# 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
6 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Question #6:

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

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

- 6 years ago

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

- 6 years ago

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

- 6 years 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 - 6 years 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 $\pi/4+4\pi$?

- 6 years 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 - 6 years ago

Yes, understood. Thanks! :)

- 6 years 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?

- 6 years ago

solution for qn 8

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

- 6 years ago

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

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

- 6 years ago

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

- 6 years ago

great note!

- 6 years 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 $S =\dfrac{\sqrt{61} \cdot \sqrt{153} \cdot 14}{2 \cdot \sqrt{1037}} \leftrightarrow \boxed{S = 21.}$

- 6 years 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

- 6 years 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....

- 6 years 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

- 5 years, 7 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 - 5 years, 7 months ago