Given a rectangle ABCD, construct a square with the same area.
CODE
A----B
| |
D----C
| |
D----C
First of all, this question was asked by my brother who was trying to compare dessert sizes at Claim Jumpers to other objects, and answered already by me. But I wanted to see if anyone still remembers their high school geometry.
Given a rectangle ABCD, construct a square with the same area.
Given a rectangle ABCD, construct a square with the same area.
1) Perform the construction with straightedge and compass.
2) Prove that the areas are the same.
QUOTE (rpenner+Jul 22 2009, 12:11 PM)
Given a rectangle ABCD, construct a square with the same area.
CODE
A----B
| |
D----C
| |
D----C
1) Perform the construction with straightedge and compass.
2) Prove that the areas are the same.
Square EFGH has the same area as Rectangle ABCD.
In order to find length EF (L), we use the following equation:
CODE
E----F
| |
G----H
B = length AB
H = length AD
L = length EF
Area of EFGH:
A1 = L^2
Area of ABCD:
A2 = BH
Given A1 = A2
L^2 = BH
L = sqrt(BH)
sqrt(BH)^2 = BH
BH = BH
| |
G----H
B = length AB
H = length AD
L = length EF
Area of EFGH:
A1 = L^2
Area of ABCD:
A2 = BH
Given A1 = A2
L^2 = BH
L = sqrt(BH)
sqrt(BH)^2 = BH
BH = BH
It's been a while, but I think that's what you were asking for?
Yes, but I wanted a straight-edge and compass Euclidean construction, and then a proof that that construction was correct.
QUOTE (rpenner+Jul 22 2009, 12:51 PM)
Yes, but I wanted a straight-edge and compass Euclidean construction, and then a proof that that construction was correct.
I must have been asleep during that part of geometry class. Then again, my school wasn't exactly top notch. They might have skipped that part.
I must have been asleep during that part of geometry class. Then again, my school wasn't exactly top notch. They might have skipped that part.
Understandable. It's a pretty neat construction and at least two neat proofs of it's correctness.
rpenner
Is it a relevant concern to ask about metric used and accuracy involved ?
QUOTE
Yes, but I wanted a straight-edge and compass Euclidean construction, and then a proof that that construction was correct.
Is it a relevant concern to ask about metric used and accuracy involved ?
QUOTE (bukh+Jul 23 2009, 11:05 AM)
Is it a relevant concern to ask about metric used and accuracy involved ?
No. For instance, you can make a 90 or 60 degree angle exactly using a compass and a straight edge or a length of size L*sqrt(2) given a length L precisely, not "To 4 decimal places", as that implies some iterative process, not a finite set construction. You can do plenty of things if you allow limits of iterative processes, such squaring the circle, which would otherwise be impossible.
No. For instance, you can make a 90 or 60 degree angle exactly using a compass and a straight edge or a length of size L*sqrt(2) given a length L precisely, not "To 4 decimal places", as that implies some iterative process, not a finite set construction. You can do plenty of things if you allow limits of iterative processes, such squaring the circle, which would otherwise be impossible.
AN
[QUOTE]You can do plenty of things if you allow limits of iterative processes, such squaring the circle, which would otherwise be impossible.[QUOTE]
Limits of iterative processes - but this is the same as asking accuracy - is it not ?
[QUOTE]You can do plenty of things if you allow limits of iterative processes, such squaring the circle, which would otherwise be impossible.[QUOTE]
Limits of iterative processes - but this is the same as asking accuracy - is it not ?
I do not ever remember confronting this one years ago in school. Of course that was a long time ago (late 1970's).
No trick sprung to mind, but if some person showed me a painting that is a rectangle ans said they want that same painting done on a square of the same area, I would only think to do this. Get a piece of paper and make that rectangle (or one of same proportion) and cut it up into small squares (fitting as neatly as one can). Then, I'd play with the squares, lining them up into rows until I got rows making a square shape. There could be one or two or so pieces left over, so one would have to perhaps make it just a tiny bit bigger to compensate (guess a little). It would not be 100% accurate, but probably good enough for the person making the request.
