Solutions_2007

This page contains solutions of former potds. You are explicitly encouraged to post your own solutions to old or new potds as comments to this site. An administrator will incorporate them into this page later on (assuming they are correct and stated in a reasonably accessible manner). Please note that in the case of essentially identical solutions only one will be incorporated into the page (usually the first one).

2007-12-28 by kmh
Consider a square ABCD with a point P inside. Prove: $\angle BAP = \angle ABP=15^\circ \Rightarrow \triangle PCD \text{ is equilateral}$ .

Solution by Sniffle
Let Q be a point inside the ABCD square so that CQD is equilateral (there is only one point with these propreties). QD = DC = CQ = AD = BC and thus the triangles AQD and BQC are isosceles triangles. <ADC = 90 and <QDC = 60 => <ADQ = <QCB = 30. <ADQ = 30 and triangle ADQ is isosceles => <DAQ = 75 (by analogy: <QBC = 75). <DAB = 90 and <DAQ = 75 => <QAB = 15 (by analogy: <QBA = 15). And thus, the triangle QAB is isosceles and the base angles are of 15 degrees, which means it satisfies the same propreties as P. Since the construction of P is unique it means that Q = P (and obviously) that P = Q and thus DPC = equilateral

2007-12-15 by yaroslav
Let A be the adjacency matrix of a graph with k triangles. Find $trace(A^3)$ . Hint, first find combinatorial interpretation of $(A^k )_{ij}$

Solution by yaroslav
$(A^k)_{ij}$ gives the number of paths of length $k$ from $i$ to $j$ in a graph with adjacency matrix $A$ . Hence to find $\text{tr}(A^k)$ we need to find total number of cycles of length 3. There are 6 possible 3-cycles for each triangle and $k$ triangles, hence $\text{tr}(A^3)=6k$

Problem 2007-12-13 by kmh
Compute the probability that in tennis tournament of 8 players the 2nd best player becomes the runner up.
(Source: F. Mosteller: 50 challenging problems in probability)

Solution by kmh
Consider all permutation of he 8 players, there are 8!. Now we count those of them who will lead to the 2nd best player becoming runner up. Every permutation that has the best player and the 2nd best player in different halfs will do that.Let’s assume the 2nd best player is in the 1st half and the best player in the 2nd half. There are $\binom{6}{3}\cdot (4!)^2$ such permutations, since we have $\binom{6}{3}$ options to choose the remaining 3 players for the 1st half and after the players are chosen, they can still permute in their assigned halfs. With the 2nd best player is 2nd half rather than in the 1st i get an identical computation, hence there are $2\cdot \binom{6}{3}\cdot (4!)^2$ permutations with the 2nd best player as runner up. This yields as probability $p=\frac{2\cdot \binom{6}{3}\cdot (4!)^2}{8!}=\frac{4}{7}$ .

Problem 2007-12-9 by evilmasterer
Consider the following construction with $|AC|=15cm,\, |AB|=10cm,\, \angle CAB=90^\circ$ . Compute the gray area.

