MHS.jpg

Welcome, friendly visitor! I'm delighted to see that you're here!!!

It certainly has been a while since I was here as a young freshman. Feel free to peruse my creations through the table contents at your right. You may notice below that a number of images are missing. This is due to the now-broken Latex equation editor I used that no longer exists. Oh, and sophomores - feel free to browse around my page. Happy times! Et bon chance!

Please be aware that many images may be missing, as the permalinks to the LaTex equation editor images no longer exist.


Links



Random Tidbits

John_Lennon.jpg Translational_motion.gif Custom_Converse.jpegChinese_Character.jpgBrian_May_Red_Special.jpg
Beatles_Persona-_Top.jpg




Quote of the Day




Xtranormal Movie Assignment


This is a video Evan and I made about rhombi for our wiki assignment that is due on Valentine's Day.




Parallelogram Wallwisher Assignment

This is our wiki assignment for January 30, 2011.
I used problem 4 from page 39 of the Exeter workbook.

Band On the Run

The diagonals of a rhombus have lengths of 18 and 24. How long are its sides?

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

Accept that since this is a rhombus, the diagonals must be perpendicular to each other and they bisect each other.

Since you know the length of each diagonal, find half of each and accept that a rhombus is made of four right triangles. Use the Pythagorean theorem to find the length of the sides.
Find half of each diagonal: external image latex2png.2.php?z=100&eq=%5Cfrac%7B1%7D%7B2%7D*18%3D9, external image latex2png.2.php?z=100&eq=%5Cfrac%7B1%7D%7B2%7D*24%3D12
Use the Pythagorean theorem to find the side lengths: external image latex2png.2.php?z=100&eq=%5Csqrt%7B9%5E2%2B12%5E2%7D%3D%5Csqrt%7B225%7D%3D15
Each side is 15 units long.
Really, the only way you can mess this one up is if you overthink it.


Parallelogram Proof Assignment

This is our GeoGebra assignment due January 23, 2011.

Wiki_Assignment_1-23-2011.JPG

Above is a screenshot of the applet I used for the assignment. Below is the accompanying screencast.





Complex Number Assignment

This is our GeoGebra assignment for winter break.

Screenshot

Wiki_Assignment_Winter_Break.JPG

Applet

Due to the size of the applet, I cannot embed it, so here is a link:

Screencast



Helpful Information

Starting Points
Name
Value
TomPetty=-2+4i
-2+4i
TheHeartbreakers=4+2i
4+2i

Calculations
Operation
Result
Point
TomPetty+TheHeartbreakers
2+6i
You'reGonnaGetIt
TomPetty*TheHeartbreakers
-16+12i
SouthernAccents
TomPetty-TheHeartbreakers
-6+2i
HardPromises
TomPetty/TheHeartbreakers
i
LetMeUp
TheHeartbreakers/TomPetty
-i
TheLastDJ
TheHeartbreakers-TomPetty
6-2i
LongAfterDark
TomPetty^2
-12-16i
IntoTheGreatWideOpen
TheHeartbreakers^2
12+16i
Echo


GeoGebra 3D Assignment

This is our GeoGebra challenge for the week of December 6, 2010.

Here is a link to the applet:


Reflection Assignment

This is our GeoGebra challenge for December 3, 2010.

Unfortunately, the file is too big for me to embed here, so below is a screenshot and a link to the applet.
Wiki_Assignment_12-3-2010.png
Here is a link to my interactive applet:

So basically, you start out by finding a perpendicular line from the point to the line. You then need to find the intersection point and create a vector from the point to the intersection. After that, you simply add the vector onto the intersection point.
I will demonstrate this method on the original points in the applet.

Here’s what you need to know to start. In the original setup, this holds true:
  • Line of reflection: y=RedSpecial*x (in this setup, RedSpecial=1)
  • JohnDeacon: (2,-5)
  • FreddieMercury: (2,-2)
  • BrianMay: (6,-2)
  • RogerTaylor: (6,-5)
(Just so you know, “perpendicular” as used below will refer to the line perpendicular to the line of reflection that goes through the given point.)

