# 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. (Martin Gardner) There are $n$ racecars on a circular track. Amongst the $n$ cars, there is enough gas for one car to make a complete loop around the track. Show that there exists a car that can make a complete loop around the track by pooling gas from every car that it passes.

2. (USAMO '73) Find all complex solutions to the following system of equations: \begin{aligned} x+y+z & = 3\\ x^2+y^2+z^2 & =3\\ x^3+y^3+z^3 & = 3.\\ \end{aligned}

3. Find all real solutions to the following system of equations: \begin{aligned} x+y+z & = 3\\ x^2 + y^2+z^2 & =3. \\ \end{aligned}

4. (Live Challenge Number Theory 5) Compute the last 3 digits of $171^{172}$.

5. (Britain '87) $f(x)$ is a polynomial with integer coefficients satisfying $f(21)=17$, $f(32)=-247$, and $f(37)=33$. If there is an integer value $N$ such that $f(N)=N+51$, show that $N=26$.

6. (Live Challenge Algebra 4) Let $x$ be a real number such that $0.170 < \log_x 2 < 0.171$ and $0.270 < \log_x 3 < 0.271$. Determine the integer that is closest in value to $\log_x 7^{100}$.

7. (IMO'68) Determine all triangles whose side lengths are consecutive positive integers such that one angle is twice of another angle.

8. (*) Calvin and Dan play a game of chance. A fair coin is flipped and the sequence of Heads and Tails is recorded. Calvin wins if the sequence $HHH$ appears first, and Dan wins if the sequence $TTHH$ appears first. Who has a higher probability of winning?

Part 2 is available here. Note by Calvin Lin
6 years, 8 months ago

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SOLUTION #2

$\left\{\begin{array}{l}x+y+z=3\\ x^2+y^2+z^2=3\\ x^3+y^3+z^3=3\end{array}\right.$

Consider the polynomial $P(x)=x^3+ax^2+bx+c$ with roots $p,q,r$. We use newton's sums to obtain that $p+q+r=-a$ $p^2+q^2+r^2=a^2-2b$ $p^3+q^3+r^3=3ab-a^3-3c$

Now let the roots of this cubic be the variables in our equation. Thus, we have $-a=3$ $a^2-2b=3$ $3ab-a^3-3c=3$

From the first equation, we have $\boxed{a=-3}$

Plugging into the second equation, we have $\boxed{b=3}$

Plugging these into the third equation, we have $\boxed{c=-1}$

Thus, our polynomial $P(x)=x^3-3x^2+3x-1$ or $P(x)=(x-1)^3$

Since the only roots are $p=q=r=1$, the only solution we have for our original system of equations is $\boxed{(x,y,z)=(1,1,1)}$

- 6 years, 6 months ago

Nice I did the same thing no need to repeat

- 6 years, 6 months ago

They did ask for all complex solutions and I have found $(x,y,z) = (e^\frac{2 \pi i}{3}, e^\frac{4 \pi i}{3}, 1)$ to be a solution as well by the formula for the roots of unity. By permuting x, y and z in this solution, we also get the following solutions: $(e^\frac{4 \pi i}{3}, e^\frac{2 \pi i}{3}, 1)$ , $(1, e^\frac{2 \pi i}{3}, e^\frac{4 \pi i}{3})$ , $(1, e^\frac{4 \pi i}{3}, e^\frac{2 \pi i}{3})$ , $(e^\frac{2 \pi i}{3}, 1, e^\frac{4 \pi i}{3})$ and $(e^\frac{2 \pi i}{3}, 1, e^\frac{4 \pi i}{3})$, which, in addition to the trivial $(1,1,1)$ solution, leaves us with all complex solutions for the given system of equations.

- 6 years, 5 months ago

Simple inspection of your answer of $\left(e^{\frac{2\pi i}{3}}, e^{\frac{4\pi i}{3}},1\right)$, we can determine that this and all of the various permutations are not solutions because they don't even satisfy $x+y+z=3$. The only solution to problem 2 is $(1,1,1)$.

- 6 years, 5 months ago

SOLUTION #1

Let us call the cars gas stations instead, to make the problem easier to understand.

First, let's suppose that the car has a large amount of gas initially. As it goes around the loop once, it begins and ends with the same amount of gas. In addition, there must be one gas station where the car arrives with the least amount of gas. But we're done: start at this gas station, and the car will never go lower than $0$ gas, thus it can make it around the track once.

- 6 years, 6 months ago

Sorry no solution here, but I do have a question:

Is there any particular place where I can learn to solve problems like these?

- 6 years, 6 months ago

My attempt for SOLUTION #3

Didn't see a similar solution to mine, thought it was worth a share :)

We are given: $a+b+c=3 \tag{1}$ $a^{2}+b^{2}+c^{2}=3 \tag{2}$

From $(1)$, $c=3-a-b$.

Plugging that into $(2)$ we obtain, $a^{2}+b^{2}+(3-a-b)^{2} = 3$

Expanding the brackets and simplifying it we attain, $a^{2}+ab-3a+b^{2}-3b+3=0$

It can be rewritten as, $a^{2} + a(b-3) + (b^{2}-3b+3) = 0$

Hence, using quadratic formula: $a = \frac{-(b-3) \pm \sqrt{(b-3)^{2}-4(b^{2}-3b+3)}}{2} \tag{3}$

The discriminant must be $\ge 0$ to have real roots so, $(b-3)^{2}-4(b^{2}-3b+3) \ge 0$

Simplifying the quadratic inequality above we obtain, $-3(b-1)^{2} \ge 0$

Solving the quadratic inequality we observe that the only solution for b is, $\boxed{b=1}$.

