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Let A, B, and C be three points on the curve xy = 1 (which is a hyperbola). Prove that the orthocenter of triangle ABC also lies on the curve xy = 1.

 

Thank you!

Best Answer 

 #1
avatar+20025 
+2

Let A, B, and C be three points on the curve xy = 1 (which is a hyperbola).
Prove that the orthocenter of triangle ABC also lies on the curve xy = 1.

 

\(\text{Let $A= (x_A,y_A) = (x_A, \frac{1}{x_A} ) $ } \\ \text{Let $B= (x_B,y_B) = (x_B, \frac{1}{x_B} ) $ } \\ \text{Let $C= (x_C,y_C) = (x_C, \frac{1}{x_C} ) $ } \)

 

1.

\(\displaystyle \text{The slope of $\overline{AB}$ is $\frac{y_B-y_A}{x_B-x_A} = \frac{ \frac{1}{x_B} -\frac{1}{x_A} }{ x_B-x_A } = \frac{x_A-x_B}{x_Ax_B(x_B-x_A)} = -\frac{1}{x_Ax_B} $} \\ \displaystyle \text{ So, the slope of the altitude, which is perpendicular to $\overline{AB}$ is $-\frac{1}{-\frac{1}{x_Ax_B}} = x_Ax_B $}\)

 

\(\displaystyle \text{The equation of the altitude from C to $\overline{AB}$ is:}\)

\(\begin{array}{|rcll|} \hline y-y_C &=& m(x-x_C) \quad & | \quad m = x_Ax_B \\ y &=& y_C + x_Ax_B(x-x_C) \\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{x_C}+ x_Ax_B(x-x_C)} \\ \hline \end{array}\)

 

2.

\(\displaystyle \text{The slope of $\overline{BC}$ is $\frac{y_C-y_B}{x_C-x_B} = \frac{ \frac{1}{x_C} -\frac{1}{x_B} }{ x_C-x_B } = \frac{x_B-x_C}{x_Bx_C(x_C-x_B)} = -\frac{1}{x_Bx_C} $} \\ \displaystyle \text{ So, the slope of the altitude, which is perpendicular to $\overline{BC}$ is $-\frac{1}{-\frac{1}{x_Bx_C}} = x_Bx_C $}\)

 

\(\displaystyle \text{The equation of the altitude from A to $\overline{BC}$ is:}\)

\(\begin{array}{|rcll|} \hline y-y_A &=& m(x-x_A) \quad & | \quad m = x_Bx_C \\ y &=& y_A + x_Bx_C(x-x_A) \\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{x_A}+ x_Bx_C(x-x_A)} \\ \hline \end{array}\)

 

Solve the equations to find the intersection point of the altitudes:

 

3. \(x_{\text{orthocenter}} = \ ?\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x_C}+ x_Ax_B(x-x_C) &=& \dfrac{1}{x_A}+ x_Bx_C(x-x_A) \\ \dfrac{1}{x_C}+ x_Ax_Bx-x_Ax_Bx_C &=& \dfrac{1}{x_A}+ x_Bx_Cx-x_Bx_Cx_A \\ \dfrac{1}{x_C}+ x_Ax_Bx &=& \dfrac{1}{x_A}+ x_Bx_Cx \\ x_Ax_Bx-x_Bx_Cx &=& \dfrac{1}{x_A}-\dfrac{1}{x_C} \\ x_Bx(x_A-x_C) &=& \dfrac{x_C-x_A}{x_Ax_C} \\ x_Bx(x_A-x_C) &=& -\dfrac{x_A-x_C}{x_Ax_C} \\ x_Bx &=& -\dfrac{1}{x_Ax_C} \\ \mathbf{x_{\text{orthocenter}}} & \mathbf{=} & \mathbf{ -\dfrac{1}{x_Ax_Bx_C} } \\ \hline \end{array}\)

 

4. \(y_{\text{orthocenter}} = \ ?\)