John Deacon
  • Perpendicular: y=-x-3
  • Intersection: (-1.5,-1.5)
  • Vector: [-3.5,3.5]
  • Applied to intersection: (-1.5,-1.5) + [-3.5,3.5] = (-5,2)
  • Reflected point: (-5,2)
FreddieMercury
  • Perpendicular: y=-x
  • Intersection: (0,0)
  • Vector: [-2,2]
  • Applied to intersection: (0,0) + [-2,2] = (-2,2)
  • Reflected point: (-2,2)
BrianMay
  • Perpendicular: y=-x+4
  • Intersection: (2,2)
  • Vector: [-4,4]
  • Applied to intersection: (2,2) + [-4,4] = (-2,6)
  • Reflected point: (-2,6)
RogerTaylor
  • Perpendicular: y=-x+1
  • Intersection: (.5,.5)
  • Vector: [-5.5,5.5]
  • Applied to intersection: (.5,.5) + [-5.5,5.5] = (-5,6)
  • Reflected point: (-5,6)


Isometry Assignment

This is our GeoGebra challenge for November 19, 2010.

Applying Isometry to Rush

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

Created with GeoGebra


As you may have noticed, I am sticking with a musical theme, so I have applied isometry to Geddy Lee, the lead singer, bassist, and keyboardist of Rush.

For the base construction, I have attached an image of Geddy Lee to three points and used a vector to keep the fourth point in proportion. I applied the transformations to these points, but I have hidden them for visual simplicity.

Here are my explanations of each isometry:
  • Reflection: I reflected the points and the image over the line y=(NeilPeart*x)+AlexLifeson, which is controllable with sliders.
  • Rotation: I rotated the image by the angle GeddyLee around the origin, which is labeled as Rush.
  • Translation: I entered (x(point) - JeffJones, y(point)) for each point on the image, which translated the point based on the slider JeffJones.
  • Glide Reflection: For each point on the image, I entered (-x(point), y(point) + AlexLifeson). This reflected everything over the y-axis and glided it based on slider AlexLifeson.


Sequence Assignment

This is our GeoGebra challenge for November 12, 2010.

Applying Sequence to The Who

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

As you may have noticed, I have constructed The Who's famous bullseye logo. I did this using circles and filling. To apply a sequence to the outer blue circle, I entered Sequence[Circle[Keith, Roger], Roger, -20, 20, 1] into the input line, which shows up as AcidQueen. The two crossing lines of dots are the following sequences: Sequence[(Pete, John), Pete, -20, 20, Kenney] (which shows up as BorisTheSpider) and Sequence[(Pete, John), John, -20, 20, Kenney] (which shows up as TheWho). I'm not really sure what AcidQueen is doing. I believe that TheWho and BorisTheSpider are showing the limits for the size of my circle, based on the limits I placed on sliders Pete and John.

Furthermore, see if you can figure out all of the references I made to The Who.

Created with GeoGebra




Triangle Assignment

This is our GeoGebra challenge for the week of November 1, 2010.
Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

Created with GeoGebra




Perpendicular Parametrics

This is our GeoGebra challenge for the week of October 25, 2010.

This is the applet I was assigned to create for the wiki this week. The idea is that the lines created by each parametric equation should be perpendicular to each other. They should also arrive at the intersection point one t apart.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

A=(2t-14,2t-12)
B=(t-8,-t+10)

Created with GeoGebra




Parametric Portfolio

A Nowhere Situation

Steve Perry, former lead singer of Journey, is running a race through Madeira along Camargo Road. This particular street happens to follow the line y=(2/3)x, assuming the intersection of Miami Avenue and Camargo Road is at the origin, where the race begins. After one unit of time from the start of the race, Steve is at the point (1,2/3). Solve the questions below.












Due to the memory required for applets that include images, I have posted a screenshot of my applet and a link to the interactive applet. Notice the Journey references throughout the page.
Applet_Screenshot.JPG
This is a screenshot of my applet.




Ask the Lonely

Try the following problems:
  1. Create a parametric equation to model Steve Perry's race.
  2. Determine Steve's pace.
  3. How far is the finishing point from the start?
  4. When is Steve 10 units from the start?
  5. Assume that Separate Ways (Worlds Apart) takes nine units of time to play. Where will Steve be at the end of the song?
  6. How long does it take Steve to finish the race?
  7. If Steve could theoretically keep this same pace for 500 units of time, how far would he have traveled?
  8. Without using a perpendicular bisector tool, determine when and where Steve is equidistant from the start and finish.

Only Solutions

Okay, so now let's go through solving these questions and how to do so.

1) x=3t, y=2t. From the point-slope form of the equation, you can determine the coefficients of t. Since the starting point is at the origin, no other information is necessary.

