Choosing Coordinates That Kill the Problem

A square has side length 4. A point $P$ is chosen so that its distances to two adjacent corners of the square are 3 and 5. What is the distance from $P$ to the corner diagonally opposite the one nearest it — or, more concretely, can $P$ even exist, and where?

Before reading on, notice your instinct: are you reaching for the Pythagorean theorem in your head, or trying to draw it and eyeball an answer? The point of this page is to replace both instincts with one deliberate first move.

The one decision that matters most

Coordinate geometry turns a picture into algebra, but the same picture can be assigned coordinates in infinitely many ways — and almost all of them produce messy expressions. The single highest-leverage decision in the entire problem is where to put the origin and how to orient the axes, made before writing a single equation. Good placements share a pattern: put the origin at a point with the most constraints touching it (a right angle, a center of symmetry, a shared vertex), and align an axis along a side or a line of symmetry so that as many coordinates as possible become 0.

For the square puzzle: put the corner nearest $P$ at the origin, with the two sides of the square running along the positive $x$- and $y$-axes. That single choice makes both adjacent corners land on the axes with clean coordinates — corner $A = (0,0)$, corner $B = (4,0)$, corner $D = (0,4)$, and the far corner $C = (4,4)$ — instead of some arbitrary rotated mess.

Solving the puzzle

With $A=(0,0)$, $B=(4,0)$, and $P=(x,y)$ the unknown point: "distance from $P$ to $A$ is 3" means $x^2+y^2 = 9$. "Distance from $P$ to $B$ is 5" means $(x-4)^2+y^2=25$.

Subtract the first equation from the second to eliminate $y^2$ in one step — this is exactly the payoff of the coordinate choice, since both equations share the same $y^2$ term:

$(x-4)^2 - x^2 = 25 - 9 \;\Rightarrow\; -8x+16 = 16 \;\Rightarrow\; x = 0$

So $x=0$, and from $x^2+y^2=9$: $y^2=9$, giving $y=3$ (taking the point inside/above the square). $P = (0,3)$ — which sits exactly on side $AD$, three units up from $A$. That's already a clean, checkable answer: $P$ is a specific point on the square's edge, not floating somewhere in open space.

Distance to the far corner $C=(4,4)$:

$PC = \sqrt{(4-0)^2+(4-3)^2} = \sqrt{16+1} = \sqrt{17}$

Answer: $\sqrt{17} \approx 4.12$. The whole solution was two equations, one subtraction, and one distance formula — no trigonometry, no case-drawing, because the coordinate choice did the hard work before the algebra started.

A second example: a triangle's median

Triangle $ABC$ has $AB=6$, $AC=8$, and the angle at $A$ is 90°. Find the length of the median from $A$ to the midpoint of $BC$.

The right angle is the most-constrained point, so put it at the origin with the legs on the axes: $A=(0,0)$, $B=(6,0)$, $C=(0,8)$ — the right angle becomes automatic (the axes are perpendicular by construction, so there's nothing left to verify). The midpoint of $BC$ is $M = \left(\frac{6+0}{2}, \frac{0+8}{2}\right) = (3,4)$.

$AM = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = \textbf{5}$

Notice this instantly confirms a classical fact — the median to the hypotenuse of a right triangle equals half the hypotenuse ($BC = \sqrt{36+64}=10$, and indeed $AM=5=10/2$) — a fact that's genuinely fiddly to prove with pure angle-chasing but falls out of coordinates in one line, because putting the right angle at the origin made the right-angle condition free instead of something to prove.

The generalizable technique

The habit — spend ten seconds choosing coordinates before writing equations, rather than defaulting to whatever orientation the picture was drawn in — is what separates a two-line coordinate solution from a three-page trigonometric slog on the same problem. It generalizes directly to physics-flavored quant puzzles (choosing a reference frame that kills a component of motion), to portfolio geometry (rotating axes to align with principal components so covariance terms vanish), and to any optimization where a smart change of variables turns a coupled system into an uncoupled one. The geometry is almost never the hard part; the coordinate choice is.