Consider the triangles with vertices A(2,1),B | vedclass

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(A) Let the vertices be $A(2,1)$,$B(0,0)$,and $C(t,4)$ for $t \in [0,4]$. The perimeter $P(t) = AB + BC + AC$. Since $AB = \sqrt{2^2 + 1^2} = \sqrt{5}

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$48$

Vedclass · Answer

(A) Let the vertices be $A(2,1)$,$B(0,0)$,and $C(t,4)$ for $t \in [0,4]$. The perimeter $P(t) = AB + BC + AC$. Since $AB = \sqrt{2^2 + 1^2} = \sqrt{5}$ is constant,we need to maximize/minimize $f(t) = BC + AC = \sqrt{t^2 + 4^2} + \sqrt{(t-2)^2 + (4-1)^2} = \sqrt{t^2 + 16} + \sqrt{(t-2)^2 + 9}$.To find the minimum,we reflect $B(0,0)$ across the line $y=4$ to get $B'(0,8)$. The minimum occurs when $A, C, B'$ are collinear. The line $AB'$ passes through $(2,1)$ and $(0,8)$,so its equation is $y - 8 = \frac{1-8}{2-0}(x - 0) \implies y = -\frac{7}{2}x + 8$. Setting $y=4$,we get $4 = -\frac{7}{2}t + 8 \implies \frac{7}{2}t = 4 \implies t = \frac{8}{7}$. Thus,$\beta = \frac{8}{7}$.To find the maximum,we examine the endpoints of the interval $t \in [0,4]$. $f(0) = \sqrt{16} + \sqrt{4+9} = 4 + \sqrt{13} \approx 7.6$. $f(4) = \sqrt{16+16} + \sqrt{4+9} = 4\sqrt{2} + \sqrt{13} \approx 5.65 + 3.6 = 9.25$. Since $f(t)$ is a convex function,the maximum occurs at one of the endpoints. Comparing $f(0)$ and $f(4)$,the maximum is at $t=4$. Thus,$\alpha = 4$.Finally,$6\alpha + 21\beta = 6(4) + 21(\frac{8}{7}) = 24 + 3(8) = 24 + 24 = 48$.

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

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