# Fallacy in showing that $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$

Given a cyclic quadrilateral $ABCD$, $AB=a, BC=b, CD=c, \angle ABC=120^\circ, \angle ABD=30^\circ$, then, show that $$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$

I tried to do it using some trig bashing. What I did was basically assigned the angle $\angle BCA = \theta$ and used trigonometry. It's not hard to see that the radius of the circle is $\frac{c}{2}$. I've squared the equation on both sides and obtained: $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$.

And using $\text {Sine Rule}$, the equation can be reduced further.

I used

• $\frac{a}{\sin\theta}=c$
• $\frac{b}{\sin(60^\circ - \theta)}=c$

And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.

Then I continued to reduce it using the addition-subtraction formulae of trigonometry.

And at the end of the day what I get is $$\boxed{\cos(30^\circ-\theta)=\frac{1}{2}}$$ after all those addition-subtraction of trigonometric equations.

And that's not true I believe.

May I get rectified?

Noticing $\triangle{ADO}$ is an equilateral triangle where $O$ is the center of the circle should be a key.

Since the both sides of $$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$ are non-negative, it is equivalent to $$(\sqrt{c+a}-\sqrt{c+b})^2=c-a-b,$$ i.e. $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$ Since the both sides are positive, it is equivalent to $$(c+2a+2b)^2=4(c+a)(c+b),$$ i.e. $$\frac{3}{4}c^2=a^2+b^2+ab\tag1$$

So, all we need is to prove $(1)$.

Applying the law of cosines to $\triangle{ABC}$, we have $$|\overline{AC}|^2=a^2+b^2-2ab\cos(120^\circ),$$ i.e. $$|\overline{AC}|^2=a^2+b^2+ab\tag2$$

Applying the law of sines to $\triangle{ADC}$, we get $$\frac{|\overline{CD}|}{\sin\angle{DAC}}=\frac{|\overline{AD}|}{\sin\angle{DCA}}$$ which implies $$|\overline{AD}|=\frac c2$$

So, $\triangle{DAO}$ is an equilateral triangle where $O$ is the center of the circle.

Applying the law of sines to $\triangle{ADC}$ gives $$\frac{|\overline{AC}|}{\sin\angle{ADC}}=\frac{|\overline{AD}|}{\sin\angle{DCA}}$$ which implies $$|\overline{AC}|^2=\frac 34c^2\tag3$$

Now, $(1)$ follows from $(2)(3)$.

You assigned the angle $\angle BCA = \theta$ and used trigonometry. You got the radius of the circle as $\frac{c}{2}$. Square the equation on both sides and obtained: $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$.

And using $\text {Sine Rule}$, the equation can be reduced further.

• $\frac{a}{\sin\theta}=c$
• $\frac{b}{\sin(60^\circ - \theta)}=c$

And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.

Reduce it using the addition-subtraction formulae of trigonometry. $$1+2(\sin\theta+\sin(60-\theta) = 2\sqrt{(1+\sin\theta)(1+\sin(60-\theta)}$$

On reduction we get $$2m+1 =2\sqrt{m+m^2+\frac{1}{4}}$$ Where $m=\cos(30-\theta)$ On squaring and simplifying we get $$4m^2+4m+1=4m^2+4m+1$$ which is always true.
Hence the statement is proved.

• Oh okay, fine. Maybe there's an error in my calculations. – Mathejunior Dec 29 '17 at 5:40
• Out of curiosity, is this the question of INMO 2006 – Rohan Shinde Dec 29 '17 at 9:09
• Yes, it's some previous INMO. I am not exactly sure of the year. It's an easy problem though but I wasn't really sure of my solution though I could solve it in other ways. – Mathejunior Dec 29 '17 at 9:12

By law of cosines $$AC=\sqrt{a^2+b^2-2ab\cos120^{\circ}}=\sqrt{a^2+b^2+ab}.$$

In another hand, since $\measuredangle CBD=90^{\circ},$ we see that $CD$ is a diameter of the circle.

Thus, $$\cos30^{\circ}=\frac{\sqrt{a^2+b^2+ab}}{c}$$ or $$3c^2=4(a^2+b^2+ab),$$ which is $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$ exactly!