If someone does find a more accurate way, that would come in handy one day.
No trick sprung to mind, but if some person showed me a painting that is a rectangle ans said they want that same painting done on a square of the same area, I would only think to do this. Get a piece of paper and make that rectangle (or one of same proportion) and cut it up into small squares (fitting as neatly as one can). Then, I'd play with the squares, lining them up into rows until I got rows making a square shape. There could be one or two or so pieces left over, so one would have to perhaps make it just a tiny bit bigger to compensate (guess a little). It would not be 100% accurate, but probably good enough for the person making the request.
If someone does find a more accurate way, that would come in handy one day.
AN
Well - you say EXACTLY - and exactly is a big word - to me exactly is always followed by a definition of the exactness involved. A magnetic field - would you define a magnetic field as a continuous - in the meaning of true continuity - points of no dimension - or would you define such a field according to points taking a certain space ?
Perhaps you are right that a magnetic needle will point exactly - in as much as the needle is following Mother Nature and not math or physic rules.
QUOTE
No. For instance, you can make a 90 or 60 degree angle exactly using a compass
Well - you say EXACTLY - and exactly is a big word - to me exactly is always followed by a definition of the exactness involved. A magnetic field - would you define a magnetic field as a continuous - in the meaning of true continuity - points of no dimension - or would you define such a field according to points taking a certain space ?
Perhaps you are right that a magnetic needle will point exactly - in as much as the needle is following Mother Nature and not math or physic rules.
WOG, let's fix AB as the not-shorter side. Use straightedge to extend AB. Set compass at B and use it to find E, further down on the line AB such that BE = BC. These are the two basic operations of straightedge and compass construction.
Now that ABE is a straight line segment, we need to find the perpendicular bisector of the line segment AE. Place the compass point at A and the drawing end at E and draw a circle. Next place the compass point at E and the drawing end at A and draw another circle. These two circles intersect each other at F and G (it doesn't matter which name goes which which point). Use to the straightedge to build the line FG which intersects AE at a point H exactly halfway between A and E.
(Triangles ▵AEF and ▵AEG are equilateral and congruent, so of course FG goes halfway between A and E -- every point on line FG is exactly the same distance from F and G.)
Assuming AB > BE, then we have AHBE as a straight line. (The alternative is that H and B are the same point.) But the important thing at this point is the distance AH = HE. Let's emphasize this by setting our compass at H and drawing a semicircle from A to E. Remember the magic thing about circles is every single point is the same distance from the center.
Now we know how to construct the perpendicular bisector to any point, we can easily use this to construct the perpendicular to a line at any chosen point. Set the compass point at B and draw a circle to find the other point, I, which is the same distance as BE and on the same line. Now construct the perpendicular bisector of IE which we call the line JK (which intersects at point B ). Finally extend JK until it hits the big semicircle we drew between A and E. This new point L forms a segment BL which is the first side of our square. Since we know how to construct a perpendicular to a line at a point, the actual square is brief tedium.
That might be too confusing without pictures. Here's some pictures (not made by me) which roughly show you most of what I was talking about:
http://planetmath.org/?method=l2h&from=obj...cMean&op=getobj
http://www.cut-the-knot.org/pythagoras/GeometricMean.shtml (Lots of methods)
http://aleph0.clarku.edu/~djoyce/java/elem...I/propII14.html
http://aleph0.clarku.edu/~djoyce/java/elem...I/propVI13.html
So we have the line AHBE, the semicircle A to E, the perpendicular line BL where L is on the semicircle. How can we prove BL×BL = AB×BE ?
Using geometry: ⊿LBE is a right triangle. ⊿ABL is a right triangle. And because it is inscribed in a semicircle, ⊿ELA is a right triangle. But since the two small angles of a right triangle are complementary (sum to a right angle), and angles ∠ELB and ∠BLA sum to ⊾ELA which is also a right angle, then all three triangles have the same three angles and are similar. And so the ratio AB to BL is the same as the ratio BL to BE.
