Assignment 6: Graphs
A 4 by 4 picture hangs on a wall such that its bottom edge is 2 ft above your eye level. How far back from the picture should you stand, directly in front of the picture, in order to view the picture under the maximum angle? Side view:
There are (hopefully) a few ways to solve this problem
1) By taking measurements in GSP, it's very easy to follow the resulting angle change from sliding point E along the x axis. Our viewing angle TEB appears to be maximized at 30 when E is around 3.5
2) This point can be found through geometry/trig. and possibly calculus.
3) There is hopefully a geometric way in GSP to find the exact point and measurement.
I can't quickly think of the method to solve using GSP geometry so lets first explore the math.
We have a number of right triangles and fixed distance measures along the y axis, so we should be able to optimize using trig.
Note that TOE is a right triangle as is BOE.
Tan (<TEO) = (4+2)OE, so <TEO = tan-1 (6/OE)
Tan (<BEO) = 2/OE and <BEO = tan-1 (2/OE)
<TEB = <TEO - <BEO
<TEB = tan-1 (6/OE) - tan-1 (2/OE)
Now we expect angle TEB to vary over the values of OE, so let's graph using graphing calculator to get a visual. See purple graph to the right.
We know that the min point of the equation can be found when the derivative = 0, as is indicated by the red graph (line). This validates the approximately 3.5 answer we suspected.
For the life of me, I haven't a clue what the derivative of atan() is anymore, so look it up and find it at:
so now we know the D[arctan(x)] = 1/(1+x^2). Since our x in this case is the 6/OE we'll need to use the chain rule as well.
If we let the length OE = x and substituting into <TEB = tan-1 (6/OE) - tan-1 (2/OE) and set the derivative = 0 yields:
1/(1+ (6/x)^2)* (-6/X^2) = 1/(1+ (2/x)^2)*(-2/X^2)
1/(1+36/x^2) * (-6/X^2) = 1/(1+4/x^2) *(-2/X^2)
1/((x^2 + 36)/x^2)) * (-6/X^2) = 1/((x^2 + 4)/x^2) *(-2/X^2)
-6/ (x^2 + 36) = -2 / (X^2 + 4)
-6x^2 -24 = -2x^2 -72
4X^2 = 48
X^2 = 12
X= Sqr (12)
So we have our answer, and sqr (12) is approx. 3.464 so the answer matches our expectations above.
If we draw a circle using three points TE and B, then we will have each segment as a chord on our circle for which we know certain properties.
We know that the chords formed by subtending another chord on the circle from a common point, have similar angles, as do <TEB and <TE'B.
Through manipulation and review of the measured angle, we see that the angle appears to be maximized when E and E' first meet when the circle is tangent
This picture is getting busy, so go to next tab for circle of Apollonius for our points T B and the eye level line.
So here we have our point P. measurements in GSP don't work, but it appears to be at 3.5 as per above results.
Construct a second triangle with the three sides having the lengths of the three medians from your first triangle. Find some relationship between the two triangles. (E.g., are they congruent? similar? have same area? same perimeter? ratio of areas? ratio or perimeters?) Prove whatever you find.
then the triangle of medians is equilateral. Will an isosceles original triangle generate and isosceles triangle of medians? Will a right triangle always generate a right triangle of medians? What if the medians triangle is a right triangle? Under what conditions will the original triangle and the medians triangle both be right triangles?
3. Medians for Triangle
hangs on a wall such that its bottom edge is 2 ft above your eye level. How far back from the picture should you stand, directly in front of the picture, in order to view the picture under the maximum angle? Side view:
5. The football rules in college football were changed a few years ago have made the uprights 5 feet narrower than previously. Many game commentators have harped about how much harder it is to kick field goals from the hash marks. Assume the field goal is attempted from the hash marks. At what yard marker does the kicker have maximum angle to the two uprights. Note: You will need to find out the width of the uprights and the width of the hash marks . . . make a sketchpad model. Is there any merit to some commentators argument to take a penalty in order to have a "better angle" on the field goal kick?
6. Given three points A, B, and C. Draw a line intersecting AC in the point X and BC in the point Y such that
AX = XY = YB
7. Construct the common tangents to two given circles. Make a script tool. Test it for all the different cases. Or, do you need different script tools for the different cases?
8. Of a triangle, given two vertices A and B, and the angle at the third vertex C (the angle opposite side AB). What is the locus of the point C? See Script # 18 in Assignment 5.
(Note: Think of this as turning a fixed angle so that its sides always rest on the two endpoints of the segment AB. Remember that the size of the angle is fixed but the sides can move along A and B simultaneously.)
9. A parabola is the set of points equidistant from a line, called the directrix, and a fixed point, called the focus. Assume the focus is not on the line. Construct a parabola given a fixed point for the focus and a line (segment) for the directrix.
a. Use an Action Button to generate the parabola from an animation and trace of a constructed point.
b. Repeat 9a with a trace of the tangent line at the constructed point.
c. Use the locus command to generated the parabola from a constructed point or the tangent line at that point.
10. Construct the locus of points equidistant from a fixed point F and a circle. In other words, repeat the parabola construction but use a circle as the "directrix." Let F be any point in the plane other than the center of the circle. Assume F is not on the circle; it can be either inside or outside.
11. Consider any triangle ABC. Find a construction for a point P such that the sum of the distances from P to each of the three vertices is a minimum.