Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.
Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and \(\angle BDC\)=3\(\angle BAC\). then the perimeter of \(\Delta ABC\) may be written in the form \(a+\sqrt{b}\) where a and b are integers, find a+b.
Integers
Triangle
Trigonometry
Answer: is 616.
AIME I, 1995, Question 9
Plane Trigonometry by Loney
Let x= \(\angle CAM\)
\(\Rightarrow \angle CDM =3x\)
\(\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}\)=11 [by trigonometry ratio property in right angled triangle]
\(\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx\)
solving we get, tanx=\(\frac{1}{2}\)
\(\Rightarrow CM=\frac{11}{2}\)
\(\Rightarrow 2(AC+CM)\) where \(AC=\frac{11\sqrt {5}}{2}\) by Pythagoras formula
=\(\sqrt{605}+11\) then a+b=605+11=616.
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