Square with same Perimeter and Area as a Triangle
Date: 10/16/96 at 1:2:14 From: Todd Edwards - a.k.a mathman Subject: Square with same perimeter and area as a triangle Dr. Math: I am perplexed! I've been hunting for a square/triangle combo with the same perimeters and areas. Is this possible? I believe this is an application of semi-perimeters (is that what it is called?). I've been fiddling around with this for 3 or 4 hours now on CABRI with no success. A solution in CABRI (as opposed to GSP) would be unbelievably appreciated by both myself (and my students)! Thank you so much, Todd Edwards Upper Arlington High School, Columbus, OH
Date: 10/17/96 at 13:53:24 From: Doctor Yiu Subject: Re: Square with same perimeter and area as a triangle Dear Todd, This is a very interesting question. It is indeed IMPOSSIBLE to find a square and a triangle with the same perimeter and the same area. Here is the reasoning. It relies on two basic facts. (1) A formula for the area of a triangle. There are many ways of finding the area of a triangle. One of these is the formula that gives the area in terms of the lengths of the sides. This is the Heron formula: if the lengths of the sides of a triangle are a, b, and c, and if the SEMIPERIMETER is denoted by s, namely, s = (a+b+c)/2, then the area A is given by A^2 = s(s-a)(s-b)(s-c). (HERON) (2) An inequality involving the arithmetic mean and geometric mean of three positive numbers. If you have three positive numbers, say X, Y, Z, there are many different ways of taking "averages". Two of the most common averages are (i) the ARITHMETIC MEAN: (X+Y+Z)/3, and (ii) the GEOMETRIC MEAN: the CUBE root of XYZ. Now, it is true that for all positive numbers X, Y, Z, the arithmetic mean is always NO LESS THAN the geometric mean. In other words, Cube root of XYZ < or = (X+Y+Z)/3, or 3(Cube root of XYZ) < or = X+Y+Z. If you take the cube (third power) of both sides, then 27XYZ < or = (X+Y+Z)^3. (AGI) Since we shall make use of this inequality below, let's label it (AGI) Now, consider a triangle with sides a, b, c, and semiperimeter s. Its area A is given by the formula (HERON) above. Suppose there is a square with the same perimeter and same area. Then, the perimeter of the square being 2s, each of its sides has length (2s)/4 = s/2. The area of the square is then (s^2)/4. This means that the area of the triangle is also A = (s^2)/4. Plugging into the formula (HERON), we now have (s^4)/16 = s(s-a)(s-b)(s-c). Cancelling a common factor s, and multiplying by 16, we have 16(s-a)(s-b)(s-c) = s^3. (******) We shall use (AGI) to see that this is indeed impossible. The numbers s-a, s-b, and s-c are POSITIVE. (In fact, since the sum of the lengths of any two sides of a triangle must exceed the third, say, a + b > c, we have a+b+c > 2c, or 2s > 2c, or s > c. This means s-c is positive. For the same reason, s-a and s-b are also positive). Note that their sum is exactly s, because (s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s. Now, using (AGI) with X = s-a, Y = s-b and Z = s-c, from (AGI): 27XYZ < or = (X+Y+Z)^3, we have 27(s-a)(s-b)(s-c) < or = s^3. and the equation (*****) above is impossible. This completes the demonstration that there is no triangle/square pair equal in perimeter and area. Sincerely, -Doctor Yiu, The Math Forum
Date: 10/17/96 at 14:19:19 From: Doctor Ken Subject: Re: Square with same perimeter and area as a triangle Hi Todd- Here's a little intuition as to why this can't work. Let's say we choose our perimeter first. Call it P. Once you've chosen P, then you've completely determined the square. It has side length P/4, and area P^2/16. The only freedom you have at this point is how the length P is distributed among the three sides of the triangle you're making. So the question is whether there's a triangle with perimeter P and area P^2/16. That's where Yiu's proof comes in. -Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: Thu, 17 Oct 1996 21:00:15 -0400 (EDT) From: Todd Edwards Subject: Re: Square with same perimeter and area as a triangle Thank you so much for your thorough proof regarding my square/triangle problem. I can't tell you enough how happy I was to hear from you. I was really excited by this problem, but quite frankly haven't had a lot of time to try and solve it. It's exciting to know that a resource such as yours exists. I told all of my students about Dr. Math and have encouraged them to visit your Web Site. I've used a good number of lessons from your site and have found them really great! Keep up the great work. You help me feel "connected" to a larger mathematics community - which is important for high school teachers! Thanks again, Todd Edwards Upper Arlington High School - Columbus, Ohio
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