[Prof. Templeton]

There was once a very forgetful landscape contractor who was constantly "misrembering" the detailed instructions of his employers. He was always forgetting essential parts of what the customer had intended and instead would substitute his own without realizing it. How he stayed in business as long as he did is a mystery. Here is one example of his forgetfulness:

The customer had a yard in the shape of a right triangle, none of the sides of equal length. He had purchased a certain amount of flower bulbs that he wanted planted in a patch in the square corner (90°) of the triangle.

"I want an angled patch of flowers that extends out from this corner 3 meters, kind of like a slice of pie. If you were to run a line from this corner so that it ended perpendicular to the long side of the yard over there in the Northeast, that would give us the correct angle between the line and Southern edge of the yard", he instructed.

"No problem", replied the contractor. "I'll get to work as soon as I can get my equipment back here".

When the contractor returned he proceeded to make a line that ran from the middle of the long side of the yard back to the square corner and then dug up the ground between that line and the Western side of the yard to 3 meters out from the corner.

The customer came out to check on the work being done and bring the boxes of flower bulbs. He said, "Well, it's not quite what I had in mind, but the angle looks about right, I suppose. I was very precise when I ordered the bulbs, though. Since the angle is not where I had originally intended, do you think I might need more or less bulbs now?"

"Ummm…bulbs?" said the contractor.

Does the customer need more or less bulbs?

I love proofs.

[Guest]

is it possible to solve without knowing which edge is longer?

a perpendicular line from the hypotenuse could be closer to either side depending on which side is longer, right?

[Prof. Templeton]

is it possible to solve without knowing which edge is longer?

a perpendicular line from the hypotenuse could be closer to either side depending on which side is longer, right?

The lengths of the side will either increase or decrease the angle. The relationship between the angles would be the same regardless on the side lengths.

[bonanova]

I'm not getting what the bounds of "an angled patch" are.

Sounds like a triangle one of whose sides is 3 m.

What are the other two sides?

[Guest]

The lengths of the side will either increase or decrease the angle. The relationship between the angles would be the same regardless on the side lengths.

Is this the picture we're looking at?

Bold lines are the midpoint lines that the landscaper used, lighter lines are what was requested (approx of course)

triangles.bmp

Edited June 5, 2009 by palmerc7

[Guest]

"I want an angled patch of flowers that extends out from this corner 3 meters, kind of like a slice of pie."

For clarification, is the extension out from the corner along the angle or along the side of the yard?

Thanks,

Solvitur Ambulando

[Prof. Templeton]

I'm not getting what the bounds of "an angled patch" are.

Sounds like a triangle one of whose sides is 3 m.

What are the other two sides?

an angled patch that starts from a point on the right angled corner (let's call that A), extends along the Southern boundry to a point (let's call it B) 3 meters away. Then follows an arc to a point (let's call it C) 3 meters from point A that lies on the line of altitude of the hypotnuese. Now we have angle CAB. The contractor started at A, went 3 meters along the Western boundry to a point (D), then followed an arc to a point (E) 3 meters from A on the median line of the hypotnuese. Angle DAE. Are the angles the same or are the areas of the sectors the same?

Is this the picture we're looking at?

Bold lines are the midpoint lines that the landscaper used, lighter lines are what was requested (approx of course)

Yes.

I apologize for not being clear.

[Prof. Templeton]

"I want an angled patch of flowers that extends out from this corner 3 meters, kind of like a slice of pie."

For clarification, is the extension out from the corner along the angle or along the side of the yard?

Thanks,

Solvitur Ambulando

Along the side of the yard. The patch would look like a sector taken from a circle.

[bonanova]

Along the side of the yard. The patch would look like a sector taken from a circle.

If so, wouldn't it be either? Both would be radii.

[Prof. Templeton]

If so, wouldn't it be either? Both would be radii.

Yes. I guess I didn't really understand the last question.

[Glycereine]

Unfortunately I just don't get what the patches would look like so I guess I'm out on this one.

[Guest]

Given: A right scalene triangle

Let ABC be the right triangle with right angle A.

Let A’ be the midpoint of BC. Then A’ is the center of the circle which circumscribes triangle ABC, so AA’ = BA’, so triangle AA’B is isosceles and angle ABA’ = angle A’AB.

Let A” exist on BC such that AA” is perpendicular to BC. Since angle CA”A = angle CAB (right angles) and angle BCA = angle A”CA (identity: same angle), then angle

A”AC = (180 – angle A”CA – angle A”AC) = (180 – angle ACB – angle BAC) = angle ABC. [angle A”AC = angle ABC].

Now, assume the customer wanted the slice along the southern side of the yard (angle A”AC with a radius of 3m), but the landscaper created the slice along the western side of the yard (angle A’AB with a radius of 3m). Since angle ABA’ (angle ABC) = angle A’AB and angle A”AC = angle ABC, then angle A’AB = angle A”AC and the area of the desired slice equals the area of the actual slice. Thus, no more and no less bulbs are needed.

[Guest]

...the angle, and hence the area of the sector, is the same in both cases. This angle is same as the angle at the northwest corner of the yard. So the answer is the customer does not need any more bulbs.

