## 2nd grade - high school - A category, Primošten, April 7, 2008.

## Solutions

Problem 1. Find all real solutions to the equation

$$
\sqrt{2 x^{2}+3 x+5}+\sqrt{2 x^{2}-3 x+5}=3 x
$$
Solution. By squaring and rearranging, we get:

$$
\begin{gathered}
2 x^{2}+3 x+5+2 \sqrt{\left(2 x^{2}+5\right)^{2}-(3 x)^{2}}+2 x^{2}-3 x+5=9 x^{2} \\
2 \sqrt{4 x^{4}+11 x^{2}+25}=5 x^{2}-10 \\
16 x^{4}+44 x^{2}+100=25 x^{4}-100 x^{2}+100 \\
144 x^{2}=9 x^{4} \\
9 x^{2}\left(x^{2}-16\right)=0
\end{gathered}
$$

The solutions to this equation are $x \in\{-4,0,4\}$. Clearly, $x \geq 0$ and $x=0$ is not a solution to the original equation. The only remaining solution is $x=4$, and it can be verified that this is indeed a solution.

Second solution.

The discriminant of the quadratic functions $f(x)=2 x^{2} \pm 3 x+5$ is: $D=( \pm 3)^{2}-$ $4 \cdot 2 \cdot 5=-310$.

Notice that

$$
\left(2 x^{2}+3 x+5\right)-\left(2 x^{2}-3 x+5\right)=6 x
$$

The left side of this equation can be written as a difference of squares:

$$
\left(\sqrt{2 x^{2}+3 x+5}-\sqrt{2 x^{2}-3 x+5}\right)\left(\sqrt{2 x^{2}+3 x+5}+\sqrt{2 x^{2}-3 x+5}\right)=6 x
$$

The second term on the left side is equal to $3 x$, so we have

$$
\sqrt{2 x^{2}+3 x+5}-\sqrt{2 x^{2}-3 x+5}=2
$$

Adding this equation to the one in the problem, we get

$$
2 \sqrt{2 x^{2}+3 x+5}=3 x+2
$$

By squaring and rearranging, we obtain the equation $x^{2}=16$, whose solutions are $x_{1}=-4$, and $x_{2}=4$. Since the solution must be positive, only $x=4$ remains. It can be directly verified that this is indeed a solution.

Problem 2. Let $a, b, c$ be positive real numbers such that $a^{2}+b^{2}+c^{2}=3$. Prove the inequality

$$
\frac{1}{1+a b}+\frac{1}{1+b c}+\frac{1}{1+c a} \geq \frac{3}{2}
$$

Solution. Using the A-H inequality, we get

$\frac{1}{1+a b}+\frac{1}{1+b c}+\frac{1}{1+c a} \geq \frac{9}{3+a b+b c+c a} \geq \frac{9}{3+a^{2}+b^{2}+c^{2}}=\frac{9}{3+3}=\frac{3}{2}$.

Here we used the inequality $a b+b c+c a \leq a^{2}+b^{2}+c^{2}$, which is equivalent to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$, and this is true.

Problem 3. Determine all integers $x$ such that $1+5 \cdot 2^{x}$ is a square of a rational number.

Solution. We need to consider the following three cases.

$1^{\circ} \mathrm{For} x=0$ we have $1+5 \cdot 2^{x}=6$, which is not a perfect square.

$2^{\circ}$ For $x>0$ we have $1+5 \cdot 2^{x} \in \mathbf{N}$, so we have

$$
1+5 \cdot 2^{x}=n^{2}
$$

for some natural number $n$. Therefore,

$$
5 \cdot 2^{x}=(n-1)(n+1)
$$

Clearly, $n$ must be odd and greater than 1. One of the numbers $n-1, n+1$ is divisible by 4. Additionally, $n^{2}-1$ is divisible by 5, so $n^{2}-1 \geq 5 \cdot 8=40$, i.e., $n \geq 7$.

One of the numbers $n-1$ or $n+1$ is divisible by 5. One of them is divisible by 2 but not by a higher power of 2; the other is divisible by $2^{x-1}$. Clearly, one must be equal to $2 \cdot 5$, and the other to $2^{x-1}$. The possibilities are: $n-1=10$ or $n+1=10$. Only the latter possibility satisfies, i.e., $n=9$. Then $5 \cdot 2^{x}=80$, so $x=4$.

$3^{\circ} \mathrm{For} x<0$ we have $1+5 \cdot 2^{x}$ is a rational number with denominator $2^{-x}$. Then $1+5 \cdot 2^{x}=$ $q^{2}, q \in \mathbf{Q}$ and the denominator of $q$ is $2^{-x / 2}$. Thus, $x$ is even. Setting $x=-2 y, y \in \mathbf{N}$, we have

$$
2^{2 y}+5=\left(q \cdot 2^{y}\right)^{2}
$$

Since $q \cdot 2^{y}=r$ is an integer, we get

$$
5=\left(r-2^{y}\right)\left(r+2^{y}\right)
$$

We must have $r-2^{y}=1$ and $r+2^{y}=5$. From this, we get $y=1$ and $x=-2$. Then $1+5 \cdot 2^{x}=\frac{9}{4}=\left(\frac{3}{2}\right)^{2}$

Thus, the desired numbers are $x=4$ and $x=-2$.

Problem 4. Given a quadrilateral $A B C D$ with angles $\alpha=60^{\circ}, \beta=90^{\circ}, \gamma=120^{\circ}$. The diagonals $\overline{A C}$ and $\overline{B D}$ intersect at point $S$, with $2|B S|=|S D|=2 d$. From the midpoint $P$ of diagonal $\overline{A C}$, a perpendicular $\overline{P M}$ is dropped to diagonal $\overline{B D}$, and from point $S$ a perpendicular $\overline{S N}$ is dropped to $\overline{P B}$.

Prove: (a) $|M S|=|N S|=\frac{d}{2}$;

(b) $|A D|=|D C|$;

(c) $P(A B C D)=\frac{9 d^{2}}{2}$.

Solution. Triangles $A B C$ and $A C D$ are right-angled with hypotenuse $\overline{A C}$, so quadrilateral $A B C D$ is cyclic. Point $P$ is the center of the circumscribed circle.

![](https://cdn.mathpix.com/cropped/2024_05_30_f40c78e39180d999fd44g-15.jpg?height=547&width=787&top_left_y=661&top_left_x=458)

(a) Point $M$ is the midpoint of diagonal $\overline{B D}$. From the given conditions, we have: $|D M|=|M B|=\frac{3 d}{2}$, $|M S|=\frac{d}{2}$.

Furthermore, from $\Varangle B P M=\frac{1}{2} \Varangle B P D=\Varangle B A D=60^{\circ}$ we get $\Varangle P B M=30^{\circ}$.

Since $N S \perp P B$, we have $|N S|=\frac{1}{2}|B S|=\frac{1}{2} d=|M S|$.

(b) Triangles $MPS$ and $NPS$ are right-angled with a common hypotenuse and congruent legs. Therefore, they are congruent and $\Varangle M P S=\Varangle S P N=30^{\circ}$, so $|S P|=2|N S|$.

Triangle $A P B$ is isosceles and $\Varangle P A B=\frac{1}{2} \Varangle S P B=15^{\circ}$.

Then $\Varangle D A C=\Varangle D A B-\Varangle C A B=60^{\circ}-15^{\circ}=45^{\circ}$. Therefore, triangle $A C D$ is an isosceles right triangle