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Right angled triangle | AIME I, 1994 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Right angled triangle.

Right angled triangle - AIME I, 1994


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.

  • is 107
  • is 450
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Right angled triangle

Pythagoras Theorem

Check the Answer


Answer: is 450.

AIME I, 1994, Question 10

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

\(\Delta ABC \sim \Delta CBD\)

\(\frac{BC}{AB}=\frac{29^{3}}{BC}\)

\(\Rightarrow {BC}^{2}=29^{3}AB\)

\(\Rightarrow 29^{2}|BC and 29|AB\)

\(\Rightarrow BC and AB are in form 29^{2}x, 29x^{2}\) where x is integer

Second Hint

\(by Pythagoras Theorem, AC^{2}+BC^{2}=AB^{2}\)

\(\Rightarrow (29^{2}x)^{2}+AC^{2}=(29x^{2})^{2}\)

\(\Rightarrow 29x|AC\)

Final Step

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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