Hence, solving for $a$ using $(3)$ we obtain, $a = \frac{-1(1-3) \pm \sqrt{-3(1-1)^{2}}}{2} =\frac{2}{2}$

$\boxed{a=1}$

Therefore, $c$ can be easily calculated using equation $(1)$, $c = 3 - 1 -1 = 1$

The only real solution to this question is:

# $\boxed{ (a,b,c) = (1,1,1) }$

- 6 years, 6 months ago

Nice. Is there an easier way to get at this?

Staff - 6 years, 6 months ago

$a+b+c=3\\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=3\\ which\quad implies,{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca=0\\ =>\frac { 1 }{ 2 } (({ a }^{ 2 }-2ab+{ b }^{ 2 })+({ a }^{ 2 }-2ac+{ c }^{ 2 })+({ b }^{ 2 }-2bc+{ c }^{ 2 }))=0\\ =>{ (a-b) }^{ 2 }+{ (b-c) }^{ 2 }+{ (a-c) }^{ 2 }=0\\ =>a=b=c=1$

- 6 years, 6 months ago

Incredible! Nice solution. Clean, short and clear.

- 6 years, 6 months ago

thanks!

- 6 years, 6 months ago

that's what i approached good job buddy !!!

- 6 years, 6 months ago

An easier way, as requested:

Let $x+y+z=3$. We have that $3(x^2+y^2+z^2)\ge (x+y+z)^2$ by Cauchy. Dividing by $3$ on both sides, we have $x^2+y^2+z^2\ge x+y+z$, where equality case is $x=y=z$. However, since $x^2+y^2+z^2=x+y+z$, then we must have $x=y=z$. Solving with this condition in mind easily gives us $x=y=z=1$ as the only solution.

- 6 years, 6 months ago

Thats a very nice way of solving it. The thing with me is i have very few knowledge of theorems heh..

- 6 years, 6 months ago

AM-GM / Cauchy is a way to do it.

The other approach is to think of the geometric interpretation. We have a plane and a sphere, which intersect at the obvious point $(1, 1, 1)$. To show that this is the unique point of intersection, it suffices to show that the plane is tangential to the sphere.

Staff - 6 years, 6 months ago

Yes, I understand what you mean. But can anyone enlighten me on how do we prove a plane is tangential to a sphere? (What field of geometry is this? 3D??)

- 6 years, 6 months ago

There is a variety of ways you can approach it.

If you are familiar with 3-d vectors, then you find the vector that is perpendicular to the plane, and show that is is parallel to the vector perpendicular to the sphere at the point of contact.

If you are familiar with multivariable calculus, you just apply the definition of tangency, which is a restatement of the above.

If you are familiar only with 2-d concepts, then take any plane and show that the intersection with this plane produces a circle and a line which are tangential.

Staff - 6 years, 6 months ago

Oh.... I am not familiar with the above! :(

- 6 years, 6 months ago

How to rigorously prove that $\displaystyle x^2+y^2+z^2=x+y+z\implies x=y=z$?

- 6 years, 4 months ago

EDIT: real solutions. Oops. (ignore the rest)

Your conclusion is incorrect. One example of another solution is $\left(0,-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i,-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\right)$ and permutations.

However, proving that $(1,1,1)$ is the only real solution is an easy task. From my solution, this is when $\sqrt{c+1}$ has only a real value which is only when $c+1=0$ or $c=-1$; then, the solution follows to be $(1,1,1)$.

- 6 years, 6 months ago

SOLUTION #4

Let's define, for convenience, a function $f(x)$ that shows the last three digits of $171^x$. We have, by grinding, $f(3)=211, f(9)=931, f(27)=491, f(54)=81, f(162)=441, f(171)=571$ and finally $f(172)=\boxed{641.}$

- 6 years, 6 months ago

sorry, I cannot understand this grinding.

- 6 years, 6 months ago

Explanation:

$171^3 = 5000211$, so the last three digits of $171^3$ are $211$ ($f(3)=211$). $171^9 = (171^3)^3 = 125015825667824393931$, which has the same last three digits as $211^3 = 9393931$, so $f(9)=931$. ...etc...

- 6 years, 6 months ago

okay! got it, well thanks!

- 6 years, 6 months ago

in the second step can we only do cubing/sq. of the last digits??

- 6 years, 1 month ago

# solution 7

Let the sides be a,a+1,a+2 Let the angle opposite side to a be A,opposite to b be a+1,and side opposite to c be a+2 By using cosine rule, cosA=a+5/2a+4,cosB=a+1/2a and cosC=a-3/2a Using cos2x=2cos^2X-1 We get a=4 for A=2C a=1 for A=2B But a cannot be 1 as difference of two sides should be less than the third side Therefore sides are 4,5 and 6

- 6 years, 6 months ago

SOLUTION #3

Similarly to solution #2, we make the polynomial $P(x)$ and obtain $a=-3$ and $b=3$. However, we cannot tell what $c$ is.

Thus, our polynomial is $P(x)=x^3-3x^2+3x+c$ or $P(x)=(x-1)^3+c+1$

Solving for $x$, we obtain $x-1=-\sqrt{c+1}$ or $x=1-\sqrt{c+1}$

Note that there are 3 distinct values of $\sqrt{c+1}$. These three values correspond to the three solutions.

- 6 years, 6 months ago

The question asks for real solutions.

Staff - 6 years, 6 months ago

Aww darn oops. But I can still generate the real solution by letting $c=-1$.

- 6 years, 6 months ago