Safe cracker riddle, 4 digit from 1 to 9 are required to open a safe. the sum of the digit is 20. the first digit is greater than the third. the second and fourth digits differ by at least 5. exactly 529 digits are square. the first and fourth digits add up to a prime number. the fourth digit is lowest. what is the 4 digit combo?

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Bryson2148

25.03.2021

Mathematics

$x=r(\theta)\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r\sin\theta$
$y=r(\theta)\sin\theta\implies\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r\cos\theta$

Since $r=1-\cos\theta$ , you have $\dfrac{\mathrm dr}{\mathrm d\theta}=\sin\theta$ . Now,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}=\dfrac{\sin^2\theta+(1-\cos\theta)\cos\theta}{\sin\theta\cos\theta-(1-\cos\theta)\sin\theta}$
$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\cos\theta-\cos2\theta}{\sin2\theta-1}$

When $t=\dfrac\pi3$ , you have

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\cos\frac\pi3-\cos\frac{2\pi}3}{\sin\frac{2\pi}3-1}=-2\sqrt3-4$

which will be the slope of the tangent line. At this point, you have $r=1-\cos\dfrac\pi3=\dfrac12$ , so now you need to find the corresponding x and y coordinates.

$x=r\cos\theta=\dfrac12\cos\dfrac\pi3=\dfrac14$
$y=r\sin\theta=\dfrac12\sin\dfrac\pi3=\dfrac{\sqrt3}4$

So the equation of the tangent line is

$y-\dfrac{\sqrt3}4=(-2\sqrt3-4)\left(x-\dfrac14\right)$
$y=-2(2+\sqrt3)x+1+\dfrac{3\sqrt3}4$

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