$ANBM$ is a cyclic quadrilateral such that $AN = a$ , $BN = b$ , and $\angle MNA= \angle MNB = 60^\circ$ .

What is the length of diagonal $MN$ ?

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Bonus
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Use geometry
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$\frac{ab( a + b )}{4 }$
$\sqrt{a^2 + b^2}$
$\frac{2(a + b)}{\sqrt 3}$
$a + b$

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Nice to use ptolemy!

Mr. India
- 2 years, 3 months ago

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thanks! .n..tbh... admirable profile name ..

nibedan mukherjee
- 2 years, 3 months ago

Since $ANBM$ is a cyclic quadrilateral , the opposite angles $\angle ANB + \angle AMB = 180^\circ$ $\implies \angle AMB = 60^\circ$ . Also $\angle MBA = MNA = 60^\circ$ and $\angle MAB = \angle MNB = 60^\circ$ . Therefore $\triangle ABM$ is equilateral and $AB=BM=MA = c$ .

Let the length of diagonal $MN$ be $x$ . Then applying Ptolemy's theorem : the product of the lengths of the two diagonals of a cyclic quadrilateral is equal to the sum of the products of opposite sides, we have $xc = ac + bc$ $\implies MN = x = \boxed{a+b}$ .

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what a coincidence sir!.... in time and approach... :P

nibedan mukherjee
- 2 years, 3 months ago

Nice approach sir

Mr. India
- 2 years, 3 months ago

Glad that you like the solution. Upvote?

Chew-Seong Cheong
- 2 years, 3 months ago

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sure! sir upvoted... what abt mine? :P

nibedan mukherjee
- 2 years, 3 months ago

Done sir :D

Sir, can you guide me to some maths books with challenging problems and good explanations??

Mr. India
- 2 years, 3 months ago

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I am sorry, I hardly read any book nowadays. I don't know any good books for math.

Chew-Seong Cheong
- 2 years, 3 months ago

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@Chew-Seong Cheong – Ok sir, no problem, can you tell how you created the figure in solution?

Mr. India
- 2 years, 3 months ago

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@Mr. India – I used Microsoft Excel spreadsheet to graph out the figure. Removed all the axes and grids then copied it to Paint to complete the figure.

Chew-Seong Cheong
- 2 years, 3 months ago

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@Chew-Seong Cheong – Oh hard work! B-) thnx btw

Mr. India
- 2 years, 3 months ago

Please ask if you don't understand something

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