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More area. I think you'll find this one interesting, and make sure
you provide a good explanation for your answer!
Two congruent 10" x 10" 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? One square is movable, as long as its vertex remains in the center of the other square. SolutionsEvery time I have used this problem I have reminded myself to word it a little differently next time - to ask not only for the maximum area, but for the minimum as well, or maybe just ask for the area. Or at least to really PROVE that you know it's the maximum area that you found. And every time I use it, I forget that I decided I should do that the next time! So I have a lot of correct answers (162) - a bunch of answers that said that it was always going to be 25 (at least 64), and a bunch more that just said that 25 was the maximum, and showed how they got 25, but didn't say anything about why they know it's a maximum. 29 folks got the problem wrong - some because they didn't explain their answer at all and some because they misunderstood the problem or made some math errors in computing area. When you are asked for a maximum area, try to explain why you know your answer is THE answer. Finding the area of the square-shaped overlap doesn't necessarily mean you found the maximum. Try some other shapes and see what you get. I still gave credit, but that's because I think the problem isn't worded very well and isn't demanding enough. A number of people came to the conclusion that "it's always 1/4 of the area," but they never explained what they saw that made them think that was true. Tell me why you said that? What did you see or do in your drawing that helped you come to that conclusion? There are a couple of ways to "prove" the general case (that it's always 1/4), and we have some super examples of them. One is to show that as the one square moves, the new area that it covers is equal to the area that it uncovers. A lot of people noticed this, but not a lot proved it. When you find something really definite and neat like this, see if you can explain it for sure! I have highlighted solutions from Lim Yin of Raffles Institution and Emma Lindsay of Georgetown Day School. Both do a super job of proving it, and Emma provides a nice picture. Another method is to prove that the area will always be 25, regardless of the shape. One way to do this is using trigonometry, and I've included solutions from Erik Van de Vreudge of Livermore High School and CG of Germantown Academy. Erik's proof comes with a picture, and CG's comes with a lengthier explanation, though you may need to draw a picture to go along with it while you're reading. But it's worth it. Perhaps the simplest way of convincing someone that the overlap is 1/4 the area of one of the squares is to extend the sides of the moving square and notice that since your two lines meet in the middle of the square at 90 degree angles, there are four equal areas showing in the square at all times. Alison Miller, who is homeschooled, did a good sketch of this, and I've included a picture as well as a Java version of her sketch for you to play with. One last thing. When you are faced with a problem like this and you are asked for the maximum area, it's good to point out that it's 1/4 of the area of one of the squares, but it's also imperative that you give a NUMBER with your answer! You're also being tested on whether you can find the area of a region. 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: C.G.
sagbad@aol.com
Grade: 9
School: Germantown Academy, Fort Washington, Pennsylvania
Subject: POW March 9-13
please see attachment
C.G.
Germantown Academy, grade 9
Fort Washington, PA
POW March 9-13
Question:
What is the largest possible value for the area where two congruent 10 in. by 10
in. squares overlap if a vertex of one square is at the center of the other?
Answer:
The largest possible value for the area where they overlap is 25 inches.
Three possible combinations were tested, and each resulted with an area of 25.
This is the first combination:
We know that each side of both squares is 10. When the second square is placed
over the first like this, the base of the first square and one side of it are
covered. The smaller square that is formed by their overlap has an area of 25
because each of it's sides is 5: the base is 5 because that is half of the base
of the first square, and a side is 5 because that is half of the side of the
first square. If two sides of a square are known, the others are known
too...each side = 5 in. The area of a square is "side squared", so 5 squared =
25.
This is the second combination:
The overlap forms an isosceles triangle. To find it's area, first we must find
the length of the diagonals of the non-overlapping square using the Pythagorean
Theorem. 10 squared + 10 squared = hypoteneuse squared. The diagonals are both
5 root 8. Now we must divide that by 2, because half of each diagonal serves as
the legs of the isosceles triangle. Each leg is 5 root 2, and since the base of
the triangle is also the base of the non-overlapping square, we know that the
hypoteneuse of the isosceles triangle is 10. The formula for area of a triangle
is (1/2)(base)(height). We know the base is 10, but to find the height, an
altitude must be drawn from the top of the triangle to the midpoint of the base.