[Guest]

Given: A right scalene triangle

angle A”AC = (180 – angle A”CA – angle A”AC)

Typo i believe.

above should read 'angle A"AC = (180 - angle A"CA - angle AA"C)

Then I believe the proof is valid

[Guest]

Well you beat me to it with a general proof. I have a specific example, though, that seems to contradict it (which probably means that it is (I am) wrong).

Consider the triangle formed by the point (0,0) (0,3) and (4,0). The midpoint of the hypotenuse is (2,1.5) by averaging. The slope of the hypotenuse is -3/4, so the slope of the line perpendicular to it must be the negative reciprocal, 4/3. Both lines pass through the origin. The question is, which line has a greater angle (think unit circle): That with a slope of 4/3 or 3/4 (=1.5/2)? Arctan each to find the former is greater (53˚, the other is 36˚).

Edited June 5, 2009 by BrainInAVat

[Guest]

Except I made a rise/run error. It would be 3/4 both times.

Or, more generally, (0,0) (0,y) and (x,0), both producing slopes of y/x.

Edited June 5, 2009 by BrainInAVat

[Guest]

There was once a very forgetful landscape contractor who was constantly "misrembering" the detailed instructions of his employers. He was always forgetting essential parts of what the customer had intended and instead would substitute his own without realizing it. How he stayed in business as long as he did is a mystery. Here is one example of his forgetfulness:

The customer had a yard in the shape of a right triangle, none of the sides of equal length. He had purchased a certain amount of flower bulbs that he wanted planted in a patch in the square corner (90°) of the triangle.

"I want an angled patch of flowers that extends out from this corner 3 meters, kind of like a slice of pie. If you were to run a line from this corner so that it ended perpendicular to the long side of the yard over there in the Northeast, that would give us the correct angle between the line and Southern edge of the yard", he instructed.

"No problem", replied the contractor. "I'll get to work as soon as I can get my equipment back here".

When the contractor returned he proceeded to make a line that ran from the middle of the long side of the yard back to the square corner and then dug up the ground between that line and the Western side of the yard to 3 meters out from the corner.

The customer came out to check on the work being done and bring the boxes of flower bulbs. He said, "Well, it's not quite what I had in mind, but the angle looks about right, I suppose. I was very precise when I ordered the bulbs, though. Since the angle is not where I had originally intended, do you think I might need more or less bulbs now?"

"Ummm…bulbs?" said the contractor.

Does the customer need more or less bulbs?

I love proofs.

Did the contractor make the shape like a piece of pie when he dug, an arc of 3 meters, or did he dig a straight line from the western edge to the bisecting line?

[Guest]

There was once a very forgetful landscape contractor who was constantly "misrembering" the detailed instructions of his employers. He was always forgetting essential parts of what the customer had intended and instead would substitute his own without realizing it. How he stayed in business as long as he did is a mystery. Here is one example of his forgetfulness:

The customer had a yard in the shape of a right triangle, none of the sides of equal length. He had purchased a certain amount of flower bulbs that he wanted planted in a patch in the square corner (90°) of the triangle.

"I want an angled patch of flowers that extends out from this corner 3 meters, kind of like a slice of pie. If you were to run a line from this corner so that it ended perpendicular to the long side of the yard over there in the Northeast, that would give us the correct angle between the line and Southern edge of the yard", he instructed.

"No problem", replied the contractor. "I'll get to work as soon as I can get my equipment back here".

When the contractor returned he proceeded to make a line that ran from the middle of the long side of the yard back to the square corner and then dug up the ground between that line and the Western side of the yard to 3 meters out from the corner.

The customer came out to check on the work being done and bring the boxes of flower bulbs. He said, "Well, it's not quite what I had in mind, but the angle looks about right, I suppose. I was very precise when I ordered the bulbs, though. Since the angle is not where I had originally intended, do you think I might need more or less bulbs now?"

"Ummm…bulbs?" said the contractor.

Does the customer need more or less bulbs?

I love proofs.

[Guest]

So the customer wanted a 90 degree angle on the long side of the yard. What the landscaper gave him is larger or smaller than he wanted,according to whether the western side is longer or shorter than the southern side, because the contractor went from the middle of the long side to the SW corner.

Do I understand this correctly?

The new triangle formed when the contractor ran the line would not have a 90 degree angle?

then if the western side is shorter

more bulbs needed. If the western side is shorter than the southern side, the angle would be smaller than the owner intended. So less bulbs would be needed.

Edited June 6, 2009 by chicory

[Guest]

There was once a very forgetful landscape contractor who was constantly "misrembering" the detailed instructions of his employers. He was always forgetting essential parts of what the customer had intended and instead would substitute his own without realizing it. How he stayed in business as long as he did is a mystery. Here is one example of his forgetfulness:

The customer had a yard in the shape of a right triangle, none of the sides of equal length. He had purchased a certain amount of flower bulbs that he wanted planted in a patch in the square corner (90°) of the triangle.

"I want an angled patch of flowers that extends out from this corner 3 meters, kind of like a slice of pie. If you were to run a line from this corner so that it ended perpendicular to the long side of the yard over there in the Northeast, that would give us the correct angle between the line and Southern edge of the yard", he instructed.