AB/BL = BL/BE => BL×BL = AB×BE
A more algebraic attack on the proof is based on the different right triangle HBL and the good-old Pythagorean theorem.
HL = HA = HE = (1/2)(AE) = (1/2)(AB + BE)
HB = HB + BE − BE = HE − BE = (1/2)(AB + BE) − BE = (1/2)(AB − BE)
BL×BL = BL² = HL² − HB² = (1/4)(AB + BE)² − (1/4)(AB − BE)² = AB×BE
Now that ABE is a straight line segment, we need to find the perpendicular bisector of the line segment AE. Place the compass point at A and the drawing end at E and draw a circle. Next place the compass point at E and the drawing end at A and draw another circle. These two circles intersect each other at F and G (it doesn't matter which name goes which which point). Use to the straightedge to build the line FG which intersects AE at a point H exactly halfway between A and E.
(Triangles ▵AEF and ▵AEG are equilateral and congruent, so of course FG goes halfway between A and E -- every point on line FG is exactly the same distance from F and G.)
Assuming AB > BE, then we have AHBE as a straight line. (The alternative is that H and B are the same point.) But the important thing at this point is the distance AH = HE. Let's emphasize this by setting our compass at H and drawing a semicircle from A to E. Remember the magic thing about circles is every single point is the same distance from the center.
Now we know how to construct the perpendicular bisector to any point, we can easily use this to construct the perpendicular to a line at any chosen point. Set the compass point at B and draw a circle to find the other point, I, which is the same distance as BE and on the same line. Now construct the perpendicular bisector of IE which we call the line JK (which intersects at point B ). Finally extend JK until it hits the big semicircle we drew between A and E. This new point L forms a segment BL which is the first side of our square. Since we know how to construct a perpendicular to a line at a point, the actual square is brief tedium.
That might be too confusing without pictures. Here's some pictures (not made by me) which roughly show you most of what I was talking about:
http://planetmath.org/?method=l2h&from=obj...cMean&op=getobj
http://www.cut-the-knot.org/pythagoras/GeometricMean.shtml (Lots of methods)
http://aleph0.clarku.edu/~djoyce/java/elem...I/propII14.html
http://aleph0.clarku.edu/~djoyce/java/elem...I/propVI13.html
So we have the line AHBE, the semicircle A to E, the perpendicular line BL where L is on the semicircle. How can we prove BL×BL = AB×BE ?
Using geometry: ⊿LBE is a right triangle. ⊿ABL is a right triangle. And because it is inscribed in a semicircle, ⊿ELA is a right triangle. But since the two small angles of a right triangle are complementary (sum to a right angle), and angles ∠ELB and ∠BLA sum to ⊾ELA which is also a right angle, then all three triangles have the same three angles and are similar. And so the ratio AB to BL is the same as the ratio BL to BE.
AB/BL = BL/BE => BL×BL = AB×BE
A more algebraic attack on the proof is based on the different right triangle HBL and the good-old Pythagorean theorem.
HL = HA = HE = (1/2)(AE) = (1/2)(AB + BE)
HB = HB + BE − BE = HE − BE = (1/2)(AB + BE) − BE = (1/2)(AB − BE)
BL×BL = BL² = HL² − HB² = (1/4)(AB + BE)² − (1/4)(AB − BE)² = AB×BE
Let the sides of the rectangle be A,B and the side of the desired square be C
Choose x so
A = C + x
B = C - x
AB = (C+x)(C-x)
= C^2 - x^2
since AB = C^2
x=0
The original rectangle has reduced itself to a square
which shows why I don't attempt maths unless I have to 
.
Choose x so
A = C + x
B = C - x
AB = (C+x)(C-x)
= C^2 - x^2
since AB = C^2
x=0
The original rectangle has reduced itself to a square
which shows why I don't attempt maths unless I have to .
QUOTE (Empress Palpatine+Jul 23 2009, 11:10 AM)
I do not ever remember confronting this one years ago in school. Of course that was a long time ago (late 1970's).
