# High School Math Part II

This is a continuation of High School Math.

The following problems can be solved using only topics and theorems one encounters in the high school (Polynomials, Calculus, Probability/Statistics, Trigonometry, Analytic Geometry, Induction, etc).

1. Given an integer $n$ and a positive rational number $r$, show that $n^r$ is a rational number if and only if $n^r$ is an integer.

2. How many values does the graph $f(x) = x^{\frac {1}{\ln x} }$ take? (Alternatively, what is the range of $f(x)$?) Draw the graph on your graphing calculator and zoom in (or zoom fit). What do you see? Why?

3. Prove Heron's formula - The area of a triangle with side lengths $a, b$ and $c$ is $\frac {1}{4} \sqrt{ (a+b+c)(a+b-c)(a-b+c)(-a+b+c) }.$

4. [Henry Dudeney] Find the area of a triangle with side lengths $\sqrt{61}, \sqrt{153}, \sqrt{388}$.

5. Evaluate $i^i$. Your answer should be a real value.

6. [Euler] Show that $\frac {\pi}{4} = 5 \tan^{-1} \frac {1}{7} + 2 \tan^{-1} \frac {3}{79}$.

7. [Morrie] Show that $\cos 20^\circ \times \cos 40^\circ \times \cos 80^\circ = \frac {1}{8}$ and $\sin 20 ^\circ \times \sin 40^\circ \times \sin 80^\circ = \frac {\sqrt{3}}{8}$

8. Find all solutions to the system of equations $x(x+1) + y(y+1) = 12,$ $(x+1)(y+1)=4.$ Hint: Both equations are symmetric (i.e. you can replace $x$ with $y$ and vice versa, and still get the same equation).

Note by Calvin Lin
7 years, 2 months ago

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Question #6:

Since $(7+i)^5(79+3i)^2=78125000+78125000i$, $\Rightarrow 5\arg (7+i)+2\arg(79+3i)=\arg(78125000+78125000i)$,

$\displaystyle \Rightarrow 5\tan^{-1} \frac{1}{7}+2\tan^{-1}\frac{3}{79}=\frac{\pi}{4}$

- 7 years, 2 months ago

Wow..How did you think of that?? Amazing method

- 7 years, 2 months ago

I have seen it before....google for Machin-like formula. :)

- 7 years, 2 months ago

Indeed, that is one of the easiest ways of dealing with these inverse tangent formulas. The only thing that you need to take note of, is that your values are accurate up to $\pi$ or $2\pi$, so another slight angle bounding method is required.

Staff - 7 years, 2 months ago

Ah yes, I didn't think of it before. I have come up with something but I am not very sure.

Since $1/7$ and $3/79$ are quite small, one can approximate $\arctan(1/7)$ and $\arctan(3/79)$ as $1/7$ and $3/79$ i.e

$\displaystyle 5\tan^{-1} \frac{1}{7}+2\tan^{-1} \frac{3}{79} \approx \frac{5}{7}+\frac{6}{79} \approx 0.79$

and this is quite close to $\pi/4$. Is this enough to show why is it ok to pick $\pi/4$ and not $\pi/4+2\pi$ or $\pi/4+4\pi$?

- 7 years, 2 months ago

Yes. More rigorously, we can show that for small angles, $x < \tan x < 2x$, and hence $\frac{\pi}{4}$ is the only valid option.

Staff - 7 years, 2 months ago

Yes, understood. Thanks! :)

- 7 years, 2 months ago

BTW @Calvin Lin , I vaguely remember that you have posted the very same problem in past on Brilliant but I can't seem to find the link. Or was it someone else?

