# Maximizing an angle based on certain constraints

$$A (0,a)$$ and $$B(0,b)\; (a,b>0)\;$$ are the vertices of $$\triangle ABC$$ where $$C(x,0)$$ is variable. Find the value of $$x$$ when angle $$ACB$$ is maximum.

Now geometry's never really been my strong point, so I decided to go with a bit of calculus. First, I used the sine rule: $$\mathrm{sinC=\frac{b-a}{2R}}$$ where R is the radius of the circumcircle. I note that for angle C to be maximum, sinC should be maximum. As such, R must be minimum. Next, I used the relation $$\mathrm{R=\frac{(b-a)\cdot\sqrt{x^2+b^2}\cdot\sqrt{x^2+a^2}}{2\Delta}}$$ where $$\mathrm{\Delta \text{ is the area of }ABC=\frac{(b-a)x}{2}}$$.

A bit of comparatively lengthy differentiation gives me the value of $$x$$ as $$\sqrt{ab}$$.

When I go through the solutions, it's simply been stated:

For angle ACB to be maximum, the circle passing through A,B will touch the X-axis at C.

Beyond this, it's been solved using the very simple $$\mathrm{OC^2=OA\cdot OB}$$, where O is the origin. So the above statement seems to be the difference between a lengthy differentiation and a one line solution.

It's getting a little difficult for me to see why the above statement should be intuitive. Could someone shed a bit more light on it for me, and possibly provide an intuitive proof?

• I couldn't settle on a concise yet lucid title. Please feel free to edit it. Oct 19 at 17:20

It is given that $$a, b \gt 0$$ so both $$A$$ and $$B$$ are on the same side of x-axis. The first point to note is that $$\triangle ABC$$ is obtuse and $$\angle ACB$$ is acute. Now we use the relationship $$AB = 2 R \sin C$$ where $$R$$ is the circumradius of $$\triangle ABC$$. As $$AB$$ is fixed, we maximize $$\angle C$$ when we minimize $$R$$ given $$\sin$$ function is strictly increasing for $$\left(0, \frac{\pi}{2} \right)$$.

Also note that $$O$$ must be on the perpendicular bisector of $$AB$$ which is parallel to x-axis. So, $$R = OC$$ is minimum when $$OC$$ is perpendicular to x-axis.

• ''Assuming both $\mathrm{A}$ and $\mathrm{B}$ are above x-axis''. Yes that's been mentioned in the question. I really like that this answer picked up my line of thinking when I first approached this problem and provided an intuitive explanation in that fashion. The other answers I've got are also very good; it's hard to settle on a 'best' answer. Oct 19 at 18:06
• @C_Lycoris good to see you have a few answers to choose from :) I was just not sure whether question said that $a, b$ were both positive so added the line about them being on the same side of x-axis. Oct 19 at 18:09
• @C_Lycoris I see that now so I will remove that line Oct 19 at 18:10
• In the OP's post $O$ is referred to as the origin. In your answer?
– ACB
Oct 20 at 7:01
• @ACB I am referring to $O$ as the center of the circumcircle of $\triangle ABC$ Oct 20 at 7:04

Let $$\omega$$ be the circumcirle of $$\triangle ABC$$ where $$C$$ is a point on the $$x$$-axis such that $$\angle ACB$$ is maximum. Assume $$\omega$$ intersects the $$x$$-axis twice, at $$C$$ and $$D$$.

Let $$F$$ be any point on the arc $${CD}$$ (not containing $$A, B$$) and define $$E$$ as the intersection of $$AF$$ and $$x$$-axis. Observe, $$\angle ACB=\angle AFB<\angle AEB$$ which contradicts the fact that $$\angle ACB$$ is maximum.

Therefore, the assumption that $$\omega$$ intersects the $$x$$-axis twice is incorrect, which implies $$\omega$$ is tangent to the $$x$$-axis at $$C$$.