No trick sprung to mind, but if some person showed me a painting that is a rectangle ans said they want that same painting done on a square of the same area, I would only think to do this. Get a piece of paper and make that rectangle (or one of same proportion) and cut it up into small squares (fitting as neatly as one can). Then, I'd play with the squares, lining them up into rows until I got rows making a square shape. There could be one or two or so pieces left over, so one would have to perhaps make it just a tiny bit bigger to compensate (guess a little). It would not be 100% accurate, but probably good enough for the person making the request.
If someone does find a more accurate way, that would come in handy one day.
Or you could figure out the area of the rectangle (let's say 4 m^2) then get the square root of that number and draw a square of those dimensions (2m).
But just like the other solutions presented, this is in terms of math. I'm basically restating what flyingbuttressman did.
No trick sprung to mind, but if some person showed me a painting that is a rectangle ans said they want that same painting done on a square of the same area, I would only think to do this. Get a piece of paper and make that rectangle (or one of same proportion) and cut it up into small squares (fitting as neatly as one can). Then, I'd play with the squares, lining them up into rows until I got rows making a square shape. There could be one or two or so pieces left over, so one would have to perhaps make it just a tiny bit bigger to compensate (guess a little). It would not be 100% accurate, but probably good enough for the person making the request.
If someone does find a more accurate way, that would come in handy one day.
Or you could figure out the area of the rectangle (let's say 4 m^2) then get the square root of that number and draw a square of those dimensions (2m).
But just like the other solutions presented, this is in terms of math. I'm basically restating what flyingbuttressman did.
I was avoiding math. My answer was non mathematical but not exactly what Rpenner had in mind. I can see that Rpenner's solution is quite complicated, more complicated than our high school geometry class ever did.
No declared winners of the prize here I guess, no winners of a trip to Tahiti.
(Just kidding)
No declared winners of the prize here I guess, no winners of a trip to Tahiti.
(Just kidding)
QUOTE (bukh+Jul 23 2009, 01:02 PM)
QUOTE
You can do plenty of things if you allow limits of iterative processes, such squaring the circle, which would otherwise be impossible.
Limits of iterative processes - but this is the same as asking accuracy - is it not ?
No. It's possible to give a method which would, in an ideal universe, give an exact 90 degree angle. It's possible to give a method which, in a finite number of steps, will bisect any given angle. However, it is not possible, even in an ideal universe, to give a method which will trisect any given angle. Nor can you give a method which will, in finitely many steps, produce a circle of area exactly that of a given square or vice versa.
If you are allowed to do infinitely many steps then you can construct iterative processes which will allow you to produce angles which get closer and closer to the trisection of any given angle. Hence the distinction. It's isn't about "How accurate can you do it with a pencil and paper" but instead about constructing methods which would, in the perfect work of geometry, do what you need in finitely many steps.
QUOTE (->
| QUOTE |
| You can do plenty of things if you allow limits of iterative processes, such squaring the circle, which would otherwise be impossible. |
Limits of iterative processes - but this is the same as asking accuracy - is it not ?
No. It's possible to give a method which would, in an ideal universe, give an exact 90 degree angle. It's possible to give a method which, in a finite number of steps, will bisect any given angle. However, it is not possible, even in an ideal universe, to give a method which will trisect any given angle. Nor can you give a method which will, in finitely many steps, produce a circle of area exactly that of a given square or vice versa.
If you are allowed to do infinitely many steps then you can construct iterative processes which will allow you to produce angles which get closer and closer to the trisection of any given angle. Hence the distinction. It's isn't about "How accurate can you do it with a pencil and paper" but instead about constructing methods which would, in the perfect work of geometry, do what you need in finitely many steps.
Well - you say EXACTLY - and exactly is a big word - to me exactly is always followed by a definition of the exactness involved. A magnetic field - would you define a magnetic field as a continuous - in the meaning of true continuity - points of no dimension - or would you define such a field according to points taking a certain space ?