\(\begin{array}{|rcll|} \hline y_{\text{orthocenter}} &=& \dfrac{1}{x_C}+ x_Ax_B(x_{\text{orthocenter}}-x_C) \\ &=& \dfrac{1}{x_C}+ x_Ax_B(-\dfrac{1}{x_Ax_Bx_C}-x_C) \\ &=& \dfrac{1}{x_C}- \dfrac{x_Ax_B}{x_Ax_Bx_C}-x_Ax_Bx_C \\ &=& \dfrac{1}{x_C}- \dfrac{1}{ x_C}-x_Ax_Bx_C \\ \mathbf{y_{\text{orthocenter}}} & \mathbf{=} & \mathbf{ -x_Ax_Bx_C } \\ \hline \end{array}\)

 

\(\text{The orthocenter of triangle ABC lies on the curve $xy = 1$, if $x_{\text{orthocenter}}\cdot y_{\text{orthocenter}} = 1$ } \)

\(\begin{array}{|rcll|} \hline x_{\text{orthocenter}}\cdot y_{\text{orthocenter}} &=& -\dfrac{1}{x_Ax_Bx_C} \cdot (-x_Ax_Bx_C) \\ &=& \dfrac{x_Ax_Bx_C}{x_Ax_Bx_C} \\ &=& 1\checkmark \\ \hline \end{array}\)

 

laugh

heureka  Jun 8, 2018
 #1
avatar+20025 
+2
Best Answer

Let A, B, and C be three points on the curve xy = 1 (which is a hyperbola).
Prove that the orthocenter of triangle ABC also lies on the curve xy = 1.

 

\(\text{Let $A= (x_A,y_A) = (x_A, \frac{1}{x_A} ) $ } \\ \text{Let $B= (x_B,y_B) = (x_B, \frac{1}{x_B} ) $ } \\ \text{Let $C= (x_C,y_C) = (x_C, \frac{1}{x_C} ) $ } \)

 

1.

\(\displaystyle \text{The slope of $\overline{AB}$ is $\frac{y_B-y_A}{x_B-x_A} = \frac{ \frac{1}{x_B} -\frac{1}{x_A} }{ x_B-x_A } = \frac{x_A-x_B}{x_Ax_B(x_B-x_A)} = -\frac{1}{x_Ax_B} $} \\ \displaystyle \text{ So, the slope of the altitude, which is perpendicular to $\overline{AB}$ is $-\frac{1}{-\frac{1}{x_Ax_B}} = x_Ax_B $}\)

 

\(\displaystyle \text{The equation of the altitude from C to $\overline{AB}$ is:}\)

\(\begin{array}{|rcll|} \hline y-y_C &=& m(x-x_C) \quad & | \quad m = x_Ax_B \\ y &=& y_C + x_Ax_B(x-x_C) \\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{x_C}+ x_Ax_B(x-x_C)} \\ \hline \end{array}\)

 

2.

\(\displaystyle \text{The slope of $\overline{BC}$ is $\frac{y_C-y_B}{x_C-x_B} = \frac{ \frac{1}{x_C} -\frac{1}{x_B} }{ x_C-x_B } = \frac{x_B-x_C}{x_Bx_C(x_C-x_B)} = -\frac{1}{x_Bx_C} $} \\ \displaystyle \text{ So, the slope of the altitude, which is perpendicular to $\overline{BC}$ is $-\frac{1}{-\frac{1}{x_Bx_C}} = x_Bx_C $}\)

 

\(\displaystyle \text{The equation of the altitude from A to $\overline{BC}$ is:}\)

\(\begin{array}{|rcll|} \hline y-y_A &=& m(x-x_A) \quad & | \quad m = x_Bx_C \\ y &=& y_A + x_Bx_C(x-x_A) \\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{1}{x_A}+ x_Bx_C(x-x_A)} \\ \hline \end{array}\)

 

Solve the equations to find the intersection point of the altitudes:

 

3. \(x_{\text{orthocenