BMAD Posted May 5, 2013 Report Share Posted May 5, 2013 Two similar triangles with integral sides have two of their sides the same. If the third sides differ by 20141, find all of the sides. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted May 8, 2013 Report Share Posted May 8, 2013 Let triangle 1 have sides a b c where a <= b <= c, and triangle 2 have sides ar br cr. Two pairs of sides are equal, so that the corresponding sides area b c ar br cr so r = b/a = c/b and the sides become:a ar ar2 ar ar2ar3 we have also ar3 - a = 20141 or a = 20141/(r3 - 1) = integer. r is rational, so we can write r = n/d where n and d are integers. A simple search on integral values of n and d that gives integral values of a finds: n = 30, d = 19, a = 6859, r = 1.578947368 ..., from which the triangle sides are. 6859 10830 17100 10830 17100 27000. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted May 6, 2013 Report Share Posted May 6, 2013 Does "two of their sides the same" imply they are isosceles? Or that two sides of one triangle are equal to two of the other triangle's sides? It seems the latter, but thought I'd confirm. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted May 6, 2013 Report Share Posted May 6, 2013 Infinite number. Note that 20,141 = 11x1831 Let the smaller triangle have a side 1831 and two equal other sides larger than 916, say 1000 and 1000. Let the larger triangle have sides 12 times as large: 21972 and 12000 and 12000. There are an infinite number of lengths for the equal sides other than 1000. Therefore I assume that is not the correct interpretation of "two sides the same." Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted May 6, 2013 Report Share Posted May 6, 2013 Let triangle 1 have sides a b c where a <= b <= c, and triangle 2 have sides ar br cr. Two pairs of sides are equal, so that the corresponding sides area b car br crso r = b/a = c/b and the sides become:a ar ar2ar ar2 ar3we have also ar3 - a = 20141 or r3 = (a + 2041)/a or r = ((a + 20141)/a)1/3.We need to solve for a and r.More later. Quote Link to comment Share on other sites More sharing options...
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BMAD
Two similar triangles with integral sides have two of their sides the same. If the third sides differ by 20141, find all of the sides.
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