Perhaps you are right that a magnetic needle will point exactly - in as much as the needle is following Mother Nature and not math or physic rules.
bukh, rather than opening your mouth in a desperate attempt to whine and nitpick about something you don't really understand in the vain attempt to seem like you're not the thick headed clod you actually are, next time why don't you spend a small amount of time doing a little bit of background reading. This thread is quite clearly a question about geometric constructions, about methods in geometry. If you were familiar with such things, such as things like angle bisection or perpendicular bisectors (thoughts taught in school to children) or such problems as 'squaring the circle' you'd know that the answer is a method, independent of the physical limitations of the real world. Mathematics is not confined by physics, if a mathematician says "Imagine a circle" you don't put your hand up and whine "But in real life you can't make a perfect circle!!", you're utterly missing the fundamental concept of what mathematics is. But then it's very clear to me that that's precisely your problem.
QUOTE (bukh+Jul 23 2009, 05:00 PM)
AN
Perhaps you are right that a magnetic needle will point exactly - in as much as the needle is following Mother Nature and not math or physic rules.
http://upload.wikimedia.org/wikipedia/comm..._(drafting).jpg
What does a magnetic needle have to do with this?
Perhaps you are right that a magnetic needle will point exactly - in as much as the needle is following Mother Nature and not math or physic rules.
http://upload.wikimedia.org/wikipedia/comm..._(drafting).jpg
What does a magnetic needle have to do with this?
buttershug
You are right - absolutely nothing - I was naive and thought a compass was a naval instrument with a magnetic needle - my stupidity.
QUOTE
What does a magnetic needle have to do with this?
You are right - absolutely nothing - I was naive and thought a compass was a naval instrument with a magnetic needle - my stupidity.
QUOTE (rpenner+Jul 22 2009, 05:11 PM)
First of all, this question was asked by my brother who was trying to compare dessert sizes at Claim Jumpers to other objects, and answered already by me. But I wanted to see if anyone still remembers their high school geometry.
Given a rectangle ABCD, construct a square with the same area.
Given a rectangle ABCD, construct a square with the same area.
CODE
A----B
| |
D----C
| |
D----C
1) Perform the construction with straightedge and compass.
2) Prove that the areas are the same.
One easy solution is the following:
1. With the center of the compass in D and the drawing point in B draw a quarter circle that intersects the line DC in C'. The length DC'=a*sqrt(2)
2. Draw the two diagonals of the square, they intersect in P.
3. With the compass center in D (again) and with the drawing tip in P draw a quarter circle that intersects DA in A'. The length DA' is a*sqrt(2)/2=a/sqrt(2)
4. Using the right edge, draw a perpendicular on DC' in C' and a perpendicular on DA' in A'. They intersect in B'.
5. The resulting figure DA'B'C' is a rectangle of area a^2
I am quite sure that there are many other solutions, this is the first one that came to mind.
QUOTE (Trout+Jul 25 2009, 05:27 PM)
One easy solution is the following:
1. With the center of the compass in D and the drawing point in B draw a quarter circle that intersects the line DC in C'. The length DC'=a*sqrt(2)
2. Draw the two diagonals of the square, they intersect in P.
3. With the compass center in D (again) and with the drawing tip in P draw a quarter circle that intersects DA in A'. The length DA' is a*sqrt(2)/2=a/sqrt(2)
4. Using the right edge, draw a perpendicular on DC' in C' and a perpendicular on DA' in A'. They intersect in B'.
5. The resulting figure DA'B'C' is a rectangle of area a^2
I am quite sure that there are many other solutions, this is the first one that came to mind.
Algebraically, we can show that there is an infinity of algebraic solutions (not the same thing as an infinity of geometric solutions).
Indeed: we make one side shorter by x and the other one longer by y with the condition:
a^2=(a-x)(a+y)
0<x<a
0<y<a
The above results into
x=(ay)/(a+y)
There is an infinity of pairs (x,y) that satisfy the condition. We have seen in the previous post y=a(sqrt(2)-1) and x=a(1-1/sqrt(2)).
Of course, only a subset can be constructed with a straight edge and a compass.
