Let m1, m2 be the slopes of two adjacent sides of a square of side a such that a² + 11a + 3(m1² + m2

Q: Wait, why is the distance from the vertex to the diagonal $a/\sqrt{2}$? Can't I just use the distance formula between two vertices?

You absolutely could, but that would require finding the coordinates of the other vertices, which is a nightmare of algebra! The property $p = a/\sqrt{2}$ is a geometric shortcut. In JEE, time is your currency. Always look for the geometric property that bypasses the coordinate calculation.

Q: Why did we get two values for $\alpha$ ($\pi/3$ and $\pi/6$)? Does it matter which one I pick?

Great observation! The expression we need to evaluate is $72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13$. Notice that $\sin^4 \alpha + \cos^4 \alpha$ is symmetric with respect to $\alpha$ and $\pi/2 - \alpha$. Whether you pick $\pi/3$ or $\pi/6$, the value of $\sin^4 \alpha + \cos^4 \alpha$ remains the same. The symmetry saves you again!

Q: I got $m_1^2 = 3$ or $1/3$. How do I know which one to use for $\tan^2 \alpha$?

Actually, both lead to the same result for the final expression. If $\tan^2 \alpha = 3$, then $\alpha = \pi/3$. If $\tan^2 \alpha = 1/3$, then $\alpha = \pi/6$. As mentioned before, the expression $72(\sin^4 \alpha + \cos^4 \alpha)$ is invariant under the swap of $\sin$ and $\cos$, so the final answer is unique regardless of your choice.

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 $\alpha \in [0, \pi/2]$ 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:

Options:

(A) $119$

(B) $128$

(C) $145$

(D) $155$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we stand before a problem that is not just a calculation, but a symphony of geometry and algebra. We are dealing with a square, a shape of perfect symmetry, placed on the coordinate plane.

To solve this, we must first unlock the secrets of the square's dimensions and orientation. We have a vertex $V$ and a diagonal $d_1$. The side of the square is $a$.

### The Geometric Insight

The perpendicular distance $p$ from any vertex to the opposite diagonal is given by the formula:

$$p = \frac{a}{\sqrt{2}}$$

Instead of trying to find the coordinates of all four vertices—which would be a tedious, error-prone journey—we use the point-to-line distance formula. We substitute the coordinates of vertex $V(10(\cos \alpha - \sin \alpha), 10(\sin \alpha + \cos \alpha))$ into the equation of the diagonal $d_1$:

$$(\cos \alpha - \sin \alpha)x + (\sin \alpha + \cos \alpha)y - 10 = 0$$

When we calculate this, the numerator simplifies beautifully. The terms involving $\sin 2\alpha$ cancel out, leaving us with a clean $10$. The denominator, the square root of the sum of squares of the coefficients, also simplifies to $\sqrt{2}$.

Thus, we find:

$$p = \frac{10}{\sqrt{2}} = 5\sqrt{2}$$

Equating this to $\frac{a}{\sqrt{2}}$, we find that the side length $a$ is exactly $10$. The geometry has spoken!

### The Algebraic Bridge

Now that we have $a=10$, the given equation $a^2 + 11a + 3(m_1^2 + m_2^2) = 220$ becomes our next target. Substituting $a=10$, we get:

$$100 + 110 + 3(m_1^2 + m_2^2) = 220$$

This simplifies to $210 + 3(m_1^2 + m_2^2) = 220$, which implies:

$$3(m_1^2 + m_2^2) = 10 \Rightarrow m_1^2 + m_2^2 = \frac{10}{3}$$

Here is where the soul of the square comes in: the adjacent sides are perpendicular. This means their slopes satisfy $m_1 m_2 = -1$. We can rewrite the sum of squares as:

$$m_1^2 + \left(-\frac{1}{m_1}ight)^2 = \frac{10}{3}$$

Let $x = m_1^2$. Then $x + \frac{1}{x} = \frac{10}{3}$, which leads us to the quadratic equation $3x^2 - 10x + 3 = 0$. Solving this, we find $x = 3$ or $x = \frac{1}{3}$. So, $m_1^2$ is either $3$ or $\frac{1}{3}$.

### The Trigonometric Finale

The slope of the diagonal is $m_d = 	an(\alpha - \frac{\pi}{4})$. Since the sides of a square make a $45^\circ$ angle with the diagonal, the slopes of the sides are $	an(\alpha - \frac{\pi}{4} \pm \frac{\pi}{4})$. This simplifies to $	an \alpha$ and $-\cot \alpha$.

Thus, $m_1^2 = 	an^2 \alpha = 3$ or $\frac{1}{3}$. For $\alpha \in [0, \frac{\pi}{2}]$, this gives us $\alpha = \frac{\pi}{3}$ or $\alpha = \frac{\pi}{6}$.

Finally, we calculate the expression $72(\sin^4 \alpha + \cos^4 \alpha) + a^2 - 3a + 13$. Using the identity $\sin^4 \alpha + \cos^4 \alpha = 1 - \frac{1}{2}\sin^2 2\alpha$, and substituting $\alpha = \frac{\pi}{3}$:

$$\sin^4 \alpha + \cos^4 \alpha = 1 - \frac{1}{2}\left(\frac{\sqrt{3}}{2}ight)^2 = 1 - \frac{3}{8} = \frac{5}{8}$$

Plugging everything into our final expression:

$$72\left(\frac{5}{8}ight) + 10^2 - 3(10) + 13 = 45 + 100 - 30 + 13 = 128$$

And there it is! The complexity melts away, leaving us with a clean, satisfying result. The final answer is **128**.

Select Answer:

Visualized Solution

Visualize the Square and Diagonal

Calculate Perpendicular Distance $p$

Simplify the Numerator

Simplify the Denominator

Find the Side Length $a$

Substitute $a$ into the Given Equation

Solve for $m_{1}^{2} + m_{2}^{2}$

Relate $m_{1}$ and $m_{2}$

Solve for $m_{1}^{2}$

Relate Slopes to $α$

Determine $α$

Calculate $sin^{4} α + cos^{4} α$

Final Evaluation

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Geometric Insight

The Algebraic Bridge

The Trigonometric Finale

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