please help
For the second problem the area of overlap is independent of the
relative orientation of the two squares. So orient them so that finding
the area of overlap is easy, and that area is the area of overlap for
any other orientation.
Below is the hint text I produced when this was the Geometry Problem
of the week on The Math Forum at Drexel (green square is the darker
square in your diagram, and the blue the lighter square:
Also I have now found the diagram that goes with this problem and I
have attached it.
=================================================
I hope you will find the following of some help
in solving this problem:
Hint text for "Squares on Squares"
1. The first thing you could do with this problem
is to produce a labelled diagram. So copy the
diagram from the problem and follow the instructions
below
2. Draw a horizontal line from P to the right
hand edge of the green square, and label the
point where the line meets the square u. Also
label the point on the right hand edge of the
green square above u, where it meets the edge
of the blue square v.
3. Draw a vertical line from P to the top
edge of the green square and label the
point where the line meets the square w.
Also label the point on the top edge of the
green square to the left of w, where it meets
the edge of the blue square x.
4. Now look at triangles Puv and Pwx. Do they
have the same area? If so, can you explain why
the areas are the same. What does this tell
us about the area of the overlap?
RonL
Problem of the Week 25
Part 1.
With N lines, each line cuts N-1 other lines, which apparently
gives us N.(N-1) points, but we have counted each point twice
in this argument so there are N.(N-1)/2 points of intersection.
Reality check:
Two lines cut each other at 1 point, and 2*1/2=1 so that's OK
Three lines cut each other at 3 points and 3*2/2=3 so that's OK
Four lines cut each other at 6 points, and 4*3/2=6 so that's OK
when N=10 our formula gives 10*9/2=45 points of intersection.
RonL
Hello,Originally Posted by kkirk909
seems to me as if your teacher likes to make some jokes.
I've thought about your 2nd post:
When the square on top is situated as I've drawn in my 1st drawing you have exactly a quarter of the square as overlayed area.
If the suare on top rotates a little bit further on you only have to show, that the triangle T1 which you "cut" from quarter of the square has the same value for the area as the triangle T2 (use cogruent rule(?) ASA on both triangles.
That means the overlayed area doesn't change it's value for the area: It has always a quarter of the square which means 64 square inches.
Bye
Problem of the Week 25
Part 2.
Suppose that we know how many regions $\displaystyle R(n-1)$ that $\displaystyle n-1$ lines satisfying
the conditions of the problem devide the plane into.
Now add a new line starting from a long way away, as it comes in it divides
the region that it was being drawn through into 2 parts when it meets the
first of the existing lines, thus increasing the region count by 1.
Similarly when it reaches the second line it adds one more region to the
region count, and so on untill it has cut the last of the existing lines. At this
point the region count has increased by $\displaystyle n-1$. As the line
moves off towards infinity it divides the last region into 2.
Thus $\displaystyle R(n)=R(n-1)+n$, which we may use to evaluate $\displaystyle R(10)$,
or we can expand this and find the closed form for $\displaystyle R(n-1)$ .
$\displaystyle R(1)=2$,
$\displaystyle R(2)=2+2=4$,
$\displaystyle R(3)=4+3=7$,
$\displaystyle R(4)=7+4=11$,
$\displaystyle R(5)=11+5=16$,
$\displaystyle R(6)=16+6=22$,
$\displaystyle R(7)=22+7=29$,
$\displaystyle R(8)=29+8=37$,
$\displaystyle R(9)=37+9=46$,
$\displaystyle R(10)=46+10=56$.
RonL