Apr

02

2014

02

2014

A uniform bar has identical frictionless wheels on each end. The distance between the axle centers is 1m. The bar is suspended by a rope attached to the lower axle and to a fixed point offset from the wall a distance that equals the wheel radius. The length from the attachment point to the axle center is 1.5m. Neglecting the weight of the rope, at what angle (a) will the system rest at equilibrium? A: a ≈ 30 degrees B: a ≈ 45 degrees C: a ≈ 50...

Mar

06

2014

06

2014

Several entrants correctly answered the Balancing Act Geek Challenge. The correct answer is B, the scale tips to the right. Correct answers were received from Tim Jager of DMC, Devon Fritz of DMC, Brandon Williams of Yaskawa, Ian Schleifer, and Gareth Meirion-Griffith of DMC. Many people answered that the scales would tip to the left. The intuitive answer points that way. Since the pool ball is suspended externally, it is easy to attribute all of its weight to the external supp...

Jan

08

2014

08

2014

December’s Infinite Snowman Geek Challenge winner is John Jacobsma of Dickson. Adnaan Velji of DMC also answered all questions correctly. The determination of best answer goes to John due to his elegant solution for the Center of Mass. The correct answer to the primary questions is B, that the snowman will be 3m tall. The extra credit answers are that its construction will not consume more than the available snow, and that the belly button at the center of mass sh...

Dec

02

2013

02

2013

Ever wanted to construct the most mathematically magnificent snowman of all time? Here’s your chance. Figure out the Geek Challenge below…and don’t forget to actually build your snowman! A snowman is to be built with an unlimited number of spherical snowballs where the diameter of each ball is 2/3 the diameter of the ball below. If the first ball is 1m diameter, how tall will the completed snowman be? A: 2m B: 3m C: 5m E: Infinitely tall. Ex...

Nov

06

2013

06

2013

Last month, we asked you to find the odds that a class of 23 students has one or more shared birthdays. The correct answer to the Baffling Birthdays question is C: 50.7%. The winner of this month’s Geek Challenge is John Jacobsma of Dickson. His answer covered the basics, as well commented on the problem’s implicit assumptions. There were several people who also correctly answered this month’s Geek Challenge. They are: Jesse Batche of DMC Joseph C. Bond ...

Oct

04

2013

04

2013

As school got back in session this fall, the first grade teachers at a large elementary school posted a list of student birthdays. The teachers noticed that of the 8 first grade classes at the school, 4 of the classes had students with duplicate birthdays. Each of the classes have just 23 students. Is their observation a statistical anomaly, or it a highly probable? What are the odds that a class of 23 students has one or more shared birthdays? A: 5.8% B: 50.0% C: 50.7% D: 75.0%...

Aug

01

2013

01

2013

Congratulations to the winner of the Matrix Mind-Boggler Geek Challenge, Jordan Kuehn of the Colex Group! Jordan wrote a program which generates random matrixes using an efficient technique. He gets extra props for including a DMC logo on his program! I added extra annotations to demonstrate how the solution works. The key to solving the problem is noticing that each row and each column is used exactly once. Therefore, if each cell was the sum of numbers representing the row, and a number rep...

Jun

11

2013

11

2013

Special thanks to Han Yang for providing this month’s Geek Challenge. A 4x4 matrix of numbers can be devised such that when any 4 cells are chosen where none of the chosen cells share a row or column with another chosen cell, the sum of the chosen cells is 25. Below is an example of such a matrix with a chosen set of cells highlighted in yellow. The sum of the yellow cells is 25. Here is the same matrix, with a different set of cells, where the chosen cells also...

May

07

2013

07

2013

Thanks for playing the Geek Challenge Crossword Conundrum. Last month's challenge had three winners: Harry Maddock of Tacoma, WA, Adnaan Velji of DMC, and Grant Anderson of DMC. The geeky answer key is shown here (click to enlarge): Submit your comments to geekchallenge@dmcinfo.com.

Apr

22

2013

22

2013

This month's Geek Challenge takes the classic quick crossword puzzle to a new nerdy level. Simply answer the clues to solve the puzzle. Submit your answers to geekchallenge@dmcinfo.com. The first person to respond with the most correct answers will be this month's winnner.

