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Area Ascertainment II (Posted on 2014-10-24) Difficulty: 3 of 5
In triangle ABC, it is known that AB = 9 and BC/AC = 40/41.

Find the maximum possible area of the triangle ABC.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Nice problem | Comment 3 of 8 |

There are many neat ways of doing it, and this is just one of them, avoiding Heron or derivatives.

Note that there are 3 points at which the equality AB = 9 and BC/AC = 40/41 clearly holds:
(1) when the height of the triangle is 0, and BC= 40/9, and AC = 41/9.
(2) when BC=40, and AC=41, and the height of the triangle is 9.
(3) when the height of the triangle is 0, and  BC= 40*9, and AC= 41*9.

These are simply 3 points on a circle the upper half of which constitutes the locus of all points with a constant and b and c in the given proportion.

Let the circle have radius r, where (ac-c/a)/2 = (ab+b/a)/2

For clarity, let a=(2n-1), then ((2n-1)c-c/(2n-1))/2 = ((2n-1)b+b/(2n-1))/2; then c(2n^2-2n) = b(2n^2-2n+1): b and c are in the ratio: b=(2(n^2-n)), c=(2(n^2-n)+1), and if n is fixed (e.g. n=5, so a=9) then the ratio is also fixed (e.g. 40:41)

Substituting: ((2n-1)(2(n^2-n)+1)-(2(n^2-n)+1)/(2n-1))/2 = (2n(n-1)(2n^2-2n+1))/(2n-1) = bc/a = r.

Call the centre of the circle E, and let D be a point on the upper half of the circle. Triangle ABD represents all the triangles whose base is a, and whose long sides are in the proportion b/c. Since a is constant, 1/2 base*height requires that the altitude of D be maximised, ie that it be directly above E.

But if so, the area of the maximised triangle is just: bc/a*1/2a = bc/2

In the given example: (40*41)/2 = 820.

In like manner it can easily be shown that if ABC is a right triangle of short side a, and whose longer side, b, and hypotenuse, c, differ by 1, then the ratio of the area, A1, of this triangle to the largest possible area, A2, of a triangle having short side a, and the two remaining sides in the same ratio b/c, is a/c; equally, area A1= 1/2ab, area A2 = 1/2bc.

It follows at once that the 'maximised' right triangle of type a=(2n-1), b=(2(n^2-n)), c=(2(n^2-n)+1) is another right-angled triangle of sides (2(n^2-n)), (2(n^2-n)+1).

Finally, distance AE = bc/a+c/a =c(c-1)/a+c/a=c^2/a, another nice result.

Edited on October 25, 2014, 3:07 am
  Posted by broll on 2014-10-24 23:53:09

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