"No problem", replied the contractor. "I'll get to work as soon as I can get my equipment back here".

When the contractor returned he proceeded to make a line that ran from the middle of the long side of the yard back to the square corner and then dug up the ground between that line and the Western side of the yard to 3 meters out from the corner.

The customer came out to check on the work being done and bring the boxes of flower bulbs. He said, "Well, it's not quite what I had in mind, but the angle looks about right, I suppose. I was very precise when I ordered the bulbs, though. Since the angle is not where I had originally intended, do you think I might need more or less bulbs now?"

"Ummm…bulbs?" said the contractor.

Does the customer need more or less bulbs?

I love proofs.

the angle wanted would be the same size as the angle in the SE corner of the yard.

The angle received would be =

[angle SE +

(90 degrees -(angle SE corner + angle NW corner))]

Edited June 7, 2009 by chicory

[Guest]

So the customer wanted a 90 degree angle on the long side of the yard. What the landscaper gave him is larger or smaller than he wanted,according to whether the western side is longer or shorter than the southern side, because the contractor went from the middle of the long side to the SW corner.

Do I understand this correctly?

The new triangle formed when the contractor ran the line would not have a 90 degree angle?

then if the western side is shorter

more bulbs needed. If the western side is shorter than the southern side, the angle would be smaller than the owner intended. So less bulbs would be needed.

This is what I'm getting. I don't see how its possible without knowing which border is longer.

[Guest]

Given: A right scalene triangle

Let A’ be the midpoint of BC. Then A’ is the center of the circle which circumscribes triangle ABC, so AA’ = BA’, so triangle AA’B is isosceles and angle ABA’ = angle A’AB.

In order for A' to be a circle which circumscribes the triangle, and AA'=BA', it would have to be a right triangle with AB=AC.

Am I wrong?

[Guest]

ok this post might be done but i didnt see an acceptable proof (i might be wrong) but here is mine (i think)

so you need to look at the picture otherwise your doomed to confusion

so by pathagerous

b^2+c^2=t^2=G^2+f^2

a^2+c^2=e^2+f^2

a^2+e^2=c^2+f^2

now subtact the second from the third or the other way around

c^2-e^2=e^2-c2

e^2=c^2 (no negatives so...)

e=c

so that might have been relatively obvious but its a proof you have to be annoyingly thorough

now the step i should have taken from the beginning is as you can notice i have shown that the bottom right and top triangles are similar. (also looking at the bottom right picture)

this is proven by

c+b+90=180 so they are compliments

similar reasoning shows c and d and a and b to be compliments

since they are similar triangle the sides have to be a fraction (i.e. 5/1,1/5 i mean i dont mean it has to be smaller then one) of eachother

since they each have hypot T then they are equal triangles

b=f

c=g

so as seen in the after some geomettry picture a^2+c^2=d^2=t^2 so d=t

thus they are all similar triangles and the angle made against the southern wall is the same as angle a

so the guy asked for

pi/4*3*3

and got

angle a(in radians)*3*3

so the ratio of the areas is the ratio of the angles

is this acceptable or did you want it expressed as a ration of sides?

Edited June 11, 2009 by final

[Prof. Templeton]

ok this post might be done but i didnt see an acceptable proof (i might be wrong) but here is mine (i think)

so you need to look at the picture otherwise your doomed to confusion

so by pathagerous

b^2+c^2=t^2=G^2+f^2

a^2+c^2=e^2+f^2

a^2+e^2=c^2+f^2

now subtact the second from the third or the other way around

c^2-e^2=e^2-c2

e^2=c^2 (no negatives so...)

e=c

so that might have been relatively obvious but its a proof you have to be annoyingly thorough

now the step i should have taken from the beginning is as you can notice i have shown that the bottom right and top triangles are similar. (also looking at the bottom right picture)

this is proven by

c+b+90=180 so they are compliments

similar reasoning shows c and d and a and b to be compliments

since they are similar triangle the sides have to be a fraction (i.e. 5/1,1/5 i mean i dont mean it has to be smaller then one) of eachother

since they each have hypot T then they are equal triangles

b=f

c=g

so as seen in the after some geomettry picture a^2+c^2=d^2=t^2 so d=t

thus they are all similar triangles and the angle made against the southern wall is the same as angle a

so the guy asked for

pi/4*3*3

and got

angle a(in radians)*3*3

so the ratio of the areas is the ratio of the angles

is this acceptable or did you want it expressed as a ration of sides?

Looks acceptable to me. The only exception I'll take is with the 45 degree angle you have on the first picture. There are several ways to prove that the two angle are the same. Yours is a version of the one I think is the easiest. Basicly taking what we know about similar triangles made from a line from the midpoint to the right angle and from an altitude off the hypotnuese and combining them.

In the first picture triangle ACD and ADB are similar so the two angles marked are the same. In the second picture triangle ACM can be proven to be isosceles so the two angles marked are the same. From that we can see angle DAB in the first triangle = angle ACB = angle CAM in the second.

Edited June 12, 2009 by Prof. Templeton