Mar

20

2013

20

2013

Last year we created a formula-based March Madness tournament picking challenge with several potential meaningful statistics to use to generate a winning formula. While fun, the exercise proved that the winning formula doesn't depend on logical inputs. This year, we challenge you to pick the best bracket using some less conventional stats. The winner will pick the best bracket by predicting the winners based on their uniform color, nickname/mascot, and the length of head c...

Feb

11

2013

11

2013

We had one correct answer to the Geek Challenge last month, and it came from Allen Hutchison of Smalley Steel Ring. The correct answer is D, that all three transformations are possible. Thanks to Han Yang for providing the content for the Malleable Mystery challenge. She found the images we used on a Chinese math blog. Allen’s solution was to do some Terminator style morphing and re-joining. However, the transformations can also be done using continuous transformation, and without...

Jan

16

2013

16

2013

January’s Geek challenge is from the field of Topology, which is the study of continuity and connectivity of shapes. This field of math gives us such gems as the Hairy Ball Theorem. A Continuous Deformation of a shape is what you can do to a piece of clay if you never sever the clay, or rejoin the clay at any point. You can squish, bend and twist all you want at any point. Using Continuous Deformation, a coffee cup can be transformed into a doughnut, as demonstrated in this animati...

Dec

13

2012

13

2012

The best use of the mystery figure (in my opinion), and the answer to last month’s Geek Challenge is B: Used as an element of a proof of the Pythagorean Theorem. The figure generates an amazingly short derivation of the theorem, as our winner Gareth Meirion-Griffith of DMC demonstrates with just 2 lines of algebra: Consider the figure above. The figure consists of: 1) A large square with sides of length C and an area of C2 2) Four right triangles with sides of lengt...

Nov

15

2012

15

2012

What is distinctive about this figure? A. It is the optimum target for focusing a monochrome cameras B. It is an element of a proof of the Pythagorean Theorem C. It is used to design a centerless spiral staircase D. It recognized as one of the most challenging shapes in Origami E. There’s an even better use, not listed above Submit your answers to geekchallenge@dmcinfo.com. As always, the answer with the best engineering content will be this month&rs...

Sep

12

2012

12

2012

Four groups of 225 random 0’s and 1’s are presented. One of these groups is perfectly random like a coin flip (although calculated from Excel’s RAND() function). The other three groups, while still random, have rules applied which make them not a pure coin flip. Can you spot the perfectly random sample? What methods do you use to spot it? Submit your responses to geekchallenge@dmcinfo.com. As always, the correct answer with the best engineering c...

Aug

15

2012

15

2012

Last month's Geek Challenge asked, "What is the quickest way to get from point A to point B?" And they’re off! Path A, the Line, starts with a commanding advantage, having the shortest distance to travel by taking a direct path to the endpoint. However, it is quickly passed by as its acceleration is just too slow. Path B, the Cycloid, takes an early lead with a vertical drop to gain speed, then a rapid change of course toward the endpoint. Paths C and D, the Circle and Parabola,...

Jul

09

2012

09

2012

This month’s Geek Challenge is about the quickest way to get from A to B. An object is released from rest at Point A. Influenced by the constant vertical acceleration of gravity, it slides on a frictionless path to Point B (if it's helpful, imagine a soap box derby car). The following paths represent the vertical profiles of the racing courses from the starting line, A, to the finish line, B. Which of the 4 paths depicted gets the object from A to B in the least time? Extra cre...

Jun

06

2012

06

2012

Dan Freve of DMC wins this month’s Geek Challenge by correctly selecting D as the correct combination! Dan saw that at the maximum radius, 2, Graph A appeared to be zero while graphs B and C were clearly negative. Upon evaluating the formulas, only Formula 2 evaluates to zero at R=2, so 2→A was a necessary match. Of the multiple choice answers, only D had the 2→A paring, so this was the correct answer. ...

May

10

2012

10

2012

This month’s Geek Challenge is in honor of Cinco de Mayo. The traditional Mexican sombrero has an elegant symmetrical shape, and is fun to express mathematically. Your challenge is to match three sombrero graphs with their corresponding functions in cylindrical coordinates. Each graph is plotted from 0<r<2. Graph A Graph B Graph C Is the correct paring? A: 1→A 2→C 3→B B: 1→C 2→...

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