Geometry Forum - Problem of the Week
Solutions - Square on a Cartesian Coordinate System, Feb. 21-25, 1994
Jin Park
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The side of the square is 7.6157. A line from the corner of the square to P divides the right angle into two angles, a and b. There are also two triangles formed , APB with sides of x, 3, 7 and APD with sides of x, 3, 5. Using the Law of Cosines, 5^2 = 3^2 + x^2 - (2)(3)(x) cos a and 7^2 = 3^2 + x^2 - (2)(3)(x) cos b. Therefore, a = arccos((5^2 - 3^2 - x^2)/((-2)(3)(x))) and b = arccos((7^2 - 3^2 - x^2)/((-2)(3)(x))). The following program for the TI-81 tests values of x that will make a + b = 90. Input x arccos((5^2 - 3^2 - x^2)/((-2)(3)(x))) -> a arccos((7^2 - 3^2 - x^2)/((-2)(3)(x))) -> b Disp a Disp b a + b -> c Disp c
Anonymous Father
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I received one very unique answer from a Dad...using Autocad(?). - Cecil If the side of the square is equal to "s" then the coordinates of the square are A(0,0); B(0,s); C(s,s); and D(s,0). The coordinates of P will be (x,y). The distance formula for PA, PB, PD yields the following: (or the general equation of a circle could be used to get the same results) PA: x^2 + y^2 = 9 PB: x^2 + (y - s)^2 = 49 PC: y^2 + (x - s)^2 = 25 Adding each pair of equations gave the following results PA + PB s^2 - 2sy = 40 y = (s^2 - 40)/2s PA + PD s^2 - 2sx = 16 x = (s^2 - 16)/2s PB + PD x - y = 12/s The top and middle equations were solved for y and for x in terms of s. These equations now were used to make a table of values. It can be shown that 4 is less than s, is less than 8, since two sides of a triangle are always greater than the third so the first tablle developed will use those integral values. The outcomes for x and y could be checked in the third equation or in one of the original (seems like x - y = 12/s or x^2 + y^2 = 9 would be the easiest) s x y x^2 +y^2 As values for s yielded the respective values for x, y and x^2 + y^2, the value of s which gave 9 for the sum of the squares was the one we were after... s = 7.616.
Gino Perrotte
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I began by plotting the points on a graph. I followed the
directions and plotted point A at the origin. I the took a compass
and opened it to 3 units on the graph. I then put the compass point
on the origin and I made the circle on the graph. Now any point on
the circle is 3 units away from point A. I picked a point on the
circle in quadrant 1. I then opened the compass 5 units. From the
point on the circle I drew the circle. It intersected the x-axis
about 7.5. I only paid attention to where it hit the x-axis because
that is the distance for point D, which is on the x-axis. I then
opened the compass to 7 units and placed the compass point on the
point that is on the 3 unit circle that I used before. When I drew
the circle, it intersected the y-axis at 7.5. I only paid attention
to the y-axis because that is the distance for point B, which is on
the y-axis. Since the figure is a square, each side must be 7.5
units if two of the sides are that length. I conclude the last
statement by the definition of a square.
8 _|
7 _|- B . C
6 _|
5 _|
4 _|
3 _|
2 _|
1 _|
__ |_|_|_|_|_|_|_|._|____________________
A 1 2 3 4 5 6 7 8
D
Chris Stromoski
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To answer the problem, I first made an X and Y axis. I then
numbered it with each square equal to .5 units. I placed the point A
at the origin, and using a compass, I struck an arc which was three
units from the point A. I then opened my compass up to five units.
In .5 unit incraments, and starting at .5 units on the X-axis, I
struck arcs through the arc which is three units from point A. I
continued to do this until the arcs no longer touched the original
arc. Then the compass was opened up to seven units, and the same
process was done from the Y-axis. The point were all three arcs
intersected was then found. Since this point met all the criteria in
the question, it was labeled point P. The arcs that formed the
intersection with the original arc were found to be at the 7.5 mark
on each of the axsis. Therefore, the square was sides with a measure
of 7.5 units in length, and point C is at ( 7.5, 7.5 )
Kelly Hester
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We start out by setting maximums and minimums for the sides.
The minimum is two. This answer is arrived by subtracting The five
units from D to P and the three units from A to P. The maximum is
eight units. You get this answer by adding the three units from A to
P and the five units between D and P. Now that you have the
maximums and the minimums you can start the problem.
You take an x- and y-axis on a cartesian plane. Now you take
a circle with the radius of three and place the center of the circle
at the origin. You then make two more circles. One with a radius of
eight and another with the radius of five. You take both and put
their centers eight units from the coordinates (0,0). You know to
start them at eight units because that is the maximum length of each
side.
After getting all three circles on the plane you can move
them to arrive at an answer. The five unit and seven unit circles
are the only ones moved. Since point A's location is already known,
point P will definitely fall on the three unit circle and hence does
not need to be moved. The five and seven unit circle should be move
in equal incraments to ensure that all four sides of the square are
equal. When all three circles meet you're done. The distance from
(0,0) to the center of either the seven or five unit circle is the
length of the square's sides. Using this method, I've determined
that the length of each side is 7.615 units long.
Stacy Hester
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This is my result for the problem of the week.
First I constructed a graph on graph paper and plotted point A at (0,
0). Then using a compass I made an a circle with a radius of 3
units. Since point P must be 5 units away from point D, and D must
be on the X axis, I plotted D at (8, 0). This would be the point
farthest away from where P would be. Then I constructed a circle
with a radius of 5 units. b must be on the Y axis and is 7 units away
from P. So the farthest away that D could be would be 10. But, since
it is a square, and B is only 8 units away, I knew that D could also be
only 8 units away. So from (8, 0) I constructed a circle with a
radius of 7. (The distance that P must be from B.) I knew that the
circles had to meet in one point, and this would be P. But the
circles didn't match, so I moved point D to (7.5, 0) and again I
constructed a circle. I did the same for point B. (That moved it to
(0, 7.5)). The circles still didn't meet, this time I knew that I had
to move D and B out. So I moved them to (0, 7.6), and (7.6, 0). Now
the circles all met at one point, point P. So that meant that the
sides of the square must be 7.6 units long. (Which means that point C
is at (7.6, 7.6).
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