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tototo
13th June 2008 - 06:49 PM
ABCD is a square with side of length 4.

X is a point of CB,Y is a point of CD such that m∠XAY=45


find

The perimeter of the Triangle AXY
Dabeer
13th June 2008 - 09:06 PM
QUOTE (tototo+Jun 13 2008, 02:49 PM)
ABCD is a square with side of length 4.

X is a point of CB,Y is a point of CD such that m∠XAY=45


find

The perimeter of the Triangle AXY

Not enough information.

Triangle AXY varies between ∠AXY=∠AYX = 67.5 and ∠AXY==>45, ∠AYX==>90

This gives perimeters between ~= 11.31 and ~= 13.66
tototo
13th June 2008 - 09:10 PM
ABCD is a square with side of length 4.

X is a point of CB,Y is a point of CD such that m∠XAY=45


find

and The perimeter of the Triangle CXY
prometheus
13th June 2008 - 09:35 PM
You're a bit of a pilchard really aren't you?
mr_homm
13th June 2008 - 11:27 PM
The first posting was probably a typographical error. The second version of the question is quite solvable, and the answer is 8. It is actually a rather nice puzzle, so I will leave it for others to enjoy. If tototo wants to see the detailed solution, besides the answer, I will post it.

--Stuart Anderson
Dabeer
14th June 2008 - 12:21 AM
QUOTE (mr_homm+Jun 13 2008, 07:27 PM)
The first posting was probably a typographical error. The second version of the question is quite solvable, and the answer is 8. It is actually a rather nice puzzle, so I will leave it for others to enjoy. If tototo wants to see the detailed solution, besides the answer, I will post it.

--Stuart Anderson

Yes, I agree the second version has an easy solution.
tototo
14th June 2008 - 05:33 PM
QUOTE (mr_homm+Jun 13 2008, 11:27 PM)
The first posting was probably a typographical error. The second version of the question is quite solvable, and the answer is 8. It is actually a rather nice puzzle, so I will leave it for others to enjoy. If tototo wants to see the detailed solution, besides the answer, I will post it.

--Stuart Anderson



mr_homm


yes i want
mr_homm
15th June 2008 - 04:30 PM
First, look at this image:

User posted image: User posted image

The triangle AC'X' sitting on top of the square is the same as ACX in the lower left.

Now notice that AX'Y is mirror congruent to AXY (because they share AY, AX = AX', and a+b = 45).

Therefore, XY = X'Y.

But BX' = CX = 4-m, so that X'Y = n+4-m.

Therefore, XY = n+4-m.

Since YD = 4-n, this gives the perimeter of XYD as XY + YD + DX = (n+4-m) + (4-n) + m = 8.

Hope that helps!

--Stuart Anderson
tototo
16th June 2008 - 06:58 PM
thank you
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