Jeweler's problem!

Geometry Level 2

Raju Bhai, a jeweler, has two parallel gold bangles of radii $R_1=2\sqrt{2}-\sqrt{3}+1,\ R_2=2\sqrt{2}+\sqrt{3}-1$ held together. He wants to fit some diamonds in between the two bangles, as shown in the image below (which is not drawn to scale).

If the cost of one diamond stone is $1400, then how much money should Raju Bhai spend on diamonds to make that bangle?

Hint: The diameter of the diamond is the distance between the two bangles, and find the number of diamonds that can be fitted in the circle of bangles.

This is an original problem and belongs to my set Raju Bhai's creations .

$8400 $14000 $16800 $33600

1 solution

Shikhar Srivastava
Dec 16, 2017

The number of diamonds between the bangles is equal to number of $\angle a$ contained around a point.

Here $A$ and $B$ are centres of two adjacent Diamonds and $O$ is the centre of bangles. Now $\angle a$ can be found using cosine formula in $\triangle OAB$

We have $OA = OB = R_1 + \frac{R_2 - R_1}{2} = (2\sqrt 2 - \sqrt 3 + 1) + (\sqrt 3 - 1) = 2\sqrt 2$ and $AB$ is the distance between the centres of two adjacent diamonds which is 2 times the radius of the diamond. So, $AB$ = diameter of diamond = $R_2 - R_1 = 2(\sqrt 3 - 1)$

$\cos a = \frac{OA^2 + OB^2 - AB^2}{2\cdot OA \cdot OB} = \frac{8 + 8 - 4(4-2\sqrt3)}{16} = \frac{\sqrt 3}{2} \Rightarrow \angle a = 30^\omicron$

Number of such angles in a circle = $\frac{360^\omicron}{30^\omicron} = 12$

There are 12 diamonds between the bangles which costs $ 1400 $\cdot 12 = \$16800$

Jeweler's problem!

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