Posts Tagged ‘Mathematics’

A Handy Way to Find the Equation of a Tangent Line to a Circle

Posted: April 4, 2024 in Mathematics
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The DSE (click here to see what HKDSE is) is close. The math core exam is less than 2 weeks time. I am going to share a really nice equation to find the tangent of a given point on the circle. Suppose that the general form of a circle is given by

x^2 + y^2 + Dx + Ey + F = 0

then the equation of a tangent line at point (x_1, y_1) will be as follows:

\displaystyle x_1 x + y_1 y + D\left(\frac{x+x_1}{2}\right) + E\left(\frac{y+y_1}{2}\right) + F = 0

Proof:

Differentiate the equation of the circle with respect to x implicitly.

2x+2yy'+D+Ey'=0

hence y’ is,

\displaystyle y' = \frac{-2x-D}{2y+E}

then the slope of the tangent at point (x_1, y_1) would be

\displaystyle m=\frac{-2x_1-D}{2y_1+E}

By point slope form of a line,

\displaystyle y-y_1 = \frac{-2x_1-D}{2y_1+E}\cdot(x-x_1)

(y-y_1)(2y_1+E) = (x-x_1)(-2x_1-D)

2y_1y-2y_1^2+Ey-Ey_1 = -2x_1x+2x_1^2-Dx+Dx_1

2x_1x+2y_1y+D(x-x_1)+E(y-y_1) = 2x_1^2+2y_1^2

\displaystyle x_1x+y_1y+\frac{D(x-x_1)}{2}+\frac{E(y-y_1)}{2} = x_1^2+y_1^2

Now, since the given point satisfies the circle, we plug (x_1,y_1) into the circle,

x_1^2 + y_1^2 + Dx_1 + Ey_1 + F = 0

x_1^2 + y_1^2 = -Dx_1 - Ey_1 - F

Hence, we have the following,

\displaystyle x_1x+y_1y+\frac{D(x-x_1)}{2}+\frac{E(y-y_1)}{2} = -Dx_1 - Ey_1 - F

Move everything to the left, therefore

\displaystyle x_1 x + y_1 y + D\left(\frac{x+x_1}{2}\right) + E\left(\frac{y+y_1}{2}\right) + F = 0

QED. Good luck.

A Simple Little Math Contest Problem

Posted: May 2, 2022 in Mathematics
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So, here is the problem I found on the internet.

\displaystyle \text{Find all natural numbers } a \text{, } b \text{, and } c \text{ such that } a>b>c \text{ and } \frac1a + \frac2b + \frac3c = 1

Solution

Let’s find some bounds on a, b, and c. Since \frac1a+\frac2b+\frac3c=1, hence

a>1, b>2, \text{ and } c>3

Since a > b > c, then

\displaystyle \frac1a + \frac2a + \frac3a < \frac1a + \frac2b + \frac3c = 1

\therefore a > 6

This doesn’t seem to help us much. Let’s consider c,

\displaystyle \frac1c + \frac2c + \frac3c > \frac1a + \frac2b + \frac3c = 1

\therefore 3 < c < 6

So, this is great! We only have 2 cases to consider.

Case c = 4,

\displaystyle \frac1a + \frac2b + \frac{3}{4} = 1 \Rightarrow \frac1a + \frac2b = \frac{1}{4}

4b + 8a = ab

0 = ab - 8a - 4b

32 = a(b-8)-4(b-8)

32 = (a-4)(b-8)

(a-4, b-8) = (32, 1) \Rightarrow (a,b) = (36, 9)

(a-4, b-8) = (16, 2) \Rightarrow (a,b) = (20, 10)

Similarly for case c = 5,

\displaystyle \frac1a+\frac2b+\frac{3}{5}=1 \Rightarrow \frac1a + \frac2b = \frac{2}{5}

5b + 10a = 2ab

0 = 2ab - 10a - 5b

25 = 2a(b-5) - 5(b-5)

25 = (2a-5)(b-5)

(2a-5, b-5) = (25, 1) \Rightarrow (a, b) = (15, 6)

Therefore, the solutions are

(a, b, c) = (36,9,4), (20,10,4), (15,6,5)

Find the Last Three Digits of the 2022nd Fibonacci Number

Posted: April 29, 2022 in Mathematics
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So, here is the problem. Given that F_n = \{1, 1, 2, 3, 5, 8, \ldots\}. We want to find F_{2022}\pmod{1000}

