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Geometry Difficulty 4.8 AIME Find the answer

4. Given the three sides of a triangle a,b,ca, b, c are integers, and a+b+c=11a+b+c=11. Then when the product abcabc takes the minimum value, the area of the triangle is \qquad .

A number or a short expression. Spacing and $ signs are ignored.

Solution

4. Since a,b,ca, b, c are the lengths of the three sides of a triangle, when a=1a=1, b=c=5b=c=5, abc=25abc=25; when a=2a=2, b=4b=4, c=5c=5, abc=40abc=40; when a=3a=3, b=3b=3, c=5c=5, abc=45abc=45, or b=c=4b=c=4, abc=48abc=48. It is evident that the product abc=25abc=25 is the minimum value. In the isosceles ABC\triangle ABC, the height from vertex AA to the base BCBC is h=52(12)2=3112h=\sqrt{5^{2}-\left(\frac{1}{2}\right)^{2}}=\frac{3 \sqrt{11}}{2}, thus the area is S=3114S=\frac{3 \sqrt{11}}{4}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.