Geometric Optimization

What you need to know

• Triangle inequality: $a + b > c$ strictly; the extreme value of a length is often approached as the triangle degenerates to a line.
• Shortest path via reflection: to minimize $AP + PB$ for $P$ on line $\ell$, reflect $B$ over $\ell$ to $B'$; the minimum is $AB'$, attained where $\overline{AB'}$ crosses $\ell$.
• For two given sides, the right triangle has maximum area since $\frac12 ab\sin C \le \frac12 ab$.

How AMC 10 tests it

• "Largest possible area of a triangle with sides $a$ and $b$?" or "...with integer sides and perimeter $n$?" (problems 8–15).
• A spider or ant on a box or cylinder; unfold and compare a few nets.
• Maximum number of intersection points of lines, circles, and polygons; the answer needs an upper bound and a construction.

Standard approaches

1. Reflect a fixed point over the constraint line (or unfold the surface) so the path becomes a straight segment.
2. Push the configuration to its degenerate extreme (collinear, right angle, tangent) and check that it is allowed.
3. For "maximum number of" questions, bound each pair's contribution, then build an example achieving the bound.

Worked example

Points $A$ and $B$ lie on the same side of line $\ell$, at distances $2$ and $6$ from $\ell$, and the feet of the perpendiculars from $A$ and $B$ to $\ell$ are $15$ units apart. Point $P$ on $\ell$ minimizes $AP + PB$. What is $AP$?

(A) $\dfrac{15}{4}$ (B) $4$ (C) $\dfrac{17}{4}$ (D) $\dfrac92$ (E) $5$

Solution. Reflect $A$ over $\ell$ to $A'$, which is $2$ units below $\ell$. For any $P$ on $\ell$, $AP = A'P$, so $AP + PB = A'P + PB \ge A'B$. The vertical separation of $A'$ and $B$ is $2 + 6 = 8$ and the horizontal separation is $15$, so $A'B = 17$. By similar triangles, $P$ splits the horizontal $15$ in ratio $2:6$, so $P$ is $\frac{15}{4}$ from the foot of $A$. Hence
$$AP = \sqrt{2^2 + \left(\tfrac{15}{4}\right)^2} = \sqrt{\tfrac{289}{16}} = \tfrac{17}{4}.$$
The answer is $\boxed{\textbf{(C)}\ \dfrac{17}{4}}$.

Pitfalls

• Reflecting both points, which leaves the path unchanged, or reflecting the wrong one.
• Reporting the total minimum ($17$) when the problem asks for a piece of it.
• Forgetting the construction half of a "maximum number" argument.

Traps that recur

• Picking choice (E), which needs 5 space diagonals, without checking that a corner can be the endpoint of only one space diagonal. (2010 AMC 10A #20)
• Assuming a triangle with the given perimeter and area exists and answering 2 or 4 by symmetry, without checking whether height 20 is reachable. (2019 AMC 10B #10)
• Answering 13 by allowing the longest side to equal the sum of the other three, which gives a degenerate flat figure. (2023 AMC 10A #4)
• Stacking the rectangles end to end (3 + 4 = 7) and answering 49, without trying the side-by-side arrangement. (2006 AMC 10B #5)