_1\ge a_2\ge \cdots \ge a_n$ .

Prove that $a_1=a_2=\cdots = a_n$ .

Solution

Define the vector $\vec{v_i}$ to equal $\cos{\left(\frac{2\pi}{n}i\right)}\vec{i}+\sin{\left(\frac{2\pi}{n}i\right)}\vec{j}$ . Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length $a_i$ is parallel to $\vec{v_i}$ . We then have that

$\sum_{i=1}^{n} a_i\vec{v_i}=\vec{0}\Rightarrow \sum_{i=1}^{n} a_i\cos{\left(\frac{2\pi}{n}i\right)} = \sum_{i=1}^{n} a_i\sin{\left(\frac{2\pi}{n}i\right)} =0$

But $a_i\geq a_{n-i}$ for all $i\leq \lfloor \frac{n}{2}\rfloor$ , so

$a_i \sin{\left(\frac{2\pi}{n}i\right)} = -a_i\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq -a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}$

for all $i\leq \lfloor \frac{n}{2}\rfloor$ . This shows that $a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)}\geq 0$ , with equality when $a_i=a_{n-i}$ . Therefore

$\sum_{i=1}^{n} a_i \sin{\left(\frac{2\pi}{n}i\right)}=\sum_{i=1}^{\lfloor \frac{n}{2}\rfloor} a_i \sin{\left(\frac{2\pi}{n}i\right)}+a_{n-i}\sin{\left(\frac{2\pi}{n}(n-i)\right)} \geq 0$

There is equality only when $a_i=a_{n-i}$ for all $i$ . This implies that $a_1=a_{n-1}$ and $a_2=a_n$ , so we have that $a_1=a_2=\cdots =a_n$ .

Problem 04

Find all solutions $x_1,x_2,x_3,x_4,x_5$ of the system $\begin{eqnarray*} x_5+x_2&=&yx_1\\ x_1+x_3&=&yx_2\\ x_2+x_4&=&yx_3\\ x_3+x_5&=&yx_4\\ x_4+x_1&=&yx_5,\end{eqnarray*}$ where $y$ is a parameter.

Problem 05

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$

Solutions

Solution 1

Let $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=S$ . We have

$S=\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\cos{\frac{\pi}{7}}+\cos{\frac{3\pi}{7}}+\cos{\frac{5\pi}{7}}$

Then, by product-sum formulae, we have

$S * 2* \sin{\frac{\pi}{7}} = \sin{\frac{2\pi}{7}}+\sin{\frac{4\pi}{7}}-\sin{\frac{2\pi}{7}}+\sin{\frac{6\pi}{7}}-\sin{\frac{4\pi}{7}}=\sin{\frac{6\pi}{7}}=\sin{\frac{\pi}{7}}$

Thus $S = 1/2$ . $\blacksquare$

Solution 2

Let $a=\sin{\frac{\pi}{7}}$ and $b=\cos{\frac{\pi}{7}}$ . From the addition formulae, we have

$S=b-[b^2-a^2]+[ b(b^2-a^2)-a(2ab) ]=b-b^2+b^3+a^2(1-3b)$

From the Trigonometric Identity, $a^2=1-b^2$ , so

$S=b-b^2+b^3+(1-b^2)(1-3b)=4b^3-2b^2-2b+1$

We must prove that $S=1/2$ . It suffices to show that $8b^3-4b^2-4b+1=0$ .

Now note that $\cos{\frac{4\pi}{7}}=-\cos{\frac{3\pi}{7}}$ . We can find these in terms of $a$ and $b$ :

$\cos{\frac{4\pi}{7}}=[b^2-a^2]^2-[2ab]^2=b^4-6a^2b^2+b^4=b^4-6(1-b^2)b^2+b^4=8b^4-8b^2+1$

$\cos{\frac{3\pi}{7}}=b[b^2-a^2]-a[2ab]=b^3-3a^2b=b^3-3(1-b^2)b=3b-4b^3$

Therefore $8b^4-8b^2+1=-(3b-4b^3)\Rightarrow 8b^4+4b^3-8b^2-3b+1=0$ . Note that this can be factored:

$8b^4+4b^3-8b^2-3b+1=(8b^3-4b^2-4b+1)(b+1)=0$

Clearly $b\neq -1$ , so $8b^3-4b^2-4b+1=0$ . This proves the result. $\blacksquare$

Problem 06

Five students, $A,B,C,D,E$ , took part in a contest. One prediction was that the contestants would finish in the order $ABCDE$ . This prediction was very poor. In fact no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order $DAECB$ . This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.

Solution

We are given that no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so in order $ABCDE$ . None of them finished in that order. Also only two of them had their actual positions in $DAECB$ . After imposing these two conditions the list of possible outcomes is: (1) $CAEBD$ , (2) $DCAEB$ , (3) $DCEBA$ , (4) $EDACB$ . One more condition is that two disjoint pairs of students predicted to finish consecutively actually did so. Out of the above four in the list, (1) and (2) have $AE$ as the correctly predicted consecutive finishers(but only 1 pair), (3) has no correctly predicted consecutive finishers. But (4) has 2 disjoint correctly predicted consecutive finishers who are $DA$ and $CB$ . Hence, order is $EDACB$ .