Nov

05

2014

05

2014

Thirteen people correctly answered the Crossing of the Chords Geek Challenge by selecting C: 70 intersections. In their explanations, two very distinct methods were demonstrated to arrive at the general equation for intersections as a function of perimeter points. This month’s winners are John Jacobsma of Dickson, Devon Fritz of DMC, Sudeep Gowrishankar of DMC, and Adnaan Velji of DMC. They used a Combinations method to arrive at the method very efficiently. To give proper contex...

Oct

01

2014

01

2014

A number of points (N) are distributed along the circumference of a circle. Every point is connected to every other point by a chord. The points are spaced unevenly such that no more than two chords intersect at a common point inside the circle. As demonstrated in these figures, when there are 4, 5 and 6 perimeter points, there are 1, 5 and 15 internal chord intersections respectively. If N=8 (circle with 8 points on the perimeter), how many internal cord intersections are...

Jul

08

2014

08

2014

Last month’s Geek Challenge was to find the limiting curve created by drawing lines on graph paper in a certain pattern. The correct answer is C, and the values for the constants were 1, 2, and 1. These three people answered the question correctly: Andrea Gotti of Milan Polytechnic, Adnaan Velji of DMC, and Dan Freve of DMC. Of these, the winner is Andrea Gotti. Andrea solved the problem as follows: Each line of the paper doodle passes through two points. Those points will ...

Jun

11

2014

11

2014

A graph-paper doodle connects each point on the x-axis with a consecutive point on the y-axis with a straight line. Within a defined boundary, the positions on the x-axis move away from the origin as the positions on the y-axis move toward the origin. The resulting shape fills a defined curving boundary with a pretty pattern. The geek challenge for this month is to define the curve generated by these straight lines. Focusing on the lines in the positive x and y quadrant, select the functi...

Apr

29

2014

29

2014

Three people successfully solved the Equilibrium Angle Geek Challenge. They were Dan Freve of DMC, Brandon Williams of Yaskawa and Jeff Winegar of DMC. The correct answer is C: a ≈ 50 degrees The interpretation of this answer is that if the bar is released from the wall at an angle <50°, it will fall back to the wall with the bar still suspended from its bottom point. If the angle is >50°, the angle will increase, the bar will invert and ultimately dangle from its t...

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...

Feb

05

2014

05

2014

An apparatus is constructed as shown below comprising equivalent buckets of water. The buckets and water were placed on the scale first, and it balanced. Then two suspended balls of equal diameter are added in the configuration shown. One is wooden, and floats. It is suspended from the bottom of the bucket. The other is a pool ball that sinks. It is suspended externally. Assuming the weight and displacement of the strings is not significant, what happens ...

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 ...

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....

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. ...

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% A...

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 ...

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 add up to 2...

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. Learn more about DMC's company culture.

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 he...

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 with...

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 anima...

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 l...

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