Saturday, June 18, 2016

Two problems I'm looking at

I was reading a fun post over @ where Lisa is brainstorming problem sets. She's inspired me to collate the material we did this year.  But in the meantime here's a teaser:

I don't usually include algebra problems but I ran into this one last night and its too good not to keep for future use when I have a group of older kids.  At first this problem seems like its missing enough information.  For example, you can't directly solve for $$x_1 .. x_7$$ since there are only 3 equations. However, if you assume that some linear combination of the 3 equations will equal the  target i.e.

$$f(x_1..x_2) = x_1 + 4x_2 + 9x_3  + 16x_4 + 25x_5 + 36x_6 + 49x_7 \\
g(x_1..x_2) = 4x_1 + 9x_2 + 16x_3  + 25x_4 + 36x_5 + 49x_6 + 64x_7 \\
h(x_1..x_2) = 9x_1 + 16x_2 + 25x_3  + 36x_4 + 49x_5 + 64x_6 + 81x_7$$


$$a \cdot f()  + b \cdot g() + c \cdot h() = 16x_1 + 25x_2 + 36x_3  + 49x_4 + 64x_5 + 81x_6 + 100x_7$$

This reduces to 7 equations in 3 unknowns The first and easiest 3 of them are:

$$(a + 4b +  9c) x_1 = 16x_1$$ $$(4a + 9b + 16c) x_2 = 25x_2$$ $$(9a + 16b + 25c) x_3 = 36x_3$$

Then these can be solved directly and it only remains to check if the solution (1,-3,3) works for the other 4 terms.

Or can you generalize and confirm:  $$ x^2  -3 (x + 1)^2 + 3(x+2)^2 = (x+3)^2$$

What's interesting is to examine the general problem afterwards for instance if there were only 2 equations is there a linear combination that works (nope)? What about four (this has multiple solutions)? I also find a hint of the third row of Pascal's triangle in the solution but haven't looked into whether that's a coincidence.

Update: its easiest to solve the general equation above  $$ ax^2  + b(x + 1)^2 + c(x+2)^2 = (x+3)^2$$ This reduces to:

$$a+b+c = 1, b+2c = 3, b + 4c = 9$$
Interestingly in 4 cubic terms the solution  to $$ ax^3 + b (x + 1)^3+ c(x+2)^3 +d(x+3)^3 = (x+4)^3$$  is {3,4,-6,4}. Once again close to the 4th term in the triangle and in 5 quad terms the solution is (1,-5,10,5). The binomial theorem is all over this problem so its not completely surprising but I'll keep thinking about it a bit more.

Folding problems are always fun. This one stands out because it uses the notion of proportionality twice in both direction i..e if you draw a line from one vertex of a triangle to somewhere in its base the ratio of areas of the two triangles are the same as the ratio of the two bases (since they share the same height).

No comments:

Post a Comment