2) Steve Perry runs at a pace of approximately 3.606 units per unit of time. This is found by determining the rise and run from the starting point to the point in question and substituting them into the Pythagorean theorem (it is assumed here that t=1).
external image latex2png.2.php?z=100&eq=%5Csqrt%7B2%5E2%2B3%5E2%20%7D%3D%5Csqrt%7B13%7D%3D3.606

3) It is approximately 21.633 units from the start to the finish of the race. This is found by substituting the x- and y-values of the finishing point into the Pythagorean theorem.
external image latex2png.2.php?z=100&eq=%5Csqrt%7B18%5E2%2B12%5E2%20%7D%3D%5Csqrt%7B468%7D%5Capprox21.633

4) Steve is ten units from the start after 2.77 units of time. He will be on the point (8.32,5.55). Below is a link to an applet that will model this and explain how to solve it.


5) Steve will be at the point (27,18). This is found by solving the parametric equation such that t is equal to nine.
external image latex2png.2.php?z=100&eq=x%3D3t%3D3*9%3D27, external image latex2png.2.php?z=100&eq=y%3D2t%3D2*9%3D18

6) Steve will finish the race after six units of time. Since the start is at the origin, you can find this by substituting the point at which the finish is located into x and y in your parametric equations.
external image latex2png.2.php?z=100&eq=18%3D3t%5Crightarrow%20t%3D6, external image latex2png.2.php?z=100&eq=12%3D2t%5Crightarrow%20t%3D6

7) In 500 units of time, Steve would travel approximately 1,802.776 units of distance. You can determine this by substituting 500 for t in your parametric equations and then substituting x and y for a and b in the Pythagorean theorem.
external image latex2png.2.php?z=100&eq=x%3D3*500%3D1500, external image latex2png.2.php?z=100&eq=y%3D2*500%3D1000
external image latex2png.2.php?z=100&eq=%5Csqrt%7B1500%5E2%2B1000%5E2%20%20%7D%3D%5Csqrt%7B3%2C250%2C000%7D%5Capprox%201802.776

8. Steve is equidistant from the start and finish after three units of time. He is at the point (9,6). This can be determined by finding the intersection point of the parametric equation and external image latex2png.2.php?z=100&eq=%5Csqrt%7Bx%5E2%2By%5E2%20%20%7D%3D%5Csqrt%7B%28x-18%29%5E2%2B%28y-12%29%5E2%20%20%7D. This equation comes from substituting the start and finish points into the distance formula. Once you determine the values of x and y, you can simply substitute them into the parametric equation.
external image latex2png.2.php?z=100&eq=9%3D3t%5Crightarrow%20t%3D3, external image latex2png.2.php?z=100&eq=6%3D2t%5Crightarrow%20t%3D3
Here is a link to an applet I built to model this question:


Don't Stop Believin'

external image 633558289059773896-StevePerryWontstopbelievingMotivator.jpg&t=1
I love motivational posters...


Paper-Folding Problem

This is from September 9, 2010.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

This is the paper folding problem we did from a worksheet during class on September 9th.

Michael Grimm, Created with GeoGebra




Defining a Kite

This is a question taken from page 8 of the Exeter workbook: If you were writing a geometry book and you had to define a mathematical figure called a kite, how would you word your definition?

I think it would help if we were to first look at what one might first think of when they hear "kite."
external image Kite.png
Now that we have that cleared up, this should make defining the shape a lot simpler. Let's make some observations first:
  • This is a quadrilateral.
  • Clearly, we have no right angles present at any joints.
  • The kite itself is symmetrical.
  • The upper two sides are both the same length, as are the bottom two sides.
  • Notice that if you connect opposite points, the intersection in the middle will form four right angles.

Taking these observations into consideration, let's write a definition. One might define a kite as a symmetrical quadrilateral in which no more than two sides are the same length and the intersection of opposite angles will form four right angles. The sides of the same length must be adjacent to each other.

Now let's see how another source defines a kite. According to the supposed experts at Wikipedia, a kite is "a quadrilateral with two disjoint pairs of congruent adjacent sides." Let's break this up a bit into more understandable language:
  • A quadrilateral is a four-sided polygon.
  • Disjointed means not connected.
  • Congruent essentially means equal.
  • Adjacent means meeting at a common vertex.

So basically what Wikipedia is telling us is that a kite is a four-sided figure with two sets of equal sides that are not connected. This doesn't seem right, does it? Furthermore, how can sides meet at a common vertex if they are not connected? Think about this a bit.



Michael G.
Class of 2014
Madeira High School