Question

The isosceles triangle below has height AQ of length 3 and base BC of length 2. A point P may be placed anywhere along the line segment AQ. What is the minimum value of the sum of the lengths of AP, BP, and C

Answer 1

The question is missing parts. Here is the complete question.

The isosceles triangle below has height AQ of length 3 and base BC of length 2. A point P may be placed anywhere along the line segment AQ.

What is the minimum value of the sum of the lengths of AP, BP and CP?

Answer: The sum is 4.73.

Step-by-step explanation: Height of a triangle is a perpendicualr line linking a vertex and its opposite side.

Because triangle ABC is isosceles, point Q divides the base in 2 equal parts:

BQ = CQ = 1

Suppose QP = x

To calculate minimum value of the sum:

AP = AQ - QP

AP = 3 - x

Since triangles BQP and CQP are congruent and right triangles, use Pythagorean Theorem to figure out the value of BP and CP:

BP = CP = $\sqrt{BQ^{2}+PQ^{2}}$

BP = $\sqrt{1+x^{2}}$

Then, sum of AP, BP and CP is

$f(x)=3-x+2\sqrt{1+x^{2}}$

The minimum value is calculated using first derivative:

$f'=-1+\frac{2x}{\sqrt{x^{2}+1} }$

The value of x is limited: it can assume value of 0, when P=A and x=3, when P=Q. So, interval is [0,3].

x has value:

$-1+\frac{2x}{\sqrt{x^{2}+1} }=0$

$2x=\sqrt{x^{2}+1}$

$4x^{2}-x^{2}-1=0$

$3x^{2}=1$

x = ± $\frac{1}{\sqrt{3} }$

x can't assume negative value because is not in the interval:

x = $\frac{1}{\sqrt{3} }$

To find the minimum value of the sum, substitute x in the function above:

f($\frac{1}{\sqrt{3} }$)=$3-\frac{1}{\sqrt{3} } +2\sqrt{1+(\frac{1}{\sqrt{3} })^{2} }$

$f(\frac{1}{\sqrt{3}} )=3-\frac{1}{\sqrt{3}}+2(\sqrt{\frac{4}{3} } )$

$f(\frac{1}{\sqrt{3}} )=3-\frac{1}{\sqrt{3}} +\frac{4}{\sqrt{3}}$

$f(\frac{1}{\sqrt{3}} )=3+\frac{3}{\sqrt{3} }$

$f(\frac{1}{\sqrt{3}} )=4.73$

The minimum value of the sum of AP, BP and CP is 4.73.

Answer 2

The minimum value of the sum of the lengths of AP, BP and CP is $3 + \frac{\sqrt{3}}{3}$ units.

According to this statement we must determine the sum of the line segment lengths so that a minimum is found. By Pythagorean theorem and given data we have the following expression:

$y = AP + BP + CP$

$y = 3-x +2\sqrt{x^{2}+1}$ (1)

Now we proceed to find the critical values by performing first and second derivative tests.

FDT

$-1 +\frac{2\cdot x}{\sqrt{x^{2}+1}} = 0$

$2\cdot x = \sqrt{x^{2}+1}$

$4\cdot x^{2}-x^{2}-1=0$

$3\cdot x^{2} = 1$

$x^{2} = \frac{1}{3}$

$x = \frac{\sqrt{3}}{3}$

SDT

$y'' = \frac{2\cdot \sqrt{x^{2}+1}-2\cdot x \cdot \left(\frac{2\cdot x}{\sqrt{x^{2}+1}} \right)}{x^{2}+1}$

$y'' = \frac{2\cdot x^{2}+2-4\cdot x^{2}}{(x^{2}+1)^{3/2}}$

$y'' = 2\cdot \frac{1-x^{2}}{(x^{2}+1)^{3/2}}$

$y'' \approx 0.866$

A minimum exists when $y'' > 0$, then we conclude that $x = \frac{\sqrt{3}}{3}$ lead to a relative minimum. And by (1) we have the minimum sum:

$y = 3-\frac{\sqrt{3}}{3}+2\sqrt{\frac{4}{3} }$

$y = 3 - \frac{\sqrt{3}}{3} +\frac{4\sqrt{3}}{3}$

$y = 3+\frac{\sqrt{3}}{3}$

The minimum value of the sum of the lengths of AP, BP and CP is $3 + \frac{\sqrt{3}}{3}$ units.

Nota - The statement reports typographical issues, the correct form is presented below:

The isosceles triangle below has height AQ of length 3 and base BC of length 2. A point P may be placed anywhere along the line segment AQ. What is the minimum value of the sum of the lengths of AP, BP and CP.

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