5. The sum of the squares of the lengths of the sides of triangle $ABC$ is 36, and its area is $5 \sqrt{3}$.

Points $P, Q$, and $R$ are the vertices of isosceles triangles with an angle of $120^{\circ}$, constructed on the sides of triangle $ABC$ as bases (externally). Find the lengths of the sides of triangle $PQR$.

## Answers and Solutions

Problem 1 Answer: $s=2 \sqrt{3}+6 \sqrt{7}$
Solution

Given $x \neq 0, y \neq 0$. Divide the left and right sides of the equations: $\frac{y^{3}}{x^{3}}=\frac{2 x+5 y}{5 x+2 y}=\frac{2+5 y / x}{5+2 y / x}$.

Substitution: $t=\frac{y}{x} \rightarrow t^{3}=\frac{5 t+2}{2 t+5} \rightarrow 2 t^{4}+5 t^{3}-5 t-2=0, t \neq-\frac{5}{2}$.

Factorize the polynomial: $(t-1)(t+1)\left(2 t^{2}+5 t+2\right)=(t-1)(t+1)(2 t+1)(t+2)=0$.

Case 1. $t=1 \rightarrow\left\{\begin{array}{l}x^{3}=5 x+2 y \\ y=x, x \neq 0\end{array} \rightarrow\left\{\begin{array}{l}x^{2}=7 \\ y=x\end{array} \rightarrow\left\{\begin{array}{c}x= \pm \sqrt{7} \\ y=x\end{array} .(A, G)\right.\right.\right.$

Case 2. $t=-1 \rightarrow\left\{\begin{array}{l}x^{3}=5 x+2 y \\ y=-x, x \neq 0\end{array} \rightarrow\left\{\begin{array}{l}x^{2}=3 \\ y=-x\end{array} \rightarrow\left\{\begin{array}{c}x= \pm \sqrt{3} \\ y=-x\end{array} .(D, F)\right.\right.\right.$

Case 3. $t=-2 \rightarrow\left\{\begin{array}{c}x^{3}=5 x+2 y \\ y=-2 x, x \neq 0\end{array} \rightarrow\left\{\begin{array}{c}x^{2}=1 \\ y=-2 x\end{array} \rightarrow\left\{\begin{array}{c}x= \pm 1 \\ y=-2 x\end{array} .(B, E)\right.\right.\right.$

Case 4. $t=-\frac{1}{2} \rightarrow\left\{\begin{array}{c}y^{3}=2 x+5 y \\ x=-2 y, y \neq 0\end{array} \rightarrow\left\{\begin{array}{c}y^{2}=1 \\ x=-2 y\end{array} \rightarrow\left\{\begin{array}{l}x={ }^{\mp} 2 \\ y= \pm 1\end{array} .(C, H)\right.\right.\right.$

The figure is a polygon with such vertices:

![](https://cdn.mathpix.com/cropped/2024_05_06_af395609edf9f7165024g-2.jpg?height=571&width=597&top_left_y=334&top_left_x=638)

The figure is symmetric with respect to the lines $y=x$ and $y=-x$. The polygon $B C F D$ is a trapezoid, $A B C$ is an isosceles triangle, $A B=A C$. The coordinates of point $A(\sqrt{7} ; \sqrt{7}) \rightarrow A O=\sqrt{14}$, The coordinates of points $B(-1 ; 2)$ and $C(2 ;-1)$, the equation of the line $B C: x+y=1$, the length $B C=3 \sqrt{2}$. The height of the trapezoid $O K=\frac{1}{\sqrt{2}}$. The length of the base $D F=2 \sqrt{6}$. The area of the trapezoid $S_{\text {BCFD }}=\frac{3+2 \sqrt{3}}{2}$.

The height of the triangle $A B C$ is $A K=A O-O K=\sqrt{14}-\frac{1}{\sqrt{2}}=\frac{2 \sqrt{7}-1}{\sqrt{2}}$.

