GYanggg

We first use \(y^3-x^3=(y−x)(x^2+xy+y^2)\)

\(x^2 + xy + y^2 = y^3 - x^3 \\ x^2 + xy + y^2=(y-x)(x^2+xy+y^2)\\ y=x+1 \)

From the second equation, we get:

\(yz + 1 = y^2 + z\\ (x+1)z+1=(x+1)^2+z\\ xz+z+1=x^2+2x+1+z\\ x^2+2x-xz=0\\ x^2+x=zx-x\\ x(x+1)=x(z-1)\\ x+1=z-1\\ y=z-1 \)

Concluding:

\(z=y+1\\ y=x+1.\\ \therefore z>y>x\)

They are also consecutive integers. 

In this list, there are \(1998-1732+1\) rows, and therefore, 267 ordered triples. 

I hope this helped,

Gavin

Euclid's proof:

Suppose that \(p_1=2 < p_2 = 3 < ... < p_r\) are all of the primes. Let \(P = p_1\cdotp_2...pr+1\) and let p be a prime dividing P;

then p can not be any of \(p_1, p_2, ..., p_r,\) otherwise p would divide the difference \(P-p_1p_2...p_r=1\), which is impossible.

So this prime p is still another prime, and \(p_1, p_2, ..., pr\) would not be all of the primes.

\(1. \text{If}\ f(x) = x^2\ \text{and}\ g(x)=\ln(x), \text{compute } f'(1) + g'(1)\)

\(\text{We compute}\ f'(x) = 2x\ \text{and}\ g'(x)=\frac1x,\ \text{so plugging in 1 for both gives us}\ \boxed3. \)

\(2. \int_0^4\frac{dx}{\sqrt{|x-2|}}\)

\(\text{Since the function } \sqrt{|x-2|}\text{ discontinues at } x=2,\\ \text{we can split the integral into two parts and compute separately.}\)

\(\int_0^2\frac{dx}{\sqrt{|x-2|}}=\int_0^2\frac{dx}{\sqrt{2-x}}=-2\sqrt{2-x}|_0^2=2\sqrt2\)

\(\int_2^4\frac{dx}{\sqrt{|x-2|}}=\int_2^4\frac{dx}{\sqrt{x-2}}=2\sqrt{x-2}|_4^2=2\sqrt2\)

\(2\sqrt2+2\sqrt2=\boxed{4\sqrt2}\)

​​​

Method 1: Logic

One the sphere, there exists a point A and its antipodal point, A'. This means that A' is the point with the farthest distance from A, the 3D version of the diameter, which means when you connect A and A', the line contains the center of the sphere. After one chooses the three points, W, X, Y, the region to place Z is the spherical traingle W'X'Y' opposite WXY. It is only then, the tetrahedron would contain the center of the sphere. This means that the probability the tetrahedron would contain the center of the sphere is the expected area of the triangle W'X'Y', which has the same expected area of WXY and 7 other triangles: WXY, W'XY, W'X'Y, W'X'Y', WX'Y, WX'Y', WXY', W'XY'. Since there are 8 such triangle, and the total area is set 1, the expected area is 1/8.  The probabilty is 1/8. 

Method 2: Mathematical Proof

We can proof the logic using the notion of barycentric coordidates: https://nicoguaro.github.io/posts/putnam_prob/ 

I think the website does a great job explaining the process, better than I can at least. If you have any questions though, about the website or my logic, ask away!

I hope this helped,

Gavin

Hey Lightning, \(a=4\)

\(4x^2+12x+9\\ =(2x+3)^2\)

I used: \((x+y)^2=x^2+2xy+y^2\)

I saw the nine, so I know \(y=3\)

\(2xy=12\), and \(y=3\), so \(x=2\)

\(2^2=4, a=4\)

I hope this helped,

Gavin

This is factorable:

\(-x^2-14x-24=0\\ =(−x−2)(x+12)=0\\ x_1=-2;x_2=-12\)

I'm assuming that there is a equal zero at the end of the equation. 

I hope this helped,

Gavin

Hey Guest,

I have slept on it, and finished the solution. 

First, we use \(x = \tan θ\)

We can rewrite the desired integral as \(\int_0^{\pi/4}\log(\tan(\theta)+1)d\theta\)

Also:

\(\log(\tan(θ) + 1) = \log(\sin(θ) + \cos(θ)) − \log(\cos(θ))\)

Note that: \(\sin(θ) + \cos(θ) = \sqrt2 \cos(π/4 − θ).\)

Rewriting the integral, we get:

\(\frac12\log(2) + \log(\cos(π/4 − θ)) − \log(\cos(θ)).\)

Over the interval \([0, π/4]\), the integrals of \(\log(\cos(θ))\ \text{and}\ \log(\cos(π/4 − θ)) \) are equal, so their contributions cancel out. The answer you seek is \( π \log(2)/8. \)

Gavin

Edit: I have done some research, and googled Putnam competition. 

This problem is A5 of the 66th annual competition. 

The full solution is here: http://math.hawaii.edu/home/pdf/putnam/2005.pdf.

Hey tertre!

You are absolutely correct if someone wants to find the equation of the line with two given points. 

However, the equation of the line is already given to us, we just need to identify the slope. 

:)

This is an equation for a line in slope-intercept form. 

In the line: \(y=kx+b\), the slope is \(k\) and the y-intercept is \(b\), make sure you understand why. :)

I believe we don't need to use a table of values. 

The question is saying, when you plug a number, \(x\), into the two functions, the two results are the same. 

Let's call the number, \(n\). 

When we plug \(n\) into \(f(x)\), we get \(n^2 - 6n + 8.\)

When we plug \(n\) into \(g(x)\), we get \(n - 2\)

The problem states that these two values are equal, so we can write the equation: 

\(n^2-6n+8=n-2\\ n^2-7n+10=0\\ (n-5)(n-2)=0\\ n_1=5\ n_2=2\)

I hope this helped,

Gavin