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Here's a few problems I need help with:
1.We are given a regular heptagon of side length 1. Let S be the set of points that are within a distance of 1 from some point on or inside the heptagon, but not including the heptagon itself. Find the area of S.
2. Let C and D be points on the semicircle with diameter \overline{AB}. If AD = BC = 3 and CD = 7, then find the radius of the semicircle.
Thx!
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I have one more problem I need help with. This isn't really circle geo but...
3. Equilateral triangle ABC has centroid G. Triangle A'B'C' is the image of triangle ABC upon a dilation with center G and scale factor -2/3. Let K be the area of the region that is within both triangles. Find K/[ABC].
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Could someone plz help? I really need help because it's due soon.
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#1) Why does your drawing have rounded points like that?
Last edited by ElainaVW (2014-08-08 10:07:51)
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I think it's supposed to. The heptagon is the white figure and then the set of points that are within a distance of 1 from some point on or inside the heptagon is the shaded part.
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I actually got the first 2 problems. Thx EVW on your help on problem 2!
I still need help on this problem though: Equilateral triangle ABC has centroid G. Triangle A'B'C' is the image of triangle ABC upon a dilation with center G and scale factor -2/3. Let K be the area of the region that is within both triangles. Find K/[ABC]. I thought K was the area of the dilated equilateral triangle but that doesn't seem to be the case (the answer I got was 4/9).
Thanks!
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For some strange reason I cannot post my help here. EDIT: This problem has been solved by bobbym. See post 9.
4/9 is right for the smaller triangle. But you want the overlap.
I have done this before but I cannot find the thread so I'll have to start again. ......................................
Right. Here we go:
Let's start with setting AH = 9a.
As the centroid is 1/3 of the way up the median you can now work out in terms of a:
GH, then AG.
Lengths in A'B'C' will be 2/3 of the lengths in ABC so you can then work out in terms of a:
A'G and hence A'H
You can also work out A'J, B'L and hence JL.
You now know the relative sizes of ABC, CJL and A'JK.
So by adding and subtracting the right amounts you can work out the overlap in terms of ABC.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob;
403 error?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Q1.
This diagram answers ElainaVW;s question.
To get points that are 1 unit from the straight sided heptagon you need straight lines completing the red bits, and parts of a circle radius 1 to go round the points.
It helps to be familiar with UK coins here. The 50p and 20p coins have this property. They have constant width and so can be correctly detected in slot machines.
So calculate 7 red bits, and 7 sectors.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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This is a problem that up to now has only affected Agnishom and I. The server does not like some word combination that you have in your post. This is localized to a single thread usually. Do not try to force the post to pass, the server will then bar you. Rather, change the wording of the post until the error disappears.
I have entered the post for you.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I'v also had that problem with the word "p.o.k.e.r".
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Exactly that did happen. I tried to post it last night and got the error. I tried again, then logged off and tried to get back on. It wouldn't let me. (ha!) Then it got worse. I couldn't even get the site up. Had to email Rod.
Thanks for your help.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi;
I have entered your work in post #9. Please read post #10 for what to do to prevent being locked out or barred.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks for the post insert. I've read and inwardly digested your advice. yum
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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What did you guys get for Q3?
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hi RandomPieKevin
I've had to re-work this. Hope I've got it right. I'm getting 11/27
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Can you explain how you got it? Thanks.
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I'm using the diagram in post 7. I need to label one more point. H' is the point where AA' and C'B' cross.
As we only want ratios of areas it doesn't matter exactly what units I use, so I've simplified my working by assuming the easiest units.
Let's say the area of ABC is 81 units.
Then (using the scale factor 2/3) the area of A'B'C' must be 81 x2/3 x2/3 = 36
And let's say that AH = 9 units.
Then AG = 6 units, GH = 3 units.
(G is the centroid of both triangles, so I don't need G')
A'H' = 2/3 of 9 = 6
So A'G = 4 (and GH' = 2)
so A'H = A'G - HG = 4 - 3 = 1
So the little triangles like A'JK are 1/9 of ABC.
area AJK = 1/9 x 1/9 x 81 = 1
So the required are is 36 - 1 - 1 - 1 = 33
So k = 33/81 = 11/27
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Can someone help for problem 1?
EDIT: Nevermind, got it
Last edited by Enshrouded_ (2015-09-04 13:15:27)
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hi Enshrouded_
To be within 1 unit of the heptagon but not including the heptagon, the locus is the 7 squares shown here in black plus the 7 sectors of a circle.
You can work out the angle of each sector like this:
Where one sector meets two squares and one internal angle of the heptagon, the angles add up to 360. The internal angle is (2n-4) right angles divided by 7 so do this
There's no need to calculate this angle as 7 of these adds up to a whole circle, radius 1 unit.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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