Consider the triangles with vertices A(2,1), B(0,0) and C(t, 4), t [0,4].

Question

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 \_\_\_\_\_\_...

Answer: 48

Accepted Answer

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 $A{B}^{'}$ intersects the line $y=4$ (which is the $y$-coordinate of point $C$) at the point which gives the minimum perimeter.

The slope of $A{B}^{'}$ is :

${m}_{A{B}^{'}}=\frac{8-1}{0-2}=\frac{-7}{2}$

The equation of the line $A{B}^{'}$ is then $y-1={m}_{A{B}^{'}}(x-2)$, or $7x+2y=16$.

Solving this equation for $y=4$ yields $x=\frac{8}{7}$. So, the minimum perimeter is achieved at point $C(\frac{8}{7},4)$, 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, $6\alpha +21\beta =48$.

Solutions · 1