1. With the center of the compass in D and the drawing point in B draw a quarter circle that intersects the line DC in C'. The length DC'=a*sqrt(2)
2. Draw the two diagonals of the square, they intersect in P.
3. With the compass center in D (again) and with the drawing tip in P draw a quarter circle that intersects DA in A'. The length DA' is a*sqrt(2)/2=a/sqrt(2)
4. Using the right edge, draw a perpendicular on DC' in C' and a perpendicular on DA' in A'. They intersect in B'.
5. The resulting figure DA'B'C' is a rectangle of area a^2
I am quite sure that there are many other solutions, this is the first one that came to mind.
Algebraically, we can show that there is an infinity of algebraic solutions (not the same thing as an infinity of geometric solutions).
Indeed: we make one side shorter by x and the other one longer by y with the condition:
a^2=(a-x)(a+y)
0<x<a
0<y<a
The above results into
x=(ay)/(a+y)
There is an infinity of pairs (x,y) that satisfy the condition. We have seen in the previous post y=a(sqrt(2)-1) and x=a(1-1/sqrt(2)).
Of course, only a subset can be constructed with a straight edge and a compass.
QUOTE (Confused2+Jul 24 2009, 07:31 AM)
Let the sides of the rectangle be A,B and the side of the desired square be C
Choose x so
A = C + x
B = C - x
AB = (C+x)(C-x)
= C^2 - x^2
since AB = C^2
x=0
The original rectangle has reduced itself to a square which shows why I don't attempt maths unless I have to .
The question and AB = C^2 both assume equal areas.
A = C + x, B = C - x, assumes equal perimeters.
So, you have restricted your original rectangle to be equal in area and perimeter to your resulting square.
Of course, the only rectangle that can possibly satisfy this is the one which is identical to the resulting square!
This means that A = B = C and x = 0, exactly as you showed...
Choose x so
A = C + x
B = C - x
AB = (C+x)(C-x)
= C^2 - x^2
since AB = C^2
x=0
The original rectangle has reduced itself to a square
which shows why I don't attempt maths unless I have to . The question and AB = C^2 both assume equal areas.
A = C + x, B = C - x, assumes equal perimeters.
So, you have restricted your original rectangle to be equal in area and perimeter to your resulting square.
Of course, the only rectangle that can possibly satisfy this is the one which is identical to the resulting square!
This means that A = B = C and x = 0, exactly as you showed...
QUOTE (Trout+Jul 25 2009, 05:27 PM)
The resulting figure DA'B'C' is a rectangle of area a^2
That is not what was asked for.
ABCD is a rectangle, with area (AD)×(CD), not a square. A'B'C'D is a rectangle with area (1/2) ( (AD)² + (CD)² ), since BD is a diagonal of the original rectangle. And you can't use the straightedge to construct a perpendicular -- you can only use it to construct a line through two given points.
That is not what was asked for.
ABCD is a rectangle, with area (AD)×(CD), not a square. A'B'C'D is a rectangle with area (1/2) ( (AD)² + (CD)² ), since BD is a diagonal of the original rectangle. And you can't use the straightedge to construct a perpendicular -- you can only use it to construct a line through two given points.
QUOTE (Trout+Jul 25 2009, 05:27 PM)
One easy solution is the following:
1. With the center of the compass in D and the drawing point in B draw a quarter circle that intersects the line DC in C'. The length DC'=a*sqrt(2)
2. Draw the two diagonals of the square, they intersect in P.
3. With the compass center in D (again) and with the drawing tip in P draw a quarter circle that intersects DA in A'. The length DA' is a*sqrt(2)/2=a/sqrt(2)
4. Using the right edge, draw a perpendicular on DC' in C' and a perpendicular on DA' in A'. They intersect in B'.
5. The resulting figure DA'B'C' is a rectangle of area a^2
I am quite sure that there are many other solutions, this is the first one that came to mind.
