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Topic: Triangular area (Read 1580 times) |
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NickH
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Triangular area
« on: Oct 8th, 2002, 3:11pm » |
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In triangle ABC, produce a line from A to BC, meeting at D, and from B to AC, meeting at E. Let AD and BE meet at X.
If area AXB is 1 sq. unit larger than AXE, one sq. unit smaller than BXD, and 1/143 the area of ABC, what is the area of ABC?
Nick
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| « Last Edit: Oct 23rd, 2003, 7:35pm by Icarus » |
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NickH
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Re: NEW PUZZLE: Triangular area
« Reply #1 on: Nov 3rd, 2002, 7:34am » |
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Nobody has yet submitted a solution for this puzzle. I'm quite sure that's not because it is too difficult! Perhaps it is not sufficiently interesting? Perhaps I introduced too many irrelevant factors (i.e., the differences and ratios of areas) that obscured the elegance of the solution?
Here is a less "dressed up" version of the puzzle.
Find the area of quadrilateral CDXE in terms of the areas of triangles AXE, AXB, and BXD.
Nick
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TimMann
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Re: NEW PUZZLE: Triangular area
« Reply #2 on: Nov 3rd, 2002, 7:11pm » |
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My guess as to why no one has worked on this one is that there isn't a diagram, and we're all too lazy to read the puzzle carefully and draw our own. 
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Garzahd
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Re: NEW PUZZLE: Triangular area
« Reply #3 on: Nov 4th, 2002, 2:40pm » |
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Well, your less-dressed-up version seems trivial; just say that AXB has an area of n, then
n + n+1 + n-1 + CDXE = 143n
So CDXE = 140 * AXB. Is that all that you wanted? Or is there only one n for which the equation works out?
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NickH
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Re: NEW PUZZLE: Triangular area
« Reply #4 on: Nov 4th, 2002, 2:55pm » |
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Garzahd,
"Find the area of quadrilateral CDXE in terms of the areas of triangles AXE, AXB, and BXD." is the complete statement of the less dressed up version. That is, the areas of the three triangles are arbitrary, say, a, b, c, respectively. The task is to find the area of the quadrilateral in terms of a, b, and c.
Having determined the area of the quadrilateral, the constraints in the original version permit only one value of n.
Nick
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NickH
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Re: NEW PUZZLE: Triangular area
« Reply #5 on: Nov 7th, 2002, 12:22pm » |
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The answer, but not the solution, is:
ac(a + 2b + c)/(bČ - ac).
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| « Last Edit: Nov 7th, 2002, 12:23pm by NickH » |
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rugga
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Re: NEW PUZZLE: Triangular area
« Reply #6 on: Nov 9th, 2002, 2:58am » |
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Nick,
I found it interesting but challenging enough that it took me a couple days to work it out, by which time I saw you'd already posted the formula. I was able to get that by doing a page of algebra and some strategic shear transformations of the triangle to make the math simpler. Your earlier post implied it wasn't supposed to be that hard? Is there some trick I missed or am I just algebraically challenged?
rugga
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NickH
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Re: NEW PUZZLE: Triangular area
« Reply #7 on: Nov 9th, 2002, 10:05am » |
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rugga,
I don't think it's a trivial problem, which is why I placed it in the medium section. But then it's not really up there with some of the hard problems.
Here's my solution.
Let triangle BXE have area a, BXC have area b, and CXD have area c.
Consider BXE and BXC, with bases EX and XC, respectively.
The triangles have common height; therefore EX/XC = a/b.
Similarly, considering BXC and CXD, with respective bases BX and XD, BX/XD = b/c.
Now draw line AX. Let triangle AXE have area x and AXD have area y.
Consider AXB and AXD, with bases BX and XD, such that BX/XD = b/c.
Since AXB and AXD have common height, we have (a+x)/y = b/c.
Similarly, considering AXE and AXC, with bases EX and XC, x/(y+c) = a/b.
Cross-multiplying gives: by = cx + ac, bx = ay + ac.
Solving: x = ac(a + b)/(bČ - ac), y = ac(b + c)/(bČ - ac). (Or derive y from x, by symmetry.)
Therefore the area of quadrilateral AEXD is ac(a + 2b + c)/(bČ - ac).
Nick
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| « Last Edit: Nov 9th, 2002, 12:37pm by NickH » |
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