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Geometry Difficulty 4.8 AIME Find the answer

6. Given that the base BCBC and the height ADAD of isosceles △ABC\triangle ABC are both integers. Then, sin⁡A\sin A and cos⁡A\cos A ( ).
(A) one is a rational number, the other is an irrational number
(C) both are rational numbers
(D) both are irrational numbers, which depends on the values of BCBC and ADAD to determine

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is C, and the solution below works it through.

Solution

6. B.

Min, ∵S△ABC\because S_{\triangle A B C}
=−12AB⋅ACsin⁡A=12⋅BC⋅AD,∴sin⁡A=BC⋅ADAB⋅AC=BC⋅ADAB2=BC⋅ADBD2+AD2=BC⋅AD14BC2+AD2=4BC⋅ADBC2+4AD2 \begin{array}{l} =\frac{-1}{2} A B \cdot A C \sin A \\ =\frac{1}{2} \cdot B C \cdot A D, \\ \therefore \sin A=\frac{B C \cdot A D}{A B \cdot A C}=\frac{B C \cdot A D}{A B^{2}}=\frac{B C \cdot A D}{B D^{2}+A D^{2}} \\ \quad=\frac{B C \cdot A D}{\frac{1}{4} B C^{2}+A D^{2}}=\frac{4 B C \cdot A D}{B C^{2}+4 A D^{2}} \end{array}

In △ABC\triangle A B C, by the cosine rule
cos⁡A=AB2+AC2−BC22AB⋅AC=2AB2−BC22AB2=1−BC22AB2=1−BC22(BD2+AD2)=1−BC22(14⋅BC2+AD2)=1−2BC2BC3+4AD2. \begin{aligned} \cos A & =\frac{A B^{2}+A C^{2}-B C^{2}}{2 A B \cdot A C}=\frac{2 A B^{2}-B C^{2}}{2 A B^{2}} \\ & =1-\frac{B C^{2}}{2 A B^{2}}=1-\frac{B C^{2}}{2\left(B D^{2}+A D^{2}\right)} \\ & =1-\frac{B C^{2}}{2\left(\frac{1}{4} \cdot B C^{2}+A D^{2}\right)}=1-\frac{2 B C^{2}}{B C^{3}+4 A D^{2}} . \end{aligned}
∵BC\because B C and ADA D are both integers,
∴sin⁡A\therefore \sin A and cos⁡A\cos A are both rational numbers.

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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.