The area $S_{A B C}=\frac{1}{2} \cdot \frac{2 \sqrt{7}-1}{\sqrt{2}} \cdot 3 \sqrt{2}=\frac{3(2 \sqrt{7}-1)}{2}$. Half the area of the polygon is:

$\frac{S}{2}=S_{A B C}+S_{B C F D}=\frac{3}{2}(2 \sqrt{7}-1)+\frac{3+2 \sqrt{3}}{2} \rightarrow S=6 \sqrt{7}+2 \sqrt{3}$

Problem 2 Answer: 1) $M(7 ; 12 ; 17), \quad N(2 ; 6 ; 18) \quad 2) M(-9 ; 12 ; 33), \quad N(18 ; 6 ; 2)$

Solution

Let $M\left(a_{1} ; a_{2} ; a_{3}\right), \quad N\left(b_{1} ; b_{2} ; b_{3}\right)$ be the given points. The conditions of the problem can be written in the form of a system:

$$
\begin{aligned}
& a_{1}+b_{1}=9 \\
& a_{2}+b_{2}=18 \rightarrow b_{2}=6 \\
& a_{3}+b_{3}=35 \\
& a_{1}+a_{2}+a_{3}=36 \rightarrow a_{2}=12 \rightarrow b_{1}+b_{2}+b_{3}=9+18+35-36=26 \rightarrow\left\{\begin{array}{l}
b_{1}+b_{3}=20 \\
b_{1} \cdot b_{3}=36
\end{array}\right. \\
& a_{1}+a_{3}=2 a_{2} \\
& b_{1}=18, b_{3}=2 \\
& b_{1} \cdot b_{3}=b_{2}^{2} \\
& b_{1}=2, b_{3}=18
\end{aligned}
$$

First case $b_{1}=18, b_{2}=6, b_{3}=2 \rightarrow a_{1}=-9, a_{2}=12, a_{3}=33$.

Second case $b_{1}=2, b_{2}=6, b_{3}=18 \rightarrow a_{1}=7, a_{2}=12, a_{3}=17$

Problem 3 Answer: $T=40 \pi, a=4$.

Solution Let $a=\frac{16 x^{2}+\left(x^{2}-1\right)^{2}}{\left(x^{2}+1\right)^{2}}$ be the value of the expression for some $x$.

Then this $x$ is a solution to the biquadratic equation $a\left(x^{2}+1\right)^{2}=16 x^{2}+\left(x^{2}-1\right)^{2} \rightarrow(a-1) x^{4}+2(a-7) x^{2}+a-1=0$. Its discriminant with respect to the variable $t=x^{2}$ is $D / 4=(a-7)^{2}-(a-1)^{2}=-6(2 a-8) \geq 0 \rightarrow a \leq 4$. The equation has positive solutions for $t$ if $\frac{a-7}{a-1} \leq 0 \rightarrow a \in(1 ; 4]$ i.e., the maximum value of $a$ is 4.

The smallest period of the function $\sin \frac{4 x}{5}$ is $\frac{5 \pi}{2}$, and the period of $\cos \frac{7 x}{4}$ is $\frac{8 \pi}{7}$.

The period of their sum $T=\frac{5 \pi m}{2}=\frac{8 \pi k}{7} \rightarrow 35 m=16 \mathrm{k} \rightarrow m=16 t, k=35 t, t \in Z$, i.e., $T=40 \pi t$ and the smallest period $T=40 \pi$.

Problem 4 Answer: 1) $x_{1}=a, x_{2}=-2, x_{3}=-a / 2$
2) $a=-2, a=0, a=4$

Problem 5 Answer: $P Q=P R=Q R=4$

Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_af395609edf9f7165024g-3.jpg?height=531&width=577&top_left_y=1002&top_left_x=157)

$\mathrm{B} \square_{P Q C} \quad P C=\frac{a}{\sqrt{3}}, Q C=\frac{b}{\sqrt{3}}, \quad \square_{P C Q}=\alpha+60^{\circ}$

By the cosine rule for $\square P Q C$ :

$P Q^{2}=\frac{a^{2}}{3}+\frac{b^{2}}{3}-\frac{2 a b}{3} \cos \left(\alpha+60^{\circ}\right)=\frac{a^{2}}{3}+\frac{b^{2}}{3}-\frac{2 a b}{3}\left(\frac{1}{2} \cos \alpha-\frac{\sqrt{3}}{2} \sin \alpha\right)=$

$=\frac{a^{2}}{3}+\frac{b^{2}}{3}-\frac{a b}{3} \cos \alpha+\frac{2 \sqrt{3}}{3} S_{A B C}$

Here $S_{A B C}$ is the area of ${ }^{\circ} A B C$.

By the cosine rule for $\square A B C: c^{2}=a^{2}