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$A (0,a)$ 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?

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  • $\begingroup$ I couldn't settle on a concise yet lucid title. Please feel free to edit it. $\endgroup$
    – C_Lycoris
    Oct 19 at 17:20
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enter image description here

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.

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    $\begingroup$ ''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. $\endgroup$
    – C_Lycoris
    Oct 19 at 18:06
  • $\begingroup$ @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. $\endgroup$
    – Math Lover
    Oct 19 at 18:09
  • $\begingroup$ @C_Lycoris I see that now so I will remove that line $\endgroup$
    – Math Lover
    Oct 19 at 18:10
  • $\begingroup$ In the OP's post $O$ is referred to as the origin. In your answer? $\endgroup$
    – ACB
    Oct 20 at 7:01
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    $\begingroup$ @ACB I am referring to $O$ as the center of the circumcircle of $\triangle ABC$ $\endgroup$
    – Math Lover
    Oct 20 at 7:04
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enter image description here

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 $C$.

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    $\begingroup$ +1. I liked this proof, it really should've occurred to me. $\endgroup$
    – C_Lycoris
    Oct 19 at 18:08
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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.

enter image description here

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. $\square$

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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.

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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

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

enter image description here

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).

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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.

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

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