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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.

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


But try the problem first…

Answer: is 450.

Source
Suggested Reading

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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