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'Given a line and two distinct points A and B on the same side of the line, find the point P on the line such that ∠APB is maximised.'

Just by drawing diagrams I think it is when the circumcircle of APB is tangent to the line at P, but I can't seem to get the algebra right. (I tried using cosine rule.) Please help!

Ginny Mothball Wong
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  • Just like in the next question you posed, assume the line is on the $x$ axis, so that $P = (x_P ,, 0)$, $A = (x_A ,, y_A)$, and $B = (x_B ,, y_B)$. Because the points are on the same side of the line, you can assume $y_A \ge 0$ and $y_B \ge 0$; and that $x_A \le x_P \le x_B$ if $x_A \lt x_B$; $x_B \le x_P \le x_A$ otherwise; you can verify this by drawing the situation. I believe the arctangent function is useful here. – Nominal Animal Nov 14 '17 at 20:59

1 Answers1

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maximum angleOf all circles through $A$, $B$, and a point on line $CD$, the circle which is tangent to $CD$ is the smallest.

Let circle $APB$ be tangent to $CD$ at $P$, and join $PA$, $PB$, and $AB$. Let $Q$ be any other point on $CD$, and and join $QA$, $QB$.

Since circle $AQB$ is greater than circle $APB$, and chord $AB$ is common to both, then arc $AFB$ subtends a smaller angle at the center of circle $AQB$ than arc $AEB$ subtends at the center of circle $APB$.

And angles at the circumference are half the angles at the center.

Therefore $$\angle AQB < \angle APB$$

Edward Porcella
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