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| A [[two-dimensional]] '''equable shape''' (or perfect shape) is one whose [[area]] is numerically equal to its [[perimeter]].<ref>{{cite book |title=Challenges in Geometry: For Mathematical Olympians Past and Present
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| |first=Christopher J.|last=Bradley|publisher=Oxford University Press|year=2005|isbn=0-19-856692-1|page=15}}.</ref> For example, a [[right triangle|right angled triangle]] with sides 5, 12 and 13 has area and perimeter both equal to 30 [[Units of measurement|units]].
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| ==Scaling and units==
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| An area cannot be equal to a length except relative to a particular unit of measurement. For example, if shape has an area of 5 square yards and a perimeter of 5 yards, then it has an area of {{convert|45|sqft|m2}} and a perimeter of 15 feet (since 3 feet = 1 yard and hence 9 square feet = 1 square yard). Moreover, contrary to what the name implies, changing the size while leaving the shape intact changes an "equable shape" into a non-equable shape. However its common use as [[GCSE]] coursework has led to its being an accepted concept. For any shape, there is a [[Similarity (geometry)|similar]] equable shape: if a shape ''S'' has perimeter ''p'' and area ''A'', then [[Scaling (geometry)|scaling]] ''S'' by a factor of ''p/A'' leads to an equable shape. Alternatively, one may find equable shapes by setting up and solving an equation in which the area equals the perimeter. In the case of the square, for instance, this equation is
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| :<math>\displaystyle x^2 = 4x.</math>
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| Solving this yields that ''x'' = 4, so a 4 × 4 square is equable.
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| ==Tangential polygons==
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| A [[tangential polygon]] is a polygon in which the sides are all tangent to a common circle. Every tangential polygon may be triangulated by drawing edges from the circle's center to the polygon's vertices, forming a collection of triangles that all have height equal to the circle's radius; it follows from this decomposition that the total area of a tangential polygon equals half the perimeter times the radius. Thus, a tangential polygon is equable if and only if its [[inradius]] is two. All triangles are tangential, so in particular the equable triangles are exactly the triangles with inradius two.<ref>{{citation
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| | last = Kilmer | first = Jean E.
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| | issue = 1
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| | journal = The Mathematics Teacher
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| | jstor = 27965678
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| | pages = 65–70
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| | title = Triangles of Equal Area and Perimeter and Inscribed Circles
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| | volume = 81}}.</ref><ref>{{citation|url=http://jwilson.coe.uga.edu/emt725/Perfect/PerTri.html|title=Perfect triangles|first=Jim|last=WIlson|publisher=University of Georgia}}. See also Wilson's list of [http://jwilson.coe.uga.edu/emt725/Perfect/sol.html solutions].</ref>
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| ==Integer dimensions==
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| Combining restrictions that a shape be equable and that its dimensions be integers is significantly more restrictive than either restriction on its own. For instance, there are infinitely many [[Pythagorean triple]]s describing integer-sided [[right triangle]]s, and there are infinitely many equable right triangles with non-integer sides; however, there are only two equable integer right triangles, with side lengths (5,12,13) and (6,8,10).<ref name="bicycle">{{citation|title=Which Way Did the Bicycle Go?: And Other Intriguing Mathematical Mysteries|volume=18|series=Dolciani Mathematical Expositions|first1=Joseph D. E.|last1=Konhauser|first2=Dan|last2=Velleman|first3=Stan|last3=Wagon|author3-link=Stan Wagon|publisher=Cambridge University Press|year=1997|isbn=9780883853252|url=http://books.google.com/books?id=ElSi5V5uS2MC&pg=PA29|page=29|contribution=95. When does the perimeter equal the area?}}</ref>
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| More generally, the problem of finding all equable triangles with integer sides (that is, equable [[Heronian triangle]]s) was considered by B. Yates in 1858.<ref>{{citation|first=B.|last=Yates|title=Quest 2019|journal=The Lady's and Gentleman's Diary|year=1858|page=83}}.</ref><ref>{{citation|title=[[History of the Theory of Numbers]], Volume Il: Diophantine Analysis|first=Leonard Eugene|last=Dickson|authorlink=Leonard Eugene Dickson|publisher=Courier Dover Publications|year=2005|isbn=9780486442334|page=195}}.</ref> As [[William Allen Whitworth|W. A. Whitworth]] and D. Biddle proved in 1904, there are exactly three solutions, beyond the right triangles already listed, with sides (6,25,29), (7,15,20), and (9,10,17).<ref>{{harvtxt|Dickson|2005}}, p. 199.</ref><ref>{{citation
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| | last = Markowitz | first = L.
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| | issue = 3
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| | journal = The Mathematics Teacher
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| | pages = 222–223
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| | title = Area = Perimeter
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| | volume = 74
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| | year = 1981
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| | zbl = 1982d.06561}}.</ref>
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| The only equable [[rectangle]]s with integer sides are the 4 × 4 square and the 3 × 6 rectangle.<ref name="bicycle"/> An integer rectangle is a special type of [[polyomino]], and more generally there exist polyominoes with equal area and perimeter for any [[even number|even]] integer area greater than or equal to 16. For smaller areas, the perimeter of a polyomino must exceed its area.<ref>{{citation|title=Geometry Labs|first=Henri|last=Picciotto|year=1999|publisher=MathEducationPage.org|page=208|url=http://books.google.com/books?id=7gTMKr7TT6gC&pg=PA208}}.</ref>
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| == Equable solids ==
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| In [[Three-dimensional space|three dimensions]], a shape is equable when its [[surface area]] is numerically equal to its [[volume]].
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| As with equable shapes in two dimensions, you may find an equable solid, in which the volume is numerically equal to the surface area, by scaling any solid by an appropriate factor.
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| ==References==
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| {{reflist}}
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| [[Category:Geometric shapes]]
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