Miller's theorem

Miller's theorem (Regiomontanus's problem) — the classical solution to the "maximum viewing angle" problem.

Universal form

$\max_{P\in\ell}\, \angle APB$

Classical model

• Moving point $P$: on a line $\ell$
• Fixed points $A, B$: two fixed points on the same side of $\ell$
• Key constraint: $A, B$ must be on the same side of $\ell$; on opposite sides there is no maximum (the angle is already $180°$ at the intersection of $AB$ with $\ell$ — a degenerate case)

Historically posed by the German mathematician Johannes Müller (latinised as Regiomontanus, 1471) and one of the earliest extremal-geometry problems in the literature.

Conclusion

Draw the circle through $A, B$ tangent to $\ell$ — there are two such circles; pick the one on the same side as $A, B$. Let $P^{*}$ be the tangent point. Then $\angle AP^{*}B$ is the maximum viewing angle, and

$P^{*} \;\text{is the unique optimal point on } \ell.$

The tangent-point location follows from [[power-of-a-point]]: if $\ell$ meets line $AB$ at $C$ (i.e. the extension of $AB$ crosses $\ell$), then

$CP^{*\,2} \;=\; CA \cdot CB.$

Proof (same-arc inscribed angle + triangle exterior angle)

Take any non-tangent $P$ on $\ell$ ($P \ne P^{*}$); we show $\angle APB < \angle AP^{*}B$.

Step 1 · Construct the secant. Let line $AP$ meet circle $\odot O$ a second time at $Q$ (since $\ell$ is tangent to $\odot O$ at $P^{*}$, it meets the circle only at $P^{*}$; for $P \ne P^{*}$, the line through $P$ and $A$ must meet the circle a second time).

Step 2 · Same-arc inscribed angles. In $\odot O$, the inscribed angles $\angle AQB$ and $\angle AP^{*}B$ subtend the same chord $AB$ (since $Q, P^{*}$ lie on $\odot O$ on the same side as $AB$). By Inscribed angles on same arc are equal; angle in a semicircle is right:

$\angle AQB \;=\; \angle AP^{*}B.$

Step 3 · Triangle exterior angle. In $\triangle PQB$, $\angle AQB$ is the exterior angle at vertex $Q$ (with $A, P, Q$ collinear; depending on the configuration $P$ lies between $Q$ and $A$ or $Q$ between $P$ and $A$, but in both cases the exterior-angle inequality Exterior angle > either non-adjacent interior angle gives the same conclusion):

$\angle AQB \;>\; \angle APB.$

Step 4 · Chain them together.

$\angle AP^{*}B \;=\; \angle AQB \;>\; \angle APB.$

Since $P \ne P^{*}$ was arbitrary, $P^{*}$ gives the unique maximum viewing angle. $\qquad\blacksquare$

Geometric intuition

Think of "the angle at which $P$ sees $AB$" as "the inscribed angle at $P$ when $P$ lies on some circle through $A, B$". Larger angle ⇔ smaller-radius circle ⇔ the circle hugs closer to $\ell$. The moment the circle is tangent to $\ell$, hugging any tighter would push the circle across $\ell$, and a larger angle's circle simply does not meet $\ell$ — so the tangent circle's angle is the supremum, attained at the tangent point.

When to use / mnemonic

• "Find a point $P$ on some line that maximizes the angle at which $P$ sees two fixed points" → Miller's theorem
• "Baseline + two observation points (e.g. base + top of a statue), find the best viewing spot" → real-world Miller
• Note the distinction: if $A, B$ are on opposite sides of $\ell$, no maximum exists (the angle hits $180°$ at the intersection of $AB$ with $\ell$ — degenerate).
• If the moving point $P$ lies on a circle instead of a line, the problem is still about inscribed angles on the same arc, but the Miller construction no longer applies — just use Inscribed angles on same arc are equal; angle in a semicircle is right to compare directly.

Applications

To be added.

Related

• Inscribed angles on same arc are equal; angle in a semicircle is right — same-arc inscribed angles, the core of the proof
• Exterior angle > either non-adjacent interior angle — triangle exterior-angle inequality
• [[power-of-a-point]] — pinpoints the tangent point via $CP^{*\,2} = CA \cdot CB$
• Hidden circle — the more general "fixed chord + fixed angle ⇒ circle" model