mersenneforum.org Geometry Puzzle #2
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 2006-03-08, 21:57 #1 davar55     May 2004 New York City 5×7×112 Posts 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.
 2006-03-09, 16:28 #2 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 80416 Posts 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
 2006-03-11, 00:16 #3 mersenne   Jan 2005 3 Posts AP+PC = sqrt2+sqrt8 = sqrt2+2sqrt2 = 3sqrt2 AC = 3sqrt2 =>P lies on AC. From the congruent triangles APB,APD we get PD=PB=sqrt5
 2006-03-14, 15:59 #4 Kees     Dec 2005 22×72 Posts 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
2006-03-14, 16:48   #5
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
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

 2006-03-14, 16:54 #6 Kees     Dec 2005 C416 Posts 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
 2006-03-14, 17:03 #7 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22·33·19 Posts Geometry Puzzle Just draw the figure roughly. The answer pops out by mere inspection! Mally
 2006-03-14, 17:33 #8 Kees     Dec 2005 19610 Posts 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...
2006-03-15, 16:13   #9
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
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?
Mally

 2006-03-15, 22:27 #10 cheesehead     "Richard B. Woods" Aug 2002 Wisconsin USA 22·3·641 Posts 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!)
 2006-03-16, 08:41 #11 Kees     Dec 2005 C416 Posts Spoler test so this should be hidden thanks for the tip

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