Let m_1, m_2 be the slopes of two adjacent sides of a square of side a such that a^2+11a+3…

Question

Let $m_1, m_2$ be the slopes of two adjacent sides of a square of side $a$ such that $a^2+11a+3(m_1^2+m_2^2)=220$. If one vertex of the square is $(10(\cos\alpha-\sin\alpha),10(\sin\alpha+\cos\alpha))$, where $\alpha\in\left(0,\dfrac{\pi}2\right)$ and the equation of one diagonal is $(\cos\alpha-\sin\alpha)x+(\sin\alpha+\cos\alpha)y=10$, then $72(\sin^4\alpha+\cos^4\alpha)+a^2-3a+13$ is equal to:

Accepted Answer

Since $m_1m_2=-1$ (adjacent sides of a square are perpendicular), and the diagonal $BD$ passes through the origin with slope related to $\alpha$: slope of $AC = \dfrac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}=\tan\theta = M$.

The two sides make $45^\circ$ with this diagonal, giving $m_1,m_2 = \tan\alpha,\cot\alpha$ (as derived in the source).

Midpoint of the given diagonal and vertex relations give $a = AB = 10$ (side of the square).

$$a^2+11a+3(m_1^2+m_2^2)=220 \implies 100+110+3(\tan^2\alpha+\cot^2\alpha)=220$$
$$\tan^2\alpha+\cot^2\alpha = \frac{10}3$$

Solving: $\tan^2\alpha=3$ or $\dfrac13$, giving $\alpha=\dfrac{\pi}3$ or $\dfrac{\pi}6$.

$$72(\sin^4\alpha+\cos^4\alpha)+a^2-3a+13 = 72\left(\frac9{16}+\frac1{16}\right)+100-30+13$$
$$=72\cdot\frac58+83 = 45+83=128$$

Solutions · 1