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Problem 9

Algebra Difficulty 1.0 Prove it CEMC Euclid · Canada · 2024

If $x=2$ , what is the value of $\dfrac{x^4 + 3x^2}{x^2}$ ?
In the diagram, $\triangle ABC$ is right-angled at $B$ . Also, $AB=10$ , $BC=t-1$ , and $AC=t+1$ . What is the value of $t$ ?

Suppose that $\dfrac{2}{y} + \dfrac{3}{2y} = 14$ .
Determine the value of $y$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

For every $x \neq 0$ , we note
that $\text{x 4}$ + 3x^2}{x^2} = x^2 +
3$.

Therefore, when $x = 2$ , we have
$\text{x 4}$ + 3x^2}{x^2} = x^2 + 3 = 2^2 + 3
= 7$.

Alternatively, when $x = 2$ , we have
$\text{x 4}$ + 3x^2}{x^2} = $\text{2 4}$ + 3
$\cdot$ 2^2}{2^2} = $\dfrac{28}{4}$ = 7$.
By the Pythagorean Theorem in $\triangle ABC$ , we have $\begin{align*} AC^2 & = AB^2 + BC^2 \\ (t+1)^2 & = 10^2 + (t-1)^2 \\ t^2 + 2t + 1 & = 100 + t^2 - 2t + 1 \\ 4t & = 100\end{align*}$ and so $t = 25$ .

Alternatively, we could remember the Pythagorean triple $5$ - $12$ - $13$ and scale this triple by a factor of
$2$ to obtain the Pythagorean triple
$10$ - $24$ - $26$ , noting that the difference between
$t+1$ and $t-1$ is $2$ as is the difference between $26$ and $24$ , which gives $t + 1 = 26$ and so $t = 25$ .
Since $\$\dfrac{2}{y} + \dfrac{3}{2y}$ =
14 $, then$ $\dfrac{4}{2y} + \dfrac{3}{2y}=14$ $or$ $\dfrac{7}{2y}$ =
14$.

Therefore, $2y = $\dfrac{7}{14} = \dfrac{1}{2}$ $and so$ y =
$\dfrac{1}{4}\$.$

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.