Geometric Optimization

What you need to know

• Triangle Inequality. Any two sides must add up to more than the third, strictly. It is enough to check the two shorter sides against the longest: $1$, $3$, $10$ fails because $1+3<10$, even though $1+10>3$. If the two shorter sides add up to exactly the longest, the "triangle" is flat. Example: with sides $5$ and $8$, the third side is strictly between $3$ and $13$. So the largest or smallest possible length is often only approached, as the triangle flattens into a line.
• Shortest path by reflection. Let $A$ and $B$ be on the same side of a line $\ell$. To make $AP + PB$ smallest with $P$ on $\ell$, reflect $B$ across $\ell$ to get $B'$. Since $PB = PB'$, the minimum is the straight distance $AB'$, reached where $AB'$ crosses $\ell$. Example: for $A = (0,3)$, $B = (8,3)$ and $\ell$ the $x$-axis, $B' = (8,-3)$, so the shortest path is $AB' = \sqrt{8^2+6^2} = 10$, through $P = (4,0)$. (If $A$ and $B$ are on opposite sides, no reflection is needed: $AB$ crosses $\ell$.)
• Largest triangle with two given sides. With sides $a$ and $b$, the area is at most $\frac12 ab$, reached when the angle between them is $90^\circ$. Why: with base $a$, the height is at most $b$ (in trig terms, $\frac12 ab\sin C \le \frac12 ab$). Example: sides $6$ and $8$ give area at most $24$.

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 the surface flat and compare a few nets.
• The largest number of crossing points of lines, circles, and polygons. You need an upper bound (why it can't be more) and a construction (a drawing that reaches it).

Standard approaches

1. Reflect a fixed point across the line the path must touch (or unfold the surface), so the path becomes a straight segment.
2. Push the figure to its extreme case (points in a line, a right angle, a tangent), then check that the problem allows it.
3. For "largest number of" questions, find the most each pair of shapes can contribute, then draw an example that reaches the total. Example: for $3$ circles and $1$ line, each of the $6$ pairs crosses at most twice, so there are at most $12$ points, and $12$ can be drawn.

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. Step 1. Reflect $A$ across $\ell$ to get $A'$, $2$ units below $\ell$. For every $P$ on $\ell$, $AP = A'P$, so $AP + PB = A'P + PB \ge A'B$.

Step 2. $A'$ and $B$ are $2 + 6 = 8$ apart vertically and $15$ apart horizontally, so $A'B = \sqrt{8^2+15^2} = 17$.

Step 3. $P$ is where $A'B$ crosses $\ell$. The right triangles on either side of $P$ (heights $2$ and $6$) are similar, so $P$ splits the $15$ in the ratio $2:6 = 1:3$. That puts $P$ at $\frac{15}{4}$ from the foot of $A$.

Step 4. $AP$ is the hypotenuse of a right triangle with legs $2$ and $\frac{15}{4}$:
$$AP = \sqrt{2^2 + \left(\tfrac{15}{4}\right)^2} = \sqrt{\tfrac{64}{16} + \tfrac{225}{16}} = \sqrt{\tfrac{289}{16}} = \tfrac{17}{4}.$$
The answer is $\boxed{\textbf{(C)}\ \dfrac{17}{4}}$.

Pitfalls

• Reflecting both points (which gives back the same problem), or reflecting the wrong one.
• Giving the whole minimum ($17$ above) when the problem asks for one piece ($AP$).
• Forgetting the construction in a "largest number" problem. Showing it can't be more than $12$ is only half; you must also show $12$ can happen.

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)