Two congruent 10cm x 10cm squares overlap. A vertex of one square is at the center of the other square. What is the largest possible value for the area where they overlap? (The one square is movable, as long as the vertex remains in the center.)Below is a picture that illustrates this problem. It's linked to the sketch itself for those of you who have Sketchpad.
- Annie Fetter
Annie says: Solutions
I love this problem. Many of you liked it too - 77 correct solutions, and only 10 wrong. Most of the ones that were wrong were actually right, in that they stated the maximum area was 25, but they didn't offer any explanation.
It's true that the "maximum" area covered is in fact the only area that can be covered. A number of people pointed this out, but only a few bothered to explain why. What made you think you were right? Can you write an explanation that would convince anyone that you are right? Better yet, can you draw a picture?
When I stated the problem, I didn't ask you to explain everything, just to say what the maximum area is. You must include some explanation, and it's a good idea if you say that "25 cm^2 is the maximum," also to say something about why you think it can't be bigger. This is a problem where I think most people think a lot more than they write. If you remember that you are always writing solutions for readers who might not be familiar with the problem or the solution, it will help you to be very thorough.
People used two ways to show that the area is a constant. The easiest to see and the least work is to show that since the square takes up 90 degrees at the center, you can put three more squares around the center and they'll cover four congruent areas. This was the method used in two of the highlighted solutions, by Brent Tworetsky of JP Taravella High School, and Thomas Kuo of Burroughs High School.
The other way to do it was used by a bunch of folks. Start with the squares in "square overlap" position. As you rotate one square, you can show that the new area you cover is equal (and congruent) to the area you uncover. Shameica Edwards of Wingate High School used this method.
Brian Gordon, one of our "adult" contingent, decided to be a little extreme with his answer, and figured he could get away with folding the movable square. Since I didn't say you couldn't, I gave him credit.
An interesting question was raised by a couple of people this week. Justin Lam asked about other shapes - will this work for triangles? Regular polygons? Any irregular ones? Robert Hughey thinks it will work for rectangles and rhombii. What do you think? (I think it sounds like a good Project of the Month!)
A list of all the people who got this problem right follows the highlighted solutions below, and most of the solutions are also available.
From: blaze13@ix.netcom.com Brent Tworetzky Grade: 10 School: JP Taravella High School, Coral Springs, Florida In this problem, we are placing the vertex of a square at the center of another square. This means we are creating a right angle at the center of the square. Seeing as 90 degrees is a quarter of 360 degrees, the area of the overlapping squares is a constant 25%. To further prove this, extend the sides of the square with its vertex in the center of the other until the lines touch the sides of the square. The result is four congruent shapes (wow!). This can be done at any angle. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Any angle you connect the two squares, four congruent shapes are formed. The overlapping area is one of these squares; 1 out of 4, or 25%. Thus, the answer is 25% of 100, being 25
From: wkuo@ridgecrest.ca.us Thomas Kuo Grade: 9 School: Sherman E. Burroughs High School, Ridgecrest, California The largest possible value is 25 sq. cm. This area is the only possible value for the overlap. (1) The overlapping area of the two squares is always one-fourth of the 10 cm x 10 cm square. If four congruent 10 cm x 10 cm squares are placed with one vertex at the center of the square and they are rotated, the four sections will always be congruent, and one section is therefore one-fourth of the original square. (2) Since the section is one-fourth of the original, the area of overlapping section is the area of a 10 cm x 10 cm square divide four: (10 * 10)/4 = 25
From: star1110@gnn.com Shameica Edwards Grade: 11 School: George Wingate High School, Brooklyn, New York1. Draw a 10x10 cm square. 2. Draw a second square of the same length (10x10) with one of its vertices at the first square which form an isosceles right triangle at the base of the first square. 3. Find the area of the triangle: 1/2 the base times height. 1/2 (10x5) = 25 (the 5 is the altitude from the center of the vertex angle to the base). 4. I rotate the second square, making 1/4 of it intersect a 1/4 of the first square. 5. Find area l x w = 5 x 5 = 25 6. I rotate the second square a little more forming a quadrilateral. 7. Extend a line from the center of the first square to the center of the side line and another line to the center of the base line forming two triangles and a trapezoid. 8. I number my triangles I and II and compare them. 9. Find x of triangle I = 30. Angle x and angle y form a right angle; therefore angle y = 60. 10. Angle y and angle x of triangle 2 form right angles; therefore angle x of triangle 2 = 30. Triangles I and 2 are congruent (AAS). 11. Find area of triangle I + trapezoid = 25. 12. If triangle I is congruent to triangle II and is combined with the same trapezoid, then the areas are the same. Therefore the overlap = 25.
From: bmgordon@ntplx.net Brian Gordon Grade: 1992 School: Dartmouth Well Annie, since this problem is a rerun, you will not be receiving the standard answer of 25 cm^2 from me this week. My answer this week is 50 square centimeters if the movable square must remain a square, and 87.5 if not. The easiest number to show is the 87.5. This is obtained by folding the square back over across the points of intersection with the fixed square: Okay, here we go. The original diagram should look like this: ------------------------ | | | | | | | | | -----------B---------- | | | | | | | | | | | | | | | | | | | | ------------A----------D | | | | | | | | | ---------------------- Let A and B be the points of intersection of the two squares. Fold the lower square over itself across segment AB. It lands right on top of the first square. You will note the only area where there is no overlap is triangle ABD, which has area 1/2 * 5 * 5 = 12.5, so the overlap is 100 - 12.5 = 87.5. The line AB is a line of symmetry for the picture. Of course, if you insist that this remain a square, simply repeat the folding on all four corners. The square would then have area 50 cm^2, or half the original square. You didn't say we couldn't move the square out of the plane. *phtphtbt* See ya! --bri
The following students submitted correct solutions this week:
Megan Bean and Alice Tran Grade 9, Smoky Hill High School, Aurora, Colorado Justin Lam Grade 8, Sequoia Middle School, Pleasant Hill, California Gavino Walker Grade 10, Sammamish High School, Bellevue, Washington Brian Gordon Grade 1992, Dartmouth Brent Tworetzky Grade 10, JP Taravella High School, Coral Springs, Florida Rick Bullock Grade 10, Golden Plains High School, Rexford, Kansas Marc A. Tuler Grade 9, Smoky Hill High School, Aurora, Colorado Erin Muehlenkamp Grade 9, Smoky Hill High School, Aurora, Colorado Ken Duisenberg Grade Hewlett Packard Engineer, Stanford University '91 Kenneth Yan Grade 10, Sammamish High School, Bellevue, Washington Ryan Salvas Grade 8, Old Saybrook Middle School Michael Martin Grade 9, Sammamish High School, Bellevue, Washington Nicole Forostoski Grade 10, Martin County High School, Stuart, Florida Christine Chung Grade 10, Smoky Hill High School, Aurora, Colorado Roger Mong Grade 8, Zion Heights Junior High School, Richmond Hill, Ontario, Canada Andrea Wahlen Grade 9, Smoky Hill High School, Aurora, Colorado Jeremy Reedy Grade 11, Redbank Valley, Pennsylvania Jess Gilburne and Eric Faden and Jared Joiner Grade 8, Georgetown Day School, Washington, DC Davia Bell Grade 8, Murray Middle School, Ridgecrest, California Stephanie Parsons Grade 9, Smoky Hill High School, Aurora, Colorado Jason Howerton Grade 10, Granada High School, Livermore, California Kurt Davies Grade 10, Smoky Hill High School, Aurora, Colorado Erin McCloskey Grade 9, Smoky Hill High School, Aurora, Colorado Nick Davis Grade , South High School, Bakersfield, California Heidi Taylor Grade 10, Smoky Hill High School, Aurora, Colorado Laura Mannino and Candice Diaz Grade 9, Roselle Park High School, Roselle Park, New Jersey Lyndsey Hayes and Gwen Dillow Grade 10, Sammamish High School, Bellevue, Washington Sandra Wai Grade 9, Smoky Hill High School, Aurora, Colorado Thierry Cote Grade 7, College Marie de France, Montreal, Quebec, Canada Jenny Kaplan Grade 6, Castilleja Middle School, Palo Alto, California Kirk Vissat Grade 10, Smoky Hill High School, Aurora, Colorado Preston Sherrell Grade 9, Young Junior High School, Arlington, Texas Michael Chin Grade 10, Smoky Hill High School, Aurora, Colorado Ralph Boleslavsky Grade 12, Council Rock High School, Newtown, Pennsylvania Katherine Walther Grade 9, John Glenn Junior High, San Angelo, Texas Jonathan Shing Grade 9, Newport High School, Bellevue, Washington Libbie Skelton Grade 9, Martin County High School, Stuart, Florida Hector Mercedes Grade 12, George Wingate High School, Brookly, New York Shameica Edwards Grade 11, George Wingate High School, Brooklyn, New York Paul Chun Grade 9, Newport High School, Bellevue, Washington Simon Park Grade 9, Newport High School, Bellevue, Washington Megan LaChance Grade 10, Cheshire High School, Cheshire, Connecticut Robert Hughey Grade 10, Cheshire High School, Cheshire, Connecticut Ashley Turner Grade 10, Cheshire High School, Cheshire, Connecticut Ally Schulthess Grade 10, Cheshire High School, Cheshire, Connecticut Peter Suchy Grade 10, Cheshire High School, Cheshire, Connecticut Diana Bojka Grade 10, Cheshire High School, Cheshire, Connectictu Matt Mansur Grade 9, Cheshire High School, Cheshire, Connecticut Leah Rosen Grade , Thomas Kuo Grade 9, Burroughs High School, Ridgecrest, California Andrea Marks Grade 8, Jackie Roth Grade , Germantown Academy, Fort Washington, Pennsylvania Jackie Wong Grade , Newport High School, Bellevue, Washington Katie Quinn-Kerins Grade 10, Germantown Academy, Fort Washington, Pennsylvania Kate Martin Grade 10, Cheshire High School, Cheshire, Connecticut Jessica LaSalle Grade 10, Cheshire High School, Cheshire, Connecticut Suzanne Davis Grade 10, Newport High School, Bellevue, Washington Giscard Pongnon Grade 12, George Wingate High School, Brooklyn, New York Ginni Wentworth Grade , College Park High School, Pleasant Hill, California Joya Guha Grade , College Park High School, Pleasant Hill, California Juri Miyamae Grade , College Park High School, Pleasant Hill, California Marisa Mannari Grade 9, Newport High School, Bellevue, Washington Krissy Wall Grade , College Park High School, Pleasant Hill, California Laurel Kaminski Grade 9, Newport High School, Bellevue, Washington Joe Fantini and Greg Mellor Grade 9, Germantown Academy, Fort Washington, Pennsylvania Adam Peyman Grade 10, Cheshire High School, Cheshire, Connecticut Tim Pawlush Grade 10, Cheshire High School, Cheshire, Connecticut Scott Gorski Grade 10, Cheshire High School, Cheshire, Connecticut Hope Langan Grade 9, Smoky Hill High School, Aurora, Colorado Jackie Roth Grade , Germantown Academy, Fort Washington, Pennsylvania Rachel Collins Grade 9, Newport High School, Bellevue, Washington Brenda Hoskinson Grade 9, Newport High School, Bellevue, Washington Chris Roe Grade 9, Newport High School, Bellevue, Washington Heather Hook Grade 9, Newport High School, Bellevue, Washington Meghan Bowen Grade , Newport High School, Bellevue, Washington Lauren Hillman Grade , Germantown Academy, Fort Washington, Pennsylvania Renee Grade 9, Smoky Hill High School, Aurora, Colorado
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3 January 1997
