Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Right angled triangle.
In \(\Delta ABC\), \(\angle C\) is a right angle and the altitude from C meets AB at D. The lengths of the sides of \(\Delta ABC\) are integers, \(BD={29}^{3}\), and \(cosB=\frac{m}{n}\), where m, n are relatively prime positive integers, find m+n.
Integers
Right angled triangle
Pythagoras Theorem
Answer: is 450.
AIME I, 1994, Question 10
Geometry Vol I to IV by Hall and Stevens
\(\Delta ABC \sim \Delta CBD\)
\(\frac{BC}{AB}=\frac{29^{3}}{BC}\)
\(\Rightarrow {BC}^{2}=29^{3}AB\)
\(\Rightarrow 29^{2}|BC and 29|AB\)
\(\Rightarrow BC and AB are in form 29^{2}x, 29x^{2}\) where x is integer
\(by Pythagoras Theorem, AC^{2}+BC^{2}=AB^{2}\)
\(\Rightarrow (29^{2}x)^{2}+AC^{2}=(29x^{2})^{2}\)
\(\Rightarrow 29x|AC\)
taking y=\(\frac{AC}{29x}\) and dividing by (29x)^{2}\)
\(\Rightarrow 29^{2}=x^{2}-y^{2}=(x-y)(x+y)\)
where x,y are integers, the factors are \((1,29^{2}),(29,29)\)
\(y=\frac{AC}{29x}\) not equals 0 \(\Rightarrow x-y=1, x+y=29\)
\(\Rightarrow x=\frac{1+29^{2}}{2}\)
=421 then\(cosB=\frac{BC}{AB}=\frac{29^{2}x}{29x^{2}}\)=\(\frac{29}{421}\)
m+n=29+421=450.
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