QJEEM23Ma3c16: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3c16 Solution: Consider the triangles with vertices and . If the maximum and…

We have a triangle with vertices $A(2,1)$ , $B(0,0)$ , and $C(t, 4)$ , where $t$ belongs to the interval $[0,4]$ .

We want to find the maximum and minimum perimeters of such triangles, which occur at $t=\alpha$ and $t=\beta$ , respectively.

To find the minimum perimeter, we use a geometric approach. Reflect point $B$ over the line $y=4$ to get $B'(0,8)$ . The line segment $AB'$ intersects the line $y=4$ (which is the $y$ -coordinate of point $C$ ) at the point which gives the minimum perimeter.


The slope of $AB'$ is :

$m_{AB'} = \frac{8 - 1}{0 - 2} = \frac{-7}{2}$

The equation of the line $AB'$ is then $y - 1 = m_{AB'}(x - 2)$ , or $7x + 2y = 16$ .

Solving this equation for $y = 4$ yields $x = \frac{8}{7}$ . So, the minimum perimeter is achieved at point $C\left(\frac{8}{7}, 4\right)$ , so $\beta = \frac{8}{7}$ .

For the maximum perimeter, we notice that it will be achieved when point $C$ is either at $(0,4)$ or at $(4,4)$ , since these are the furthest points from $A(2,1)$ within the range of $t$ . By calculating the perimeters at these points, we find that the maximum perimeter is achieved at $\alpha = 4$ .

Finally, we calculate $6\alpha + 21\beta = 6 \cdot 4 + 21 \cdot \frac{8}{7} = 24 + 24 = 48$ .

Therefore, $6\alpha + 21\beta = 48$ .