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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

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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