By Binet’s formula (click here for explanation), we have

\displaystyle F_n = \frac{1}{\sqrt5} \left(\left(\frac{1+\sqrt5}{2}\right)^n - \left(\frac{1-\sqrt5}{2}\right)^n\right)

Hence, the 2022nd Fibonacci number would be

\displaystyle F_{2022} = \frac{1}{2^{2022}\sqrt5} \left(\left(1+\sqrt5\right)^{2022}-\left(1-\sqrt5\right)^{2022}\right)

Expand it using Binomial Theorem and take mod 125, we get

\displaystyle 2^{2021} F_{2022} \equiv \binom{2022}{1} + \binom{2022}{3}5 + \binom{2022}{5}25 \pmod{125}

Since 2 and 5 are relatively prime, we apply the Euler’s theorem to reduce 2^{2021}.

\phi(125) = 125 \cdot \frac45 = 100

\therefore 2^{2021} \equiv 2^{21} \equiv (2^{10})^2\cdot2 \equiv 24^2\cdot2 \equiv 576 \cdot 2 \equiv 152 \equiv 27 \pmod{125}

Calculate the rest of the terms on the right,

2022 \equiv 22 \pmod{125}

\displaystyle \binom{2022}{3}5 \equiv \frac{2022\cdot2021\cdot2020\cdot5}{6} \equiv 337 \cdot 21 \cdot 20 \cdot 5 \equiv 8700 \equiv 75 \pmod{125}

\displaystyle \binom{2022}{5}25 \equiv \frac{22\cdot21\cdot20\cdot19\cdot18\cdot25}{5\cdot4\cdot3\cdot2} \equiv 88 \cdot 450 \equiv 88\cdot75 \equiv 100 \pmod{125}

Hence,

27 F_{2022} \equiv 22+75+100 \equiv 72 \pmod{125}

3 F_{2022} \equiv 8 \pmod{125}

So, there exists an integer k such that

3F_{2022} = 125k + 8

Then take mod 3 to find k,

0 \equiv 2k + 2 \pmod{3}

k \equiv 2 \pmod{3}

So,

3F_{2022} = 258

F_{2022} \equiv 86 \pmod{125}

Next, we have to find F_{2022} \mod{8}

F_n = \{1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, \ldots \} \pmod{8}

So, the period is 12 and 2022 \equiv 6 \mod{12} hence

F_{2022} \equiv 0 \pmod{8} \equiv 86 \pmod{125}

For some integers a and b we have

F_{2022} \equiv 8a \equiv 125b + 86 \pmod{1000}

0 \equiv 5b + 6 \pmod{8}

b\equiv 2 \pmod{8}

Therefore,

F_{2022} \equiv 125b + 86 \equiv 125\cdot2+86 \equiv 336 \pmod{1000}

So the last three digits of the 2022nd Fibonacci number is 336.

Problem of the Year 2022

Posted: March 28, 2022 in Mathematics
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So, this is the problem.

\displaystyle \text{Given that } a+b+c=2022 \text{ and } \frac1a+\frac1b+\frac1c=\frac{1}{2022}\text{.}

\displaystyle \text{Find } \frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}} \text{.}

And here is the solution.

Since

\displaystyle \frac{1}{a+b+c}=\frac{1}{2022}=\frac1a+\frac1b+\frac1c

hence,

\displaystyle 0 = \frac1a + \frac1b + \frac1c - \frac{1}{a+b+c} = \frac{a+b}{ab} + \frac{a+b}{c(a+b+c)} = (a+b)\left(\frac{c(a+b+c)+ab}{abc(a+b+c)}\right)

\displaystyle = (a+b)\left(\frac{c(b+c)+ac+ab}{abc(a+b+c)}\right) = \frac{(a+b)(c(b+c)+a(b+c))}{abc(a+b+c)} = \frac{(a+b)(b+c)(c+a)}{abc(a+b+c)}

So, there are three cases.

a+b=0 \quad\text{or}\quad b+c=0 \quad\text{or}\quad c+a=0

WLOG, assume a + b = 0. Then we have c = 2022.

\displaystyle \Rightarrow a = -b \Rightarrow \frac{1}{a^{2023}} + \frac{1}{b^{2023}} = 0

\displaystyle \therefore \frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}} = \frac{1}{2022^{2023}}.