- 7 years, 2 months ago

solution for qn 8

$x(x+1)+y(y+1)=12\quad \quad .....(1)\\ (x+1)(y+1)=4\quad \quad \quad .....(2)\\ therefore,\quad (1)+2(2)=>\quad { x }^{ 2 }+2xy+{ y }^{ 2 }+3(x+y)+2=20\\ =>\quad { (x+y) }^{ 2 }+3(x+y)-18=0\\ put,\quad t=x+y\\ =>\quad { t }^{ 2 }+3t-18=0\\ (t+6)(t-3)=0\\ therefore,\quad t=-6\quad (or)\quad t=3\\ =>\quad x+y=\quad -6\quad (or)\quad 3\\ sub\quad y=-6-x\quad in\quad (2)\\ (x+1)(-5-x)\quad =\quad -{ x }^{ 2 }-6x-5=4\\ =>{ (x+3) }^{ 2 }=0\\ =>\quad x=-3\quad and\quad y=-3\\ \\ sub,\quad y=3-x\quad in\quad (2)\\ =>\quad (x+1)(4-x)\quad =\quad 3x-{ x }^{ 2 }+4\quad =\quad 4\\ =>\quad x(3-x)=0\\ =>\quad x=0\quad \quad (or)\quad x=3\\ \quad \quad \quad y=3\quad \quad \quad \quad \quad \quad \quad y=0\\ hence\quad the\quad solution\quad set\quad is\quad \{ (-3,-3),(0,3),(3,0)\}$

- 7 years, 2 months ago

solution for qn 5 (i dont know if it is correct)

$first\quad lets\quad consider\quad the\quad equation\quad \\ { e }^{ i\pi }=-1\\ therefore,\quad i\pi =\log _{ e }{ (-1) } \\ =>\quad i\pi =\log _{ e }{ { i }^{ 2 } } \\ =>\log _{ e }{ i } =\frac { i\pi }{ 2 } \\ now\quad lets\quad get\quad back\quad to\quad the\quad question.\\ let\quad ,\quad { i }^{ i }=t\\ =>\quad \log _{ e }{ t } =i\log _{ e }{ i } =i(\frac { i\pi }{ 2 } )=-\frac { \pi }{ 2 } \\ therefore,\quad t={ i }^{ i }={ e }^{ -\frac { \pi }{ 2 } }=0.2078795$

- 7 years, 2 months ago

Its correct..Check out my note on the same.

- 7 years, 2 months ago

great note!

- 7 years, 2 months ago

SOLUTION #4

Using the Law of Cosines, we have $388 = 61 + 153 - 2 \sqrt{61} \cdot \sqrt{153} \cdot \cos{\alpha} \leftrightarrow \cos{\alpha} = - \frac{29}{\sqrt{1037}}$.

By the unit circle definition, $\sin^2 {\alpha} + \frac{841}{1037} = 1 \leftrightarrow \sin{\alpha} = \frac{14}{\sqrt{1037}}$. Note $0 < \alpha < \pi$.

Finally, by the Side-Side-Angle area formula for the triangle, we have $S =\dfrac{\sqrt{61} \cdot \sqrt{153} \cdot 14}{2 \cdot \sqrt{1037}} \leftrightarrow \boxed{S = 21.}$

- 7 years, 2 months ago

Question 7 cos20\times cos40\times cos80 Multiply and divide with sin20.then multiply and divide with 2. We get 2sin20cos20 which is equal to sin40. Again multiplying and dividing by 2.we get sin80. Again multiplying and dividing by 2 we get sin160. We get it in the form,sin160/8sin20. But sin20=sin160 Therefore cos20\times cos 40\times cos 80=1/8 Next sin20sin40sin80 sin40=sin(60-20),sin80=sin(60+20) sin(A+B)sin(A-B)=sin^2(A)-sin^2(B) Therefore sin40sin80=3/4-sin^2(20) Multiplying sin 20 We get 3sin20-4sin^3(20)/4 Which is in the form 3sinA-4sin^3A=sin3A Therefore sin60/4=3^0.5/8

- 7 years, 2 months ago

Ans to Q5 : i^i......i = e^i(pi/2)......i^i = (e^i(pi/2))^i = e^ii(pi/2) = e^-(pi/2) ans....

- 7 years, 1 month ago

Question#2

$\frac{1}{logx} = log_x (10 or e)$

I don"t know at which base it is

$x^{log_x(10)} = f(x)$

or

$x^{log_x(e)} = f(x)$

thus

$f(x) = 10$

or

$f(x) = e$

its a constant function

therefore range = 10 or e

I see just a horizontal line parallel to x axis

- 6 years, 9 months ago

The point of the question is to use a (older model) graphing calculator. Because they calculate exponents with a lot of rounding error, and zoom into the y-axis to display the entire range , you actually end up with a "mess of dots all over", and it is not clear that the graph is a straight line.

Staff - 6 years, 9 months ago