MHB Solving for $x+y$ in $\triangle ABC$

[Albert1]

$\triangle ABC ,\angle C=90^o, \overline{AB}=c, \overline{BC}=a, \overline{CA}=b, x=\dfrac{a}{c}, y=\dfrac{b}{c}$

and satisfying: $13xy=15(x+y)-15,$ find $x+y$

 

[johng1]

Albert,
This is a challenge for high school students?

Spoiler

Clearly $x^2+y^2=1$ and $13xy=15(x+y)-15$ or easily
$$13(x+y)^2=30(x+y)-17$$
Hence by the quadratic formula $x+y=1$ or $x+y={17\over13}$

 

[kaliprasad]

$\triangle ABC ,\angle C=90^o, \overline{AB}=c, \overline{BC}=a, \overline{CA}=b, x=\dfrac{a}{c}, y=\dfrac{b}{c}$

and satisfying: $13xy=15(x+y)-15,$ find $x+y$

Spoiler

because $\angle C = 90^\circ$ hence $c^2= a^2 + b^2$ or $x^2+y^2 = 1$
hence $(x+y)^2 = 1 + 2xy\cdots(1)$
now
$13xy= 15(x+y) - 15$
or $15(x+y) = 15 + 13xy$
square both sides
$15^2 ( 1+ 2xy) = 225 + 2* 15 * 13 xy + 169 x^2y^2$ using (1)
or $ 60xy = 169 x^2y^2$
or $60 = 169 xy$ as xy is not 0
so $(x+y)^2 = 1 + 2xy = 1 + 2 * \frac{60}{169} = \frac{289}{169}= (\frac{17}{13})^2$
or $(x+y) = \dfrac{17}{13}$

 

[kaliprasad]

Albert,
This is a challenge for high school students?

Spoiler

Clearly $x^2+y^2=1$ and $13xy=15(x+y)-15$ or easily
$$13(x+y)^2=30(x+y)-17$$
Hence by the quadratic formula $x+y=1$ or $x+y={17\over13}$

Spoiler

good approach but x + y cannot be 1 because the x or y = 0 which cannot be true in a triangle