Two right triangles are now formed, with bases of 5 (half of 10) and
hypoteneuses of 5 root 2. Using the Pythagorean Theorem, the altitude can be
found to be 5: 5 squared + altitude squared = (5 root 2)squared. Now we can
put the 5 into the area equation as the height. So, (1/2)(10)(5) = area = 25
inches!
This is the third combination:
In order to find the area of this overlap, a line must be drawn from the center
of the non-overlapping square (point A) to the midpoint of the left side of it,
which forms a new right triangle with the base 5 (half the 10 inch median of the
square). Call this triangle #1. To find the other leg of this triangle, we can
use the tangent of A which equals it's opposite side divided by it's adjacent
side, which is 5. So, the unknown opposite leg is 5 times the tangent of A.
Now it is nessary to seperate the overlap into a right trapezoid and a right
triangle, which we'll call triangle #2. Side "b" of the trapezoid equals 5 -
5(tanA) because the entire section equals 5 (half the side of the non-
overlapping square), and we are subtracting the part we know from 5 to find the
part we do not know. The area of a trapezoid is (height)(base1+base2)/2, and we
know that the height would be 5 because that is half the base of the non-
overlapping square, and that the bases are 5 (half the vertical median) and 5
times the tangent of 5. 5(5+5tanA)/2 = area of the trapezoid. Now we have to
find the area of right triangle #2. We know that triangle #1 and triangle #2
are congruent because of the Angle-Side-Angle Postulate, so the base of triangle
#2 is also 5 times the tangent of A. The area of a traingle is (1/
2)(base)(height), so the area of triangle #2 = (1/2)(5-tanA)(5) = 25tanA/2.
Finally, to get the area of the overlap, we can add the area of the trapezoid to
the area of triangle #2. (1/2)(50 - 25tanA + 25tanA) = 25. The area of the
overlap of the two congruent squares is 25!!!
Clearly, the biggest possible area for the overlap is 25 inches.
From: Alison Miller
mary_okeeffe@classic.msn.com
Grade: 6
School: homeschooled, Niskayuna, New York
Subject: March 9 Geometry POW
Dear Annie,
Amazingly, the only possible (and obviously the largest) value for the area of
the overlap is 25 square inches. My picture shows a case of how you can
divide the square up into four equal regions (which contain two triangles, one
of each color) that are (or at least look) congruent and one of the regions is
the overlap. If you aren't convinced that the four regions are congruent,
notice that the triangles that have the same color are congruent. Also, an
isosceles right triangle whose hypotenuse is a side of the square contains one
triangle of both colors. One of those isosceles right triangles (obviously)
has an area of 25 square inches. The overlap also contains one triangle of
each color, so it also has an area of 25 square inches.
From: Erik Van De Vreugde
jlvand@jps.net
Grade: 10
School: Livermore High School, Livermore, California
Subject: Problem of the week solution for the area of overlaping squares
As the second square is rotated around the vertex, the overlapping area is
always 25 square inches.
[two examples in picture]
In example #1 where the square is in a normal position the calculated area is 25
square inches. In example #2 where the square is rotated 45 degrees the
overlapping area is a triangle with an area of 25 square inches.
The following table represents the calculation of the overlapping area as one
square is rotated between 0 and 45 degrees. Base is the length of the base of
the triangle formed as the square is rotated. The base is calculated by using
the formula Tan (angle)=base/5. Area A represents the area of the triangle A.
Triangle A and B are congruent. Area C is the area of rectangle C. The total
area is the sum of triangle A, B and rectangle C.
