JEE MAIN - Mathematics (2023 - 15th April Morning Shift - No. 19)
Consider the triangles with vertices $A(2,1), B(0,0)$ and $C(t, 4), t \in[0,4]$.
If the maximum and the minimum perimeters of such triangles are obtained at
$t=\alpha$ and $t=\beta$ respectively, then $6 \alpha+21 \beta$ is equal to ___________.
If the maximum and the minimum perimeters of such triangles are obtained at
$t=\alpha$ and $t=\beta$ respectively, then $6 \alpha+21 \beta$ is equal to ___________.
Answer
48
Explanation
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.
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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, $6\alpha + 21\beta = 48$.
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, $6\alpha + 21\beta = 48$.
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