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Old 2006-03-08, 21:57   #1
davar55
 
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Default Geometry Puzzle #2

Within a 3x3 square ABCD (corners) there is a point P
whose distances from A,B,C, and D respectively are
sqrt2, sqrt5, sqrt8, and x. Find the value of x.
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Old 2006-03-09, 16:28   #2
mfgoode
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Thumbs up Geometry Puzzle

Excellent problem davar
Ans: sq. rt. 5 Hint : Dont try Heron's formula.
Mally

Last fiddled with by mfgoode on 2006-03-09 at 16:30
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Old 2006-03-11, 00:16   #3
mersenne
 
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AP+PC = sqrt2+sqrt8 = sqrt2+2sqrt2 = 3sqrt2
AC = 3sqrt2 =>P lies on AC.
From the congruent triangles APB,APD we get PD=PB=sqrt5
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Old 2006-03-14, 15:59   #4
Kees
 
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Default A little trick

Playing with Pythagoras: Let P be the given point, PX its orthogonal projection on AB and PY its orthogonal projection on AD. Let x=d(A,PX), y=d(A,PY).

Then we have

x^2+y^2 = 2 ( I)
x^2+(3-y)^2 = 5 ( II)
(3-x)^2 + (3-y)^2 = 8 (III)

and we want to find

(3-x)^2+ y^2 (IV)

We have the following formula: III=IV+II-I which leads to

IV=III-II+I = 8-5+2 =5. Taking square root gives the result


sorry for the format, do not know how to blacken text so wrote in white
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Old 2006-03-14, 16:48   #5
mfgoode
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Smile Geometry Puzzle

Quote:
Originally Posted by Kees
Playing with Pythagoras: Let P be the given point, PX its orthogonal projection on AB and PY its orthogonal projection on AD. Let x=d(A,PX), y=d(A,PY).

Then we have

x^2+y^2 = 2 ( I)
x^2+(3-y)^2 = 5 ( II)
(3-x)^2 + (3-y)^2 = 8 (III)

and we want to find

(3-x)^2+ y^2 (IV)

We have the following formula: III=IV+II-I which leads to

IV=III-II+I = 8-5+2 =5. Taking square root gives the result


sorry for the format, do not know how to blacken text so wrote in white
A neat trick but too long a solution.
Well lets say you are not as green as you're cabbage looking!
Mally
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Old 2006-03-14, 16:54   #6
Kees
 
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Default More generally speaking

we have the following result:

PA^2+PD^2=PB^2+PC^2

which follows directly from the given formula, but I suppose there might be a geometrical argument
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Old 2006-03-14, 17:03   #7
mfgoode
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Default Geometry Puzzle

Just draw the figure roughly. The answer pops out by mere inspection!
Mally
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Old 2006-03-14, 17:33   #8
Kees
 
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Default The general figure

I am not seeing it, I place a point somewhere in a rectangle and then I can just see by drawing the lines from the vertices to this point that this solves it all ?
Drawing lines parallel to AB and AD through P helps seeing the solution, but if that is considered too long...
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Old 2006-03-15, 16:13   #9
mfgoode
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Default Geometry Puzzle#2

Quote:
Originally Posted by Kees
I am not seeing it, I place a point somewhere in a rectangle and then I can just see by drawing the lines from the vertices to this point that this solves it all ?
Drawing lines parallel to AB and AD through P helps seeing the solution, but if that is considered too long...

Seriously Kees you are quite right. // lines do help. Maybe the problem is with your vision- using three 'seeings' in one para ! Perhaps you used the invisible colour and could not 'see' the problem at all?
Just joking Kees. How about a hint to your number?
Mally
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Old 2006-03-15, 22:27   #10
cheesehead
 
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Default

Kees,

Put " [ spoiler ] " (without the spaces!) before the text you wish to blackout, and " [ / spoiler ] " (without the spaces!) after that text.

Then, those of us using off-white backgrounds will not unwillingly see your text. :)

(P.S., nice avatar!)
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Old 2006-03-16, 08:41   #11
Kees
 
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Default Spoler test


so this should be hidden


thanks for the tip
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