# Entries for the 'Geek Challenge' Category

May
07
2013

## Geek Challenge Results: Crossword Conundrum

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

## Geek Challenge: Crossword Conundrum

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

## Mascot Madness 2013 Geek Challenge

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

## Geek Challenge Results: Malleable Mystery

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

## Geek Challenge: Malleable Mystery

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

## Geek Challenge Results: Baffling Figure

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

## Geek Challenge: Baffling Figure

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

## Geek Challenge: Perfectly Random Puzzle

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

## Geek Challenge Results: Gravity Race Riddle

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

## Geek Challenge: Gravity Race Riddle

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

## Geek Challenge Results: Sombrero Stumper

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

## Geek Challenge: Sombrero Stumper

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→...

May
09
2012

## Geek Challenge Results: March Madness Part II

A big question that lingered after looking at the results of the geek challenge was how much luck played into the outcome against how strong any specific entry was over any other.  Most entries relied heavily on seeding and used other factors to tweak the effect of the overall seeding. Only Dan Freve chose a different approach, weighting strength of schedule heavily and actually using a negative coefficient on the tournament seed. His formula is reasonable, somewhat mimicking the RPI rating...

Apr
11
2012

## Geek Challenge Results: March Madness Part I

Congratulations to Dean Schmitz, who earned victory in the March Madness Geek Challenge thanks to Kentucky’s win over Kansas in the final. He used his superior average bracket score to best Dan Freve, who turned in the tournament's highest individual bracket. Six of our seven entrants centered their choices primarily around tournament seed, while DMC employee Dan Freve chose to highlight a combination of Win-Loss percentage and strength of schedule. Both methods proved effective as Dan, Dean, a...

Mar
07
2012

This month’s Geek Challenge was created in conjunction with  Kevin Ferrigno.  The challenge is to pick the winners of this year’s NCAA tournament, not by choosing teams, but by picking the winning formula.  You can use this spreadsheet to practice with based on last year’s games. We’ve simplified the process of picking a bracket from selecting the outcome of 63 games down to selecting five numbers. We will create 10 randomly generated brackets based on your ...

Feb
02
2012

## Geek Challenge Results: New Year's Beeramid Part II

The following is a supplement to Geek Challenge Results: New Year's Beeramid by the Numbers. The partial summation of xy This function can be solved to a polynomial of order y+1 by the method outlined here. Below is a graphical representation of xy.  In this graph, y=2, but this is representative of any integer value of y>0.  The total area under the curve is ∫ xy .  Specifically, for the range shown, the total area is: The area shown in blue is what we ...

Feb
02
2012

## Geek Challenge Results: New Year's Beeramid by the Numbers

January's Geek Challenge questioned the construction of a New Year's Eve party beer can pyramid. I first encountered this challenge in my freshman year at Marquette. I was building the beeramid with five other students at a band party, some were engineers and math majors and we couldn't figure it out. I didn't sleep until I found the answer at 4am! The correct answer for an 8 layer Beeramid is C: 120 cans. Congratulations to this month’s winner, Harry Maddock of the City of Tac...

Jan
10
2012

## Geek Challenge: New Year's Beeramid by the Numbers

During a New Year's Eve party, a simple pyramid was constructed by standing 3 beer cans on a table in a triangle, and then placing a single can on the top center of the three. The pyramid was expanded by expanding the triangular base to 3 cans on each side, adding 2 more cans to the second layer and adding a single can on top as the third layer. How many cans are in the pyramid when it is expanded to 8 layers? A. 70 B. 85 C. 120 D. 256 Extra credit:  How many ...

Dec
08
2011

## Geek Challenge Results: Deer Dilemma

We didn’t receive a lot of reader-feedback to last month's Geek Challenge.  Some questioned the assumptions involved, and some guesses were taken.  Here is my analysis of the problem: The Deer Dilemma asks about the probability of collision between a car traveling an unspecified speed and a number of deer traveling at unspecified speed and route.  The primary challenge is figuring out how to estimate the chance of collision of a single deer.  To solve this, conside...

Oct
27
2011

## Geek Challenge: Deer Dilemma

A state forest is square and 6 miles on each side, and known to be home to a population of 1,000 deer.  A road runs straight through the forest (not diagonally) and is silently traveled by an electric car whose driver is oblivious to the deer.  Since it is mating season, the deer are constantly on the move.  They show no more or less affinity for the road than for any of the many deer paths through the forest, and won’t notice the car.  What is the probability of the car hit...