• +1. I liked this proof, it really should've occurred to me. Oct 19 at 18:08

Let $$\bigcirc K$$ through $$A$$ and $$B$$ be tangent to the $$x$$-axis at $$D$$. For $$C$$ on the $$x$$-axis (and on the same side of the $$y$$-axis as $$D$$), let $$A'$$ and $$B'$$ be the "other" points where $$\overleftrightarrow{AC}$$ and $$\overleftrightarrow{BC}$$ meet this circle.

A corollary to the Inscribed Angle Theorem states that we can write $$\angle C = \frac12 \left(\;\angle AKB - \angle A'KB'\;\right)$$ Since $$\angle AKB$$ is fixed, maximizing $$\angle C$$ amounts to minimizing $$\angle A'KB'$$. This happens when (and only when) $$A'$$ and $$B'$$ coincide; hence, when $$C$$ and $$D$$ coincide. $$\square$$

We use the fact that if $$A$$ and $$B$$ are on a given circle, then if you have $$C$$ on the circle and $$C_0$$ (strictly) inside the circle (and $$C$$, $$C_0$$ are on the same side of $$\overline{AB}$$) $$\angle AC_0B > \angle ACB$$

You can see this by extending $$AC_0$$ to the circle at $$C'_0$$, in which case $$\angle AC_0B > \angle AC'_0B$$, but because of the inscribed angle theorem, $$\angle AC'_0B = \angle ACB$$.

Now, write the desired angle $$\angle ACB$$ as $$f(x)$$ in terms of $$x$$; we want to maximize $$f(x)$$.

The circumcircle of $$\triangle ABC$$ always intersects the $$x$$-axis at $$C = (x_0, 0)$$. Now, say for contradiction that $$f(x_0)$$ is maximal, and that the circumcircle of $$\triangle ABC$$ also intersects the $$x$$-axis at $$C' = (x_0', 0) \neq C$$. Then the midpoint $$M$$ of $$\overline{CC'}$$ is inside the circle, so $$\angle AMB > \angle ACB$$; and $$M = \left(\frac{x_0 + x'_0}{2}, 0 \right)$$ is on the $$x$$-axis, so $$f\left( \frac{x_0+x_0'}{2} \right) > f(x_0),$$ contradicting the fact that $$f(x_0)$$ is maximal.

Thus the circumcircle of $$\triangle ABC$$ must intersect the $$x$$-axis at exactly one point.

There are two ways of looking at it.

Circles such as those that pass through non red sides meeting on x-axis at D,E subtend the same angle from segment AB ( Angles in a segment are equal property). In order that there be a unique point, these points should be drawn together to make them into a repeated point. A repeated point is in fact a point of tangentcy at C.

By means of the Circle's property that the product of segments be constant (this being is the power of the Circle ) we have

$$OA \cdot OB= OC^2= x^2 \to \; x = \sqrt {ab} \tag1$$

Next way is direct confirmation with differential calculus, maxima/minima.

The "look angle " or subtended angle is

$$\tan^{-1}\frac{a}{x}-\tan^{-1}\frac{b}{x}$$

Differentiate w.r.t. $$x$$ arctan and Chain Rule

$$\dfrac{-a/x^2}{1+a^2/x^2} + \dfrac{-b/x^2}{1+b^2/x^2} =0$$

When simplified, we get the same result as (1).

Another way to solve it using calculus and geometry is to notice that $$\angle C=\frac{\pi}{2}-(\angle ACO+\angle CBO)$$ (where $$O$$ is the origin). Minimizing $$\angle ACO+\angle CBO$$ is equivalent to minimizing $$\tan \angle ACO+\tan \angle CBO=\frac{a}{x}+\frac{x}{b}$$ which is easy to differentiate.

• I'm sorry, but what does point 'O' refer to in your answer? If it's the origin, shouldn't $\angle C=\frac{\pi}{2}-(\angle ACO+\angle CBO)$? Oct 20 at 9:37
• Thanks! I did have a typo, but the function expressing sum of tangents is correct. Point $O$ is the origin. Oct 21 at 2:04
• Yes. I saw that the final expression was correct. Also a very quick differentiation to yield the answer. Thank you! Oct 21 at 2:18