# Tag Info

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Find Terry Tao's blog. He talks about his experience of learning at different levels of mathematics education. Among other insights, he writes how patterns from competition problems he later discovered to be examples of more general, deep and beautiful results. What I took away from all that was that while solving competition problems isn't directly ...

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I am not one thousandth the mathematician that Terry Tao is, but my own feeling is rather different. I had a college classmate who was far better at competition mathematics than I was, and when we went to grad school (together), he seemed good at following a prescribed path, but not so good at striking out on his own. In later professional life, he made no ...

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Being successful at Olympiad mathematics is certainly correlated with being successful in later studies and research, but there is no implication in either direction. This is what you would expect a priori: coming up with creative ideas is a part of the work of a research mathematician, but by no means the only (and arguably not the most important) part. ...

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You can always add up to 100 if 8 are removed. To add up to 100 you can use 4 pairs of numbers that add to 25. There are 12 such pairs. If you remove 8 numbers you can only eliminate at most 8 such pairs. Thus at least 4 such pairs will always be left. Thus 4 pairs will always add to 100.

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If two functions have the same output and the same domain, they are the same function. You know that $g(x)=kx+n$, and that $g^{-1}(x) = g(x)=kx+n$. Now, using the properties of inverse functions, you also know that for each $x$, you have $x = \mathrm{id}(x) = (g\circ g^{-1})(x) = (g\circ g)(x)=g(g(x))$, and this equation should give you a lot of ...

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Indeed, no polynomial with degree $3$ can satisfy this property because it has to have a negative value at some point. Also, it is clear that (by taking very large $x$), the degree of this polynomial must be $2,3$ or $4$. So this polynomial has to have a degree of $2$ or $4$. Case 1. deg$p$ is $2$. Let $p(x)=ax^2+bx+c$. Since $0 \le p(0)=c \le 1$, we ...

3

Hint: If you consider your equation $x+3y+9z+27w=95$ modulo $3$. You will get $$x\equiv 2 \pmod{3}.$$ Based on the fact that $x \in \{1,2,3\}$, you get $x=2$. For $y$ try modulo $9$ and so on. Hopefully you can handle the rest.

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Here's another approach. First write out what $h(h(x))$ is $$h(h(x))=h\left(\frac{ax+b}{bx+c}\right)=\frac{a\left(\tfrac{ax+b}{bx+c}\right)+b}{b\left(\frac{ax+b}{bx+c}\right)+c}=\frac{a(ax+b)+b(bx+c)}{b(ax+b)+c(bx+c)}.$$ Next set this equal to $x$ and clear out the numerator to get $$a(ax+b)+b(bx+c)=x(b(ax+b)+c(bx+c))=b(a+c)x^2+(b^2+c^2)x,$$ and moving ...