Solution by kmh

solution 1:
D.A,C,I,E and F,A,B,J,G form intercept theorem configurations or alternatively similar triangles ( $\triangle ACI \sim \triangle DCE, \, \triangle BAJ \sim \triangle BFG$ ). This allows to compute $AI|=\frac{|CA|}{|CD|}\cdot |ED|=6$ , $|AJ|=\frac{|BA|}{|BF|}\cdot |FG|=6$ , $|BI|=|BA|-|AI|=9$ , $|JC|=|AC|-|AJ|=4$ Note now that $\triangle EHB,\, \triangle HJC$ are similar and so are $\triangle BIH,\, \triangle HGC$ (or alternatively they are an intercept theorem configuration). This yields the following equations for the area: $\frac{10^2}{9^2} =\frac{y+\frac{4\cdot 10}{2}}{x}$ , $\frac{4^2}{15^2} =\frac{y}{x+\frac{9\cdot 15}{2}}$ Solving the system of equations yields x=21.32 for the wanted area.
solution 2:
Note that $x=|\triangle ABC| -|\triangle ACI|-|\triangle HCB|$ . First compute $|AI|,\,|IB|,\,|JC|,\,|AJ|$ as in solution 1, then compute $|IC|=\frac{|CA|}{|CD|} \cdot \sqrt{|ED|^2+|CD|^2} =2\sqrt{13}$ ( $\triangle ACI \sim \triangle DCE$ ). Now we get a system of 2 equations to compute $|HC|$ : $|IH|+|HC|=|IC|=2\sqrt{13}$ , $\frac{|IH|}{|HC|}=\frac{|IB|}{|CG|}=\frac{9}{10}$ This yields $|HC|=6.138$ . Now we compte $\angle HCB=\angle ACB - \angle ACI=arctan(\frac{15}{10})-arctan(\frac{6}{10})=25.346^\circ$ . Now we got all the data we need for the area computation from the beginning: $x =|\triangle ABC| -|\triangle ACI|-|\triangle HCB|$ $=\frac{10\cdot 15}{2}-\frac{6\cdot 10}{2}-\sqrt{10^2+15^2}\cdot 6.138 \cdot \sin(25.346^\circ)=21.32$
solution 3:
Place A in the origin of a coordinatesysten, then we have B=(0,15), C=(10,0), G=(10,-10), E=(-15,15) and H is the intersection of the line through C,E ( $y-0=\frac{15-0}{-15-10}\cdot (x-10)$ ) and the line through B,G ( $y-15=\frac{-10-15}{10-0}\cdot(x-0)$ ). This yields H=(4.737,3.158) and hence the area of the triangle is: $\frac{1}{2} \begin{vmatrix} 0 &15 &1\\ 0 &6 &1\\ 4.737 &3.158 &1 \end{vmatrix} =21.32$
solution 4:
Compute B,I,H and the lines through C,E and B,G as in solution 3, then the area is $\int_0^{4.737} \frac{-10-15}{10-0}\cdot x+15 -(\frac{15-0}{-15-10}\cdot (x-10))\,dx=\int_0^{4.737}9-1.9x \,dx= 21.32$

Problem 2007-12-4 by R^^n
Compute $\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n exp(\frac{i}{n})$

Solution by kmh
$\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n exp(\frac{i}{n})$ $={\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n exp(\frac{1}{n})^i}$ $={\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n} \frac{exp(\frac{1}{n})^{n+1}-1}{exp(\frac{1}{n})-1}}$ $={\displaystyle \lim_{n\rightarrow \infty} \frac{exp(\frac{n+1}{n})-1}{\frac{exp(\frac{1}{n})-1} {\frac{1}{n}} }}$ $= \frac{\displaystyle \lim_{n\rightarrow \infty} exp(\frac{n+1}{n})-1 }{ \displaystyle \lim_{n\rightarrow \infty} \frac{exp(\frac{1}{n})-1}{\frac{1}{n}}} =\frac{e-1}{1}=e-1$

Problem 2007-11-17 Scottish Math Special by Karlo
I once knew a gal from Dundee Whose age had the last digit 3. The square of the first Was her whole age reversed. So what would the lady’s age be?

Solution by kmh inspired by Rainer Rosenthal
As 3x is known to contain only one square
the solution is not too hard but rather fair.
One computes without further tricks
that x can only be six.
Makes her sixty three – for all those who care.

Solution by jackal
Reversing the number representing the age should give us the square of a digit, which means the age either has 1 or 2 digits. Should it have one digits, the age should be 3, however, 3 is not a perfect square. For two digits, the age is of the form: a3; and we know that 3a = a^2. The only two digit perfect square starting with 3 is 36 and thus a = 6 (verifies: 36 = 6^2). Age: 63.

Problem 2007-11-11 by kmh
A drawer contains black and red socks. When 2 socks are drawn at random the probability for both of them being red is $\frac{1}{2}$ a) How many socks must the drawer contain at least ? b) How many socks must the drawer contain at least, if the number of black socks is even? (Source: F. Mosteller: 50 challenging problems in probability)

Solution by jackal
Let: r = the number of Red socks; b = the number of Black socks. Obviously: $r\cdot b \neq 0$ . Then we have the following: $\frac{{r \choose 2}}{{b+r \choose 2}} = \frac{1}{2}$ Equivalent to (after some computation): $\frac{r\cdot (r-1)}{(b+r)\cdot(b+r-1)} = \frac{1}{2}$ Step 1: Prove that r > b: Assume r <= b, then we have: $\frac{r\cdot (r-1)}{(b+r)\cdot (b+r-1)} \leq \frac{r\cdot(r-1)}{2\cdot r\cdot(2\cdot r-1)} < \frac{1}{2}$ False. And thus $r > b$ . Step 2: $\frac{r (r-1)}{(b+r) (b+r-1)} = \frac{1}{2}$ $\Leftrightarrow 2r^2-2 r =b^2+2 b r+r^2-b-r$ $\Leftrightarrow r^2-r(2b+1)-(b^2-b)=0$ and thus we have: $\Delta = (2 b+1)^2 + 4 (b^2-b) = 8b^2+1$ and: $r_{1,2} = \frac{2b+1 \pm \sqrt{8b^2+1}}{2}$ Step 3: Let’s assume $r = \frac{2b+1 - \sqrt{8b^2+1}}{2}$ It is easy to show that in this case r < b which contradicts the result from “Step 1”. This means that: $r = \frac{2b+1 + \sqrt{8b^2+1}}{2}$ Which clearly shows that r is directly proportional to b. Step 4: By looking at the expression of r, and keeping in mind r and b are natural numbers, it is trivial to conclude that $\sqrt{8b^2+1}$ is a natural number, which is equivalent to the fact that $8b^2+1$ is a perfect square. a) Minimum number of socks: For b = 1 => $8b^2+1 = 9$ which is a perfect square, and we find r = 3, which verify the initial condition and thus the minimum number of socks is 3 + 1 = 4. b) Minimum number of socks for b = even: b = 6 is the smallest even number for which $8b^2+1$ is a perfect square, and in this case: r = 15, and these verify the initial conditions. For b > 6, because r is directly proportional to b, r will be > 15 and thus the number of socks will be higher than 15+6=21. Answer: 21 (15 red, 6 black).

Problem 2007-11-9 by kmh
Show that for every prime p with p > 17 3 divides $p^2+2$ .

Solution by rafno
By fermat’s little theorem a^(p-1) = 1(mod p) for a not divisible by p… then for p=3 a^2 = 1 (mod 3) or a^2 = -2 (mod 3) for every a not divisible by 3, in particular for a prime, bigger than 17… QED

Solution by dmhouse
Working mod 3, we see that n^2 + 2 is 0 for both 1 and 2, so in general 3 | n^2 + 2 for all integers n that aren’t multiples of 3, and in particular all primes > 17.

Problem 2007-11-3 by evilmasterer
Consider 3 identical circles with radius r placed in such a way, that the center of each circle lies on the 2 other circles (see picture). Determine the red area in the center.

Solution by kmh:
By connecting the 3 circle centers we’ll get an equilateral traingle with a side length of r. The area A we are looking for can now be obtained by adding the areas of 3 $60^\circ$ circle sectors (one from each center). Doing that however we’ve added the area of the equilateral triangle thrice rather than once, therefore we have to deduct 2 triangles. Together we have $A=3\cdot\pi r^2\cdot \frac{60}{360}-2\cdot\frac{\sqrt{3}}{4}r^2=\frac{r^2(\pi-\sqrt{3})}{2}$

Problem 2007-11-2 by an unkown chatter
Show : $|x-2|\leq0.02\quad \Rightarrow \quad |\sqrt{x}-\sqrt{2}|\leq 0.01$

Solution by kmh:
Since $x-2=(\sqrt{x}-\sqrt{2})\cdot(\sqrt{x}+\sqrt{2})$ we have $|\sqrt{x}-\sqrt{2}|=\frac{|x-2|}{|\sqrt{x}+\sqrt{2}|}\leq\frac{0.02}{|\sqrt{1.98}+\sqrt{2}|}\leq\frac{0.02}{2}=0.01$

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