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

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Discussion Overview

The discussion centers around solving for the sum $x+y$ in a right triangle $\triangle ABC$ where $\angle C=90^\circ$. Participants explore the relationship defined by the equation $13xy=15(x+y)-15$, with $x$ and $y$ expressed in terms of the triangle's sides.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents the problem involving the right triangle and the equation to solve for $x+y$.
  • Another participant questions whether this problem is appropriate as a challenge for high school students.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as one participant questions the difficulty level of the problem without further elaboration from others.

Contextual Notes

There is no indication of assumptions or limitations discussed regarding the problem or its context.

Who May Find This Useful

Students and educators interested in high school level geometry and algebraic problem-solving may find this discussion relevant.

Albert1
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$\triangle ABC ,\angle C=90^o, \overline{AB}=c, \overline{BC}=a, \overline{CA}=b, x=\dfrac{a}{c}, y=\dfrac{b}{c}$

and satisfying: $13xy=15(x+y)-15,$ find $x+y$
 
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Albert,
This is a challenge for high school students?
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 $x+y={17\over13}$
 
Albert said:
$\triangle ABC ,\angle C=90^o, \overline{AB}=c, \overline{BC}=a, \overline{CA}=b, x=\dfrac{a}{c}, y=\dfrac{b}{c}$

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

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 $(x+y) = \dfrac{17}{13}$
 
johng said:
Albert,
This is a challenge for high school students?
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 $x+y={17\over13}$
good approach but x + y cannot be 1 because the x or y = 0 which cannot be true in a triangle
 

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