3

We have three cases: $a = b$, $a > b$, $a < b$. If $a = b$, we have $a ^ {a ^ 2} = a ^ a$, which gives $a ^ 2 = a$ for $a \geqslant 1$, that is, $a = 1$ so $b = 1$: $(a, b) = (1, 1)$. Observe now the lemma below: Lemma 1. $a$ and $b$ are made up of the same primes. Proof. Let a = p_1 ^ {r_1} p_2 ^ {r_2} \cdots p_n ^ {r_n} = \prod_{i = 1}^{n} p_i ^ ... 2 Let's say that you guess a=c, e.g. by observing equations (2) and (3) as you have them above. Then from equations (5) and (6) you may see that x=y works. For a=c and x=y equation (2) becomes equal to equation (3) and (5) equal to (6), and so you get the system of equations: \begin{align*}2a^2+b^2+2x^2&=1\\ ax^2+bx^2+a\lambda&=0\\ ... 2 For the first this is the solution 2 For even n=2k, take a shape formed by a column of k+1 squares next to one of k-1 squares; for odd n=2k+1, k+1 next to k. Assemble two copies to form a rectangle of size 2\times n. Assemble n\times2 such pairs to form a square of size 2n\times2n. Then arrange n squares to form the original shape, for a total of 4n^3 copies. 2 Hint: The 13 coins from bags 1,2,3 weigh from 13 through to 39 ounces so the 27 coins from bag 4 weigh from ... through to ... and so must weigh ... each and ... in total, leaving ... for the 13 coins. The 4 coins from bags 1,2 weigh from 4 through to 12 ounces so the 9 coins from bag 3 weigh from ... through to ... and so must ... 2 Hint : No polynomial with degree 3 can have this property because it has negative values. It is easy to find out the polynomials with degree 2 and 4 satisfying the given bounds. Note that x^2>x^4 for 0<x<1 1 Since \gcd{(3n + 2, 3)} = 1 we conclude that b^2=3n+2 and a^2−ab=3. There is another possibility: b^2=-(3n+2) and a^2−ab=-3. And these conditions can be satisfied, for example, with a=1, b=4 and n=-6. So this proof is incorrect. 1 [I found this solution collaboratively with someone else offline.] \def\nn{\mathbb{N}} \def\rr{\mathbb{R}} Let T(n) = ( \text{The theorem is true for any length-n sequence from \rr} ), for any n \in \nn. If T(n) is false for some n \in \nn: Let m \in \nn be the minimum such that T(m) is false [by well-ordering]. Let ... 1 You get a tractable recurrence if, instead of considering where the largest number is, you look at the smallest number. The indexing will seem strange here, but bear with me. Let f(n,l,r) denote the number of permutations of \{0,\ldots,n\} where l+1 elements are visible from the left and r+1 are visible from the right. Clearly f(n,l,r)=0 if ... 1 Starting from one of such points on the plane draw a rectangle whose sides are m and n. Let R_1 be the total area of black squares and R_2 be that of white squares, in the rectangle. Since by drawing a diagonal we can divide the rectangle into two identical right-angled triangles as described, we have R_1 = 2S_1 and R_2 = 2S_2. Then, |R_1 - R_2| ... 1 Observe that k f (2k) < f (4k) \leqslant 2k f (2k) for k \geqslant 4 a power of 2. We use induction on n to prove the inequality. If n = 3, then it is obvious that2^{3^2/4} < f(8) = 10 < 2^{3^2/2}.$$Suppose the inequality holds for n = m, we now prove it for n = m + 1. Since f (2^m) < 2^{m^2/2} and using f (4k) \leqslant 2k ... 1 If you set x' = x - 1, y' = y - 1, z' = z - 1, and w' = w - 1, then x', y', z', w' \in \{0, 1, 2\}, and x+3y+9z+27w=95 implies that$$ 55 = 3^3 w' + 3^2 z' + 3y' + x'.$$So basically you are being asked to convert the decimal number 55 to base three. The answer is$$ 55_{10} = 2001_3, $$which implies that (w', z', y', x') = (2, 0, 0, 1) ... 1 There are some great solutions here, but there is a far simpler way of approaching this problem. Address this problem starting with the variables of highest coefficient. Thus, we clearly start with w. Notice that if you plug in 2 for w, you have 54 oz. Plugging in 3 for the remaining variables, your total weight comes up short at only 93 oz. ... 1 Here is some R code that solves your problem: #Matrix for saving results mat <- matrix(NA,ncol = 1,nrow = 4) colnames(mat) <- c("x","y","z","w") #Loop through all possible values for (x in c(1,2,3)){ for (y in c(1,2,3)){ for (z in c(1,2,3)){ for (w in c(1,2,3)){ if(x+3*y+9*z+27*w == 95){ mat <- cbind(mat,c(x,y,z,w)) ... 1 HINT:$$h(h(x))={1 \left( {\frac {a \left( ax+b \right) }{bx+c}}+b \right) \left( { \frac {b \left( ax+b \right) }{bx+c}}+c \right) ^{-1}} $$1 Let A(0,10),B(0,0),C(15,0),P(p,q) where 0\lt q. Then, solving$$p^2+q^2=12^2\quad \text{and}\quad (p-15)^2+q^2=9^2$$gives$$p=\frac{48}{5},\quad q=\frac{36}{5}.$$So,$$PA=\sqrt{\left(\frac{48}{5}\right)^2+\left(10-\frac{36}{5}\right)^2}=10.$$1 The trick is the following. Let ON \cap AB = P'. It suffices to show that P' lie on the circumcircle of BOM. After that, we will have P'=P, and then we will have P lie on ON. Let \angle BAM = a. We begin angle chasing. We have$$\angle BP'O + \angle BMO = (a+ \angle ANO) + (\angle BMA - \angle OMA) = a+90+(180- \angle B - a) -(90 - \angle ...

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