Find the minimum of $\overline{AP}+\overline{AQ}$.

Question

My niece asked me a math question: There is a point $A(-3,2)$ on the $x$-$y$ plane. A line $L$ is perpendicular to the line $x=y$. Line $L$ and $y=f(x)=2^x$ has an intersection point $P$. Line $L$ and $y=g(x)=\log_2 x$ has an intersection point $Q$. Find the minimum of $\overline{AP}+\overline{AQ}$.

Note: My niece is a freshman in a senior high school. She have not learned calculus. Therefore, I can't use calculus to solve this question.

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The following is my try. First, let the coordinates of $P$ be $(a,2^a)$. Because $f(x)$ and $g(x)$ are inverse functions of each other, the coordinates of $Q$ is $(2^a,a)$. Then $$ \overline{AP}+\overline{AQ} = \sqrt{(-3-a)^2+(2-2^a)^2} + \sqrt{(-3-2^a)^2+(2-a)^2}. \tag{1} $$ Because the two terms in $(1)$ are both positive numbers, we can use the inequality of arithmetic and geometric means (AM-GM inequality) to obtain $$ \begin{align} \overline{AP}+\overline{AQ} & \geq 2\sqrt{ \sqrt{(-3-a)^2+(2-2^a)^2} \cdot \sqrt{(-3-2^a)^2+(2-a)^2}} \\ & = 2 \{ [ (-3-a)^2+(2-2^a)^2][(-3-2^a)^2+(2-a)^2]\}^{1/4} \\ & = 2 [(a^2+6a+9+4-4\cdot 2^a+2^{2a})(2^{2a}+6\cdot 2^a+9+a^2-4a+4)]^{1/4} \\ & = 2 [(2^{2a}-2^{a+2}+a^2+6a+13)(a^{2a}+3\cdot 2^{a+1}+a^2-4a+13)]^{1/4}. \tag{2} \end{align} $$ I don't know how to continue.

Answer 1

Sorry, but AM-GM is unable to be applied in this case as $\overline{AP} \neq \overline {AQ}$ for all $P$, $Q$.

However, the triangle inequality can be applied nicely in this case. First, we must reformulate the points.

Let us name the points $A(-3,2)$, $B(2,-3)$, $P(a,2^a)$, $Q(2^a,a)$, Note that $$ \overline{AP}+\overline{AQ}=\sqrt{(2^a-2)^2+(a+3)^2} + \sqrt{(2^a+3)^2+(a-2)^2}=\overline{BP}+\overline{AP} $$

As seen in equation $(1)$ in your question. So the question becomes equivalent to minimizing the value of $\overline{BP}+\overline{AP}$. However, we have that $$\overline{BP}+\overline{AP} \ge \overline{AB}=5 \sqrt{2}$$ From the triangle inequality. Note the equality is true when $P$ is the intersection of $x+y=-1$ and $y=2^{x}$. The existance of this intersection can be confirmed through IVT.

As @amd has guessed, this is when $A, P, Q$ are colinear.