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riddles / easy / Re: Number of triangles 
Aug 27^{th}, 2021, 9:05am 
Started by navdeep1771  Last post by rmsgrey 
Do you count the degenerate traingle with sides 1004/502/502? If so, then I make it 87 Reasoning: hidden:  a right isosceles triangle has sides in ratio 2^{1/2}:1:1 and 2008/(2+2^{1/2}) is just over 588. Since any integer length for the matching sides will give an integer length for the remaining side, any integer from 502 to 588 will give an isosceles triangle with angle greater than a right angle.  
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3 
riddles / medium / Re: Maximum Separation 
Aug 13^{th}, 2021, 7:29am 
Started by navdeep1771  Last post by rmsgrey 
Answer: L_{max}=(v_{1}+v_{2})^{2}/(2(a_{1}+a[s ub]2[/sub])) Working: ::hidden:  Working in a frame where one of the bodies is stationary throughout (and ignoring relativistic corrections), the other body has initial velocity, u=v_{1}+v_{2}, and constant acceleration in the opposite direction, a=a_{1}+a_{2}. The change in separation between the two bodies, s, at a given time, t, is given by standard SUVAT formulae: s=utat^{2}/2 Since s reaches a maximum when the closing velocity, v=uat, is 0, we get: uat=0 t=u/a Substituting for t in the equation for s gives: s=u(u/a)  a(u/a)^{2}/2 =u^{2}/a  u^{2}/2a =u^{2}/2a and substituting back for u and a gives the answer above: L_{max}=(v_{1}+v_{2})^{2}/(2(a_{1}+a[s ub]2[/sub]))  :: edit: the subscript tags are behaving weirdly, and I can't get them to work properly 
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4 
riddles / medium / Re: Covering square with 2x1 rectangles 
Aug 11^{th}, 2021, 12:54pm 
Started by rmsgrey  Last post by Grimbal 
1 rectangle: obvious. I.e. the square must be covered by one rectangle. That rectangle must have area 1 and ratio 2:1. The rectangle must have dimensions sqrt(2) by sqrt(0.5). The diagonal of the rectangle is sqrt(2+0.5) = sqrt(2.5). This diagonal doesn't fit in the square with diagonal sqrt(2). So a rectangle with sufficient area doesn't fit in the square. (Or more simpler: the square isn't a 2:1 rectangle.) 3 rectangles: There are 4 corners and only 3 rectangles. One rectangle must fill two corners. That rectangle must have size is 1 x 0.5. The remaining area is also a 1 x 0.5 rectangle. And it must be filled by wo rectangles. Again, one rectangle must fill 2 corners of that area. This leaves only 2 ways, 1 x 0.5 or 0.5 x 0.25. One way doesn't give space for a third rectangle, the other leaves a 3:2 space which is not the right ratio. 4 rectangles: If one rectangle fills 2 corners, there is always one rectangle that must fill a whole side. That leaves few possibilities. The first rectangle would be 1 x 1/2. The second can only be 1/2 x 1/4. The third can be 1/2 x 1/4 leaving a 1/2 x 1/2 space or 3/4 x 3/8 leaving a narrow 3/4 x 1/8 rectangle. Both are the wrong ratio. So remains the case where each corner of the square is filled by a different rectangle. In that case, consider a rectangle that touches the center of the square. There must be one. That rectangle must reach from a corner to the center. It must have at least 1/2 in both dimensions, which actually means it must have dimensions 1 x 1/2. We are back to the previous case which didn't work. (Or in fact we contradict the hypothesis that each corner is filled by a different rectangle). I think that covers it. No pun intended. 
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