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Geometry Difficulty 5.6 AIME, harder Prove it

Example 1: Does there exist an isosceles triangle with integer sides, an integer area, and a perimeter of 2003? Prove your conclusion.

If there exists such a triangle, let the two equal sides be aa and the base be bb. The perimeter condition gives us:
2a+b=2003 2a + b = 2003

The area AA of the isosceles triangle can be expressed using Heron's formula. The semi-perimeter ss is:
s=2a+b2=20032 s = \frac{2a + b}{2} = \frac{2003}{2}

The area AA is:
A=s(s−a)(s−a)(s−b)=20032(20032−a)(20032−a)(20032−b) A = \sqrt{s(s-a)(s-a)(s-b)} = \sqrt{\frac{2003}{2} \left(\frac{2003}{2} - a\right) \left(\frac{2003}{2} - a\right) \left(\frac{2003}{2} - b\right)}

Simplifying the terms inside the square root:
A=20032(2003−2a2)(2003−2a2)(2003−2b2) A = \sqrt{\frac{2003}{2} \left(\frac{2003 - 2a}{2}\right) \left(\frac{2003 - 2a}{2}\right) \left(\frac{2003 - 2b}{2}\right)}
A=20032(2003−2a2)2(2003−2b2) A = \sqrt{\frac{2003}{2} \left(\frac{2003 - 2a}{2}\right)^2 \left(\frac{2003 - 2b}{2}\right)}
A=20032(2003−2a2)2(2003−2(2003−2a)2) A = \sqrt{\frac{2003}{2} \left(\frac{2003 - 2a}{2}\right)^2 \left(\frac{2003 - 2(2003 - 2a)}{2}\right)}
A=20032(2003−2a2)2(4a−20032) A = \sqrt{\frac{2003}{2} \left(\frac{2003 - 2a}{2}\right)^2 \left(\frac{4a - 2003}{2}\right)}
A=20032(2003−2a2)2(4a−20032) A = \sqrt{\frac{2003}{2} \left(\frac{2003 - 2a}{2}\right)^2 \left(\frac{4a - 2003}{2}\right)}
A=20032(2003−2a2)2(4a−20032) A = \sqrt{\frac{2003}{2} \left(\frac{2003 - 2a}{2}\right)^2 \left(\frac{4a - 2003}{2}\right)}
A=20032(2003−2a2)2(4a−20032) A = \sqrt{\frac{2003}{2} \left(\frac{2003 - 2a}{2}\right)^2 \left(\frac{4a - 2003}{2}\right)}

For AA to be an integer, the expression inside the square root must be a perfect square. Let's analyze the conditions for this to happen.

First, note that 20032003 is a prime number. For the expression inside the square root to be a perfect square, the terms 2003−2a2\frac{2003 - 2a}{2} and 4a−20032\frac{4a - 2003}{2} must be such that their product, when multiplied by 20032\frac{2003}{2}, results in a perfect square.

However, since 20032003 is prime, the only way for the product to be a perfect square is if each term is a multiple of 20032003. This is not possible because 2003−2a2003 - 2a and 4a−20034a - 2003 cannot both be multiples of 20032003 for integer values of aa.

Therefore, there does not exist an isosceles triangle with integer sides, an integer area, and a perimeter of 2003.

Solution

Explanation: First, express the area of a triangle using its three side lengths, then analyze whether the area is an integer given that the perimeter is 2003 and the side lengths are integers.

Assume such a triangle exists, with the base and the equal sides being aa and bb (where aa and bb are integers), then the height on the base is
h=b2−(a2)2=124b2−a2. Therefore, S△=14a4b2−a2=14a(2b+a)(2b−a). \begin{array}{c} h=\sqrt{b^{2}-\left(\frac{a}{2}\right)^{2}}=\frac{1}{2} \sqrt{4 b^{2}-a^{2}}. \text{ Therefore, } \\ S_{\triangle}=\frac{1}{4} a \sqrt{4 b^{2}-a^{2}} \\ =\frac{1}{4} a \sqrt{(2 b+a)(2 b-a)} . \end{array}

Since 2b+a2 b+a and 2b−a2 b-a have the same parity, and 2b+a=20032 b+a=2003 is odd, then aa and 2b−a2 b-a are also odd.
Thus, (2b+a)(2b−a)(2 b+a)(2 b-a) is odd.
Therefore, a(2b+a)(2b−a)a \sqrt{(2 b+a)(2 b-a)} cannot be divisible by 4, meaning S△S_{\triangle} cannot be an integer.
Hence, such a triangle does not exist.
Note: This problem is easier to set up using geometric properties to establish the relationship of the triangle's area, the key is to determine that aa and (2b+a)(2b−a)(2 b+a)(2 b-a) are both odd.

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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.