Here is an even simpler solution:
1. With the center of the compass in C and the tip in B draw a quarter circle that intersects DC in C'. DC'=2a
2. The diagonals of the square intersect in P
3. Draw two equal semicircles centered in B and C, they intersect in Q
4. The line PQ intersects DA in A'. DA'=a/2
1. With the center of the compass in D and the drawing point in B draw a quarter circle that intersects the line DC in C'. The length DC'=a*sqrt(2)
2. Draw the two diagonals of the square, they intersect in P.
3. With the compass center in D (again) and with the drawing tip in P draw a quarter circle that intersects DA in A'. The length DA' is a*sqrt(2)/2=a/sqrt(2)
4. Using the right edge, draw a perpendicular on DC' in C' and a perpendicular on DA' in A'. They intersect in B'.
5. The resulting figure DA'B'C' is a rectangle of area a^2
I am quite sure that there are many other solutions, this is the first one that came to mind.
Here is an even simpler solution:
1. With the center of the compass in C and the tip in B draw a quarter circle that intersects DC in C'. DC'=2a
2. The diagonals of the square intersect in P
3. Draw two equal semicircles centered in B and C, they intersect in Q
4. The line PQ intersects DA in A'. DA'=a/2
QUOTE (rpenner+Jul 26 2009, 01:15 AM)
That is not what was asked for.
ABCD is a rectangle, with area (AD)×(CD), not a square. A'B'C'D is a rectangle with area (1/2) ( (AD)² + (CD)² ), since BD is a diagonal of the original rectangle. And you can't use the straightedge to construct a perpendicular -- you can only use it to construct a line through two given points.
OK, I had a look at the picture and I did not read the text. Looked like a square to me. So, I solved a different problem , sorry.
ABCD is a rectangle, with area (AD)×(CD), not a square. A'B'C'D is a rectangle with area (1/2) ( (AD)² + (CD)² ), since BD is a diagonal of the original rectangle. And you can't use the straightedge to construct a perpendicular -- you can only use it to construct a line through two given points.
OK, I had a look at the picture and I did not read the text. Looked like a square to me. So, I solved a different problem , sorry.
QUOTE (rpenner+Jul 26 2009, 01:15 AM)
That is not what was asked for.
ABCD is a rectangle, with area (AD)×(CD), not a square. A'B'C'D is a rectangle with area (1/2) ( (AD)² + (CD)² ), since BD is a diagonal of the original rectangle. And you can't use the straightedge to construct a perpendicular -- you can only use it to construct a line through two given points.
Easy to fix:
1. The diagonal of the rectangle is sqrt(a^2+b^2)
2. Construct the segment a+b (this is very easy)
3. Construct ( a+b )/sqrt(2) via an isosceles right handed triangle that has the hypotenuse=a+b
4. Construct the right handed triangle that has one side equal to half the diagonal (sqrt(a^2+b^2)/2) and the hypothenuse (a+b )/sqrt(2)
5. The other side of this triangle is sqrt(ab).
6. We reduced the problem to fing (x,y) such that:
(a-x)(b+y)=ab (assume a>b
a-x=b+y
7. The solution is :
x=a-sqrt(ab)
y=sqrt(ab)-b
8. We obtained a square with sides sqrt(ab).
ABCD is a rectangle, with area (AD)×(CD), not a square. A'B'C'D is a rectangle with area (1/2) ( (AD)² + (CD)² ), since BD is a diagonal of the original rectangle. And you can't use the straightedge to construct a perpendicular -- you can only use it to construct a line through two given points.
Easy to fix:
1. The diagonal of the rectangle is sqrt(a^2+b^2)
2. Construct the segment a+b (this is very easy)
3. Construct ( a+b )/sqrt(2) via an isosceles right handed triangle that has the hypotenuse=a+b
4. Construct the right handed triangle that has one side equal to half the diagonal (sqrt(a^2+b^2)/2) and the hypothenuse (a+b )/sqrt(2)
5. The other side of this triangle is sqrt(ab).
6. We reduced the problem to fing (x,y) such that:
(a-x)(b+y)=ab (assume a>b
a-x=b+y
7. The solution is :
x=a-sqrt(ab)
y=sqrt(ab)-b
8. We obtained a square with sides sqrt(ab).
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