From: Emma Lindsay
elindsay02@gds.org
Grade: 8
School: Georgetown Day School, Washington, DC
Subject: POW 3/13
Emma Lindsay
Grade 8
Georgetown day School
Washington DC
Paul Nass
Dear Annie,
the largest are of overlap will always be 25 cm squared because the section of
overlap is always 25 cm squared
The following students submitted correct solutions this week:Hui Shen, Grade 11, Killara High School, AustraliaLim Yin, Grade 10, Raffles Institution, Singapore Nick Edwards, Grade 9, Skyview High School, Vancouver, Washington Chrissy Beaulieu, Grade 10, Marple Newtown Senior High School, Newtown Square, Pennsylvania Milan Fillmore, Grade 9, Smoky Hill High School, Aurora, Colorado Leona Reeves, Grade 10, Neah-Kah-Nie High School Natasha , Grade 9, Bexley High School, Bexley, Ohio Bradley Jennings, Grade 9, Chetek High School, Chetek, Wiconsin Melissa Branfman, Grade 8, Georgetown Day School, Washington, DC Joe Larkin, Grade 10, Marple Newtown Senior High School, Newtown Square, Pennsylvania Melissa Branfman, Grade 8, Georgetown Day School, Washington, DC Kathleen Moore, Grade 9, Southeast Raleigh High School, Raleigh, South Carolina Emily Castor, Grade 9, Granada High School, Livermore, California Emily O'Brien, Grade 10, School Without Walls Senior High School, Washington, D.C. Jonathan Emmons and Karen Altice, Grade 9, Franklin County High School, Rocky Mount, Virginia Amey Adkins, Grade 9, Franklin County High School, Rocky Mount, Virginia John Clemson and Aaron Diamond, Grade 11 & 10, Sam Barlow High School, Gresham, Oregon Anne Marie Abell, Grade 10, Washington County High School, Springfield, Kentucky Ginger King, Grade 9, Washington County High School, Springfield, Kentucky Ken Duisenberg, Grade Stanford University '93, Hewlett Packard Engineer, Santa Rosa, California Jacob Ornelas, Grade , Smoky Hill High School, Aurora, Colorado Shanon Moore, Grade 11, Livermore High School, Livermore, California Mina Ashkannejhad, Grade 9, Skyview High School, Vancouver, Washington Marta Williams, Grade 9, Skyview High School, Vancouver, Washington Megan Nowak, Grade 8, Odle Middle School, Bellevue, Washington Diana Tyson, Grade 10, Academy of Allied Health and Science, Chattanooga, Tennessee Katie Anthony, Grade 9, Casady School, Oklahoma City, Oklahoma Betsy Feldman, Grade 9, Marple Newtown Senior High School, Newtown Square, Pennsylvania Renee Cohen, Grade 9, Bexley High School, Bexley, Ohio Kevin Palmer, Grade 9, Granada High School, Livermore, California Matthew Hiatt, Grade 8, Smoky Hill High School, Aurora, Colorado Helen Wong, Grade 7, Odle Middle School, Bellevue, Washington Brandi Tsama Marsh, Grade 9, Ethel Walker School, Simsbury, Connecticut Richard Balsley, Grade 10, Granada High School, Livermore, California mike Johannsen, Grade 11, Granada High School, Livermore, California Darren Kerstien and Rob Eagle and Mark VanArsthdale and John Yi, Grade 7 & 8, Challenge School, Denver, Colorado Brandon Brown, Grade 10, School for Young Performers, New York, New York Victoria, Grade 10, Inchelium High School, Inchelium, Washington Sara Kaiser, Grade 11, Clatskanie High School, Clatskanie, Oregon Gordon Bockus Jr., Grade 9, Eastern Oklahoma State College, Wilburton, Oklahoma Dan Peischl, Grade 9, Marple Newtown Senior High School, Newtown Square, Pennsylvania Dan Chambers, Grade 9, Granada High School, Livermore, California Saurabh Sarkar, Grade 8, Albright Middle School, Houston, Texas Matthew Shaw, Grade 10, Skyview High School, Vancouver, Washington Bob Kneck III, Grade 10, Skyview High School, Vancouver, Washington Kristen Kamerdula, Grade 11, Livermore High School, Livermore, California Catherine Mangasi, Grade 12, Wilburton High School, Oklahoma City, Oklahoma Megan Ross, Grade 9, East Mecklenburg High School, Charlotte, North Carolina Julia Tarlo, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Carol Kim, Grade 10, Marple Newtown Senior High School, Newtown Square, Pennsylvania Arielle Cohen, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Theresa Perry, Grade 10, Granada High School, Livermore, California Sarah Nowak, Grade 8, Odle Middle School, Bellevue, Washington Matthew Macumber, Grade 9, Campbell High School, Smyrna, Georgia Sherry Snell, Grade 10, Swartz High School Dan McCoy , Grade 9 , Marple Newtown Senior High School, Newtown Square, Pennsylvania Megan Via, Grade 9, Franklin County High School, Rocky Mount, Virginia David Choi, Grade 9, Marple Newtown Senior High School, Newtown Square, Pennsylvania DeDe DiDonato, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Justin McVicker, Grade 9, Livermore High School, Livermore, California Stephen Nolen, Grade 9, Smoky Hill High School, Aurora, Colorado Carol Vivaldelli, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Jim Nguyen, Grade 9, Smoky Hill High School, Aurora, Colorado Chris Jones, Grade 10, Shelby County High School, Columbiana, Alabama Adam Donovan and Dan Adams, Grade 7, Brooks School, Lincoln, Massachusetts Jessica Arendal, Grade 8, Georgetown Day School, Washington, DC Bridgette Ellis and Crystal Cannaday, Grade 9, Franklin County High School, Rocky Mount, Virginia Maureen Brady, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Michelle Wagner, Grade , Mount Saint Joseph Academy, Flourtown, Pennsylvania Mary Diamond and Mary Kenney, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Tanya Colburn, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Priscilla Hamilton, Grade 10, Washington County High School, Springfield, Kentucky Brie Compton and Abbie Rowley, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Amber Benedict, Grade 10, Washington County High School, Springfield, Kentucky Jon Gantman, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Heather Whalley, Grade 11, Livermore High School, Livermore, California Noam Abrams, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Avrum Tilman, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Hannah Margoles, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Gina Christenson, Grade 9, Southeast Raleigh High School, Raleigh, North Carolina Zach Moore, Grade 9, Smoky Hill High School, Aurora, Colorado Mathew Steadman, Grade 8, Odle Middle School, Bellevue, Washington Frannie Laks, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania June Lin, Grade 5, Sahuaro Elementary School, Tuscon, Arizona Jake Nelson, Grade 9, Smoky Hill High School, Aurora, Colorado Alex Morgovsky, Grade 11, Akiba Hebrew Academy, Merion, Pennsylvani Sarah Kress, Grade 9, Mount Saint Joseph Academy, Flourtown, Pennsylvania ?????, Grade 10, Bishop O'Connell Teodora Niculce and Alexandra Niculce, Grade 10, Reading High School, Reading, Pennsylvania Jenny Kaplan, Grade 7, Castilleja Middle School, Palo Alto, California Catalina Anghel, Grade , Mackenzie High School, Deep River, Ontario, Canada Sonia Teas, Grade 10, Oak Park and River Forest High School, Oak Park, Illinois Chris Mize, Grade 9, Southeast Raleigh High School, Raleigh, North Carolina Andrew Krawczel, Grade 9, Franklin County High School, Rocky Mount, Virginia Greg Rogers, Grade 11, Granada High School, Livermore, California Ashley Monroe, Grade 9, Casady School, Oklahoma City, Oklahoma Allison Yang, Grade 5, Bartle Elementary School, Highland Park, New Jersey Lawrence Lombard, Grade 12, Rosamond High School, Rosamond, California Mallory VanderKooy, Grade 9, Franklin County High School, Rocky Mount, Virginia Sarah Hesson, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Scott, Grade 9, East Mecklenburg High School, Charlotte, North Carolina Lindsey Fangman, Grade , Redmond High School, Redmond, Oregon Raun Atkinson, Grade , Redmond High School, Redmond, Oregon Rob Johnson and Bryce Withers, Grade , Redmond High School, Redmond, Oregon Doug Yoder, Grade 12, Highland Park Senior High School, St. Paul, Minnesota Tiffanie Lam, Grade 8, Sequoia Middle School, Pleasant Hill, California Matt Niederst, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Steve Platt, Grade , Redmond High School, Redmond, Oregon Katy Wilde and Nathan Walker and Haley Zwicker, Grade , Redmond High School, Redmond, Oregon Thomas Po, Grade 8, Odle Middle School, Bellevue, Washington Neha Dalal, Grade 10, Roselle Park High School, Roselle Park, New Jersey Clint Soose, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Travis Guy, Grade , Redmond High School, Redmond, Oregon Brian Porch, Grade , Redmond High School, Redmond, Oregon Peter Hoover, Grade , Redmond High School, Redmond, Oregon Andy Keen, Grade , Redmond High School, Redmond, Oregon Mark Goodman and Heather Westendorf, Grade , Redmond High School, Redmond, Oregon Cari Arsenault, Grade , Redmond High School, Redmond, Oregon Santa Goleme and Jessica McCann, Grade 12, Roselle Park High School, Roselle Park, New Jersey Emily Chisholm, Grade , Redmond High School, Redmond, Oregon Jessie Nelson and Ryan Tesconi and Dallas Witty, Grade , Redmond High School, Redmond, Oregon Ryan Holcomb, Grade , Redmond High School, Redmond, Oregon Lisa Oakland and Michelle Peterson and Lyrica Hubbard, Grade , Redmond High School, Redmond, Oregon Laura Roos, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania C.G., Grade 9, Germantown Academy, Fort Washington, Pennsylvania Alison Miller, Grade 6, homeschooled, Niskayuna, New York Erik Van De Vreugde, Grade 10, Livermore High School, Livermore, California Jen Erhart, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Lauren Moser, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Anh Nguyen, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Elizabeth Levine, Grade 8, Georgetown Day School, Washington, DC Mac VerStandig, Grade 8, Georgetown Day School, Washington, DC Grant Bramwell, Grade 8, Georgetown Day School, Washington, DC Ben Reed, Grade 8, Georgetown Day School, Washington, DC Ariel Berenstein, Grade 8, Georgetown Day School, Washington, DC David Zax, Grade 8, Georgetown Day School, Washington, DC Sonya Kharas, Grade 8, Georgetown Day School, Washington, DC Emma Lindsay, Grade 8, Georgetown Day School, Washington, DC Thao Vuong, Grade 9, Highland Park Senior High School, St. Paul, Minnesota My Trinh, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Aaron Mertz, Grade , Plum Grove Junior High School, Rolling Meadows, Illinois Kristy Dalrymple, Grade , Granada High School, Livermore, California Mary Park, Grade 8, Odle Middle School, Bellevue, Washington Alex Chernyavsky, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Maria Volpe, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Greg Seppertt, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Jim Gugger, Grade , Roselle Park High School, Roselle Park, New Jersey Adam Lovas, Grade 10, Roselle Park High School, Roselle Park, New Jersey Thomas Kuo, Grade 10, Burroughs High School, Ridgecrest, California Dave Espenshade, Grade teacher, Olympic High School, Bremerton, Washington Zach Dillon, Grade 7, Odle Middle School, Bellevue, Washington Jennifer Liang, Grade 8, Odle Middle School, Bellevue, Washington Joel Jorgensen, Grade , Redmond High School, Redmond, Oregon Kristin Lee, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Vu Nguyen, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Joe Johnson, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Giai Tran, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Maurice Her, Grade 9, Highland Park Senior High School, St. Paul, Minnesota My Nhia Lee, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Melissa Hackel, Grade , Granada High School, Livermore, California Lydia Wang, Grade 9, Smoky Hill High School, Aurora, Colorado Teneal Dollar, Grade 10, Shelby County High School, Columbiana, Alabama Janelle Benterou, Grade 9, Granada High School, Livermore, California |
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This is true no matter what the overlap looks like, as long as a vertex
remains in the center. You can see this by dragging a vertex adjacent to the
vertex in the center in my sketch.
Two cases are shown in the attached JPEG diagram. In the first case (left hand
column) the sides of square EFGH meet at the sides of square ABCD at 90 degree
angles. Angle BCD is 90 degrees, therefore HEF must be 90 degrees because the
sum of the angles is 360. A quadrilateral with four right angles is a square.
We know that side EO is 5 cm because it is a perpendicular line from the center
of square ABCD, therefore making it half of one side of ABCD, which is 5 cm
If we then rotate square EFCH by x degrees, we get square EKIJ. The are of
overlap here is also 25 cm squared, which we can prove as follows. We know
that the angle LEN and angle OEM are congruent. THis is because they are both
90 degrees less then angle JEF because they they are both adjacent to right
angles in angle JEF. We know that the angle EOM and ELN are both 90 degrees
because they are both corners of the given squares. We also know EO and EL are
congruent because they are both 5 cm.
We now know that triangle NEL is congruent to triangle MEO because of the
angle-side-angle theorem. We know that quadrilateral NCME is always the same
as EOCL because any area that is lost through triangle MEO is gained through
triangle NEL
This works in all cases beacause the triangles only have to be congruent to
each other, not anything else. This is shown in the right hand column of the
diagram. Rotating in the opposite direction, we can go through all of the same
steps and get the same answer.