p1. Nurbaya's rectangular courtyard will be covered by a number of paving blocks in the form of a regular hexagon or its pieces like the picture below. The length of the side of the hexagon is $ 12$ cm. Installation of other paving blocks or pieces thereof so that all fully covered page surface. To cover the entire surface The courtyard of the house required $603$ paving blocks. How many paving blocks must be cut into models $A, B, C$, and $D$ for the purposes of closing. If $17$ pieces of model $A$ paving blocks are needed, how many the length and width of Nurbaya's yard? Count how much how many pieces of each model $B, C$, and $D$ paving blocks are used. p2. Given the square $PQRS$. If one side lies on the line $y = 2x - 17$ and its two vertices lie on the parabola $y = x^2$, find the maximum area of possible squares $PQRS$ . p3. In the triangular pyramid $T.ABC$, the points $E, F, G$, and $H$ lie at , respectively $AB$, $AC$, $TC$, and $TB$ so that $EA : EB = FA : FC = HB : HT = GC : GT = 2:1$. Determine the ratio of the volumes of the two halves of the divided triangular pyramid by the plane $EFGH$. p4. We know that $x$ is a non-negative integer and $y$ is an integer. Define all pair $(x, y)$ that satisfy $1 + 2^x + 2^{2x + 1} = y^2$. p5. The coach of the Indonesian basketball national team will select the players for become a member of the core team. The coach will judge five players $A, B, C, D$ and $E$ in one simulation (or trial) match with total time $80$ minute match. At any time there is only one in five players that is playing. There is no limit to the number of substitutions during the match. Total playing time for each player $A, B$, and $C$ are multiples of $5$ minutes, while the total playing time of each players $D$ and $E$ are multiples of $7$ minutes. How many ways each player on the field based on total playing time?
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Tags: algebra, geometry, number theory, combinatorics, indonesia juniors
04.11.2021 23:48
4 is the IMO 2006 problem.
05.11.2021 03:12
S4 Let $2^x=Y\Longrightarrow 1+2^x+2^{2x+1}=K^2\Longrightarrow 2Y^2+Y+1-K^2=0\Longrightarrow$ $$Y=\frac{-1+\sqrt{1-8(1-K^2)}}{4}=\frac{-1+\sqrt{8K^2-7}}{4}\Longrightarrow$$ $K=2 \Longrightarrow Y=\frac{-1+\sqrt{8.4-7}}{4}=1 \Longrightarrow x=0$ works $(1+1+2=4)$ However, it is no "non-negative integer". $K=6 \Longrightarrow Y=\frac{-1+\sqrt{8.6^2-7}}{4}=4$ but $x=2 $ does not work $(1+2^2+2^5=37)$ $K=23 \Longrightarrow Y=\frac{-1+\sqrt{8.23^2-7}}{4}=\frac{-1+\sqrt{65^2}}{4}=16 \Longrightarrow Y=2^x=2^4\Longrightarrow x=4$ works $(1+2^4+2^9=23^2)$ The only pair is thus $(4,23)$. There could be other solutions, which can be found by solving the Pell“s equation $8x^2-y^2=7$ or $(4x)^2-2y=14$ and testing $Y$ (if it is a power of 2) in the title expression. However, doubts arise that IMO participants could arrive at the pair $(4,23)$ without the help of a calculator. There might be a simpler way .
05.11.2021 03:39
07.11.2021 00:07
07.11.2021 01:28
parmenides51 wrote: p2. Given the square $PQRS$. If one side lies on the line $y = 2x - 17$ and its two vertices lie on the parabola $y = x^2$, find the maximum area of possible squares $PQRS$ .
22.07.2022 11:13
BackToSchool wrote:
how about the permutation $5!$ ?