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

Q: Wait, why do we reflect $B$ and not $A$? Does it change the result?

That is a fantastic question! The reflection principle works regardless of which point you reflect. You could reflect $A(2, 1)$ across $y = 4$ to get $A'(2, 7)$ and then find the line connecting $A'$ and $B$. You would arrive at the exact same value for $\beta$. We reflect $B$ simply because its coordinates are $(0, 0)$, making the reflection $B'(0, 8)$ very easy to calculate. It is all about choosing the path of least resistance!

Q: Why can't we just use calculus to find the maximum? Can't we just take the derivative and set it to zero?

You absolutely could! However, if you take the derivative of $f(t) = \sqrt{t^2 + 16} + \sqrt{(t-2)^2 + 9}$, you will find that the derivative is never zero within the interval $(0, 4)$. This is because the function is convex. In a convex function, the 'turning point' is a minimum, not a maximum. Therefore, the maximum must occur at the boundaries of the domain. Always check the nature of your function before diving into derivatives!

Q: How do I know for sure that the maximum is at $t = 4$ and not $t = 0$?

Great attention to detail! We simply evaluate the function $f(t) = BC + CA$ at both endpoints. At $t = 0$, $f(0) = \sqrt{0^2 + 4^2} + \sqrt{(0-2)^2 + (4-1)^2} = 4 + \sqrt{4 + 9} = 4 + \sqrt{13}$. At $t = 4$, $f(4) = \sqrt{4^2 + 4^2} + \sqrt{(4-2)^2 + (4-1)^2} = \sqrt{32} + \sqrt{4 + 9} = 4\sqrt{2} + \sqrt{13}$. Since $4\sqrt{2} \approx 5.65$ is greater than $4$, the value at $t = 4$ is clearly larger. Comparing the two values is the most robust way to be certain.

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 __________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have two anchors: point $A$ at $(2, 1)$ and point $B$ at the origin $(0, 0)$. These are fixed, unmoving.

Point $C$ is a wanderer, restricted to the horizontal line $y = 4$, with its $x$-coordinate $t$ trapped between $0$ and $4$. As $C$ slides along this line, the triangle $ABC$ changes shape, stretching and compressing.

### The Perimeter Function

The perimeter $P(t)$ is defined as $AB + BC + CA$. The distance between $(2, 1)$ and $(0, 0)$ is:

$$AB = \sqrt{2^2 + 1^2} = \sqrt{5}$$

This is a constant. To maximize or minimize the perimeter, we only need to focus on the variable part: $f(t) = BC + CA$.

### The Reflection Principle

To minimize $BC + CA$, we use the reflection principle. Imagine the line $y = 4$ is a mirror. If we reflect point $B(0, 0)$ across this mirror, we get a new point $B'(0, 8)$.

By the laws of reflection, the distance $BC$ is exactly equal to $B'C$. Now, our sum $BC + CA$ becomes $B'C + CA$.

The shortest path between two points, $B'$ and $A$, is a straight line. Thus, the minimum occurs when $B'$, $C$, and $A$ are perfectly collinear.

### Finding the Minimum

With the points $B'(0, 8)$ and $A(2, 1)$ identified, we find the slope $m$ of the line connecting them:

$$m = \frac{1 - 8}{2 - 0} = -\frac{7}{2}$$

Using the point-slope form, the equation of the line is $y - 8 = -\frac{7}{2}(x - 0)$. Since point $C$ lies on this line and has a $y$-coordinate of $4$, we substitute $y = 4$:

$$4 - 8 = -\frac{7}{2}\beta \Rightarrow -4 = -\frac{7}{2}\beta$$

Solving for $\beta$, we get $\beta = \frac{8}{7}$. This is the exact moment of minimum perimeter.

### The Maximum

The function $f(t) = \sqrt{t^2 + 16} + \sqrt{(t-2)^2 + 9}$ represents the sum of distances to two fixed points. This is a convex function, which behaves like a valley and does not have a peak in the middle.

The highest points must be at the edges of the domain $[0, 4]$. We test the endpoints:

At $t = 0$:
$$f(0) = 4 + \sqrt{13}$$

At $t = 4$:
$$f(4) = \sqrt{4^2 + 4^2} + \sqrt{(4-2)^2 + 3^2} = 4\sqrt{2} + \sqrt{13}$$

Since $4\sqrt{2} > 4$, the maximum is clearly at $t = 4$. Thus, $\alpha = 4$.

### The Grand Finale

We have $\alpha = 4$ and $\beta = \frac{8}{7}$. The problem asks for $6\alpha + 21\beta$. Substituting our values:

$$6(4) + 21\left(\frac{8}{7}\right) = 24 + 3(8) = 24 + 24 = 48$$

The final result is **48**.

Animated Solution for Mathematics - Straight Lines: 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 $t = α$ and $t = β$ respectively, then $6 α + 21 β$ is equal to __________.

Enter Numerical Value:

Visualized Solution

Setting up the Coordinate Geometry

The Perimeter Function

Minimizing $B C + C A$

Collinearity for Minimum

Equation of Line $B^{'} A$

Finding $β$ (Minimum Perimeter)

Maximizing $B C + C A$

Evaluating Endpoints

Final Calculation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Perimeter Function

The Reflection Principle

Finding the Minimum

The Maximum

The Grand Finale

Similar Questions

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Enter Numerical Value:

Visualized Solution

Setting up the Coordinate Geometry

The Perimeter Function

Minimizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC+CA

Collinearity for Minimum

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B′A

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β (Minimum Perimeter)

Maximizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC+CA

Evaluating Endpoints

Final Calculation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Perimeter Function

The Reflection Principle

Finding the Minimum

The Maximum

The Grand Finale

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0,0),(0,21) and (21,0), is

Minimizing $B C + C A$

Equation of Line $B^{'} A$

Finding $β$ (Minimum Perimeter)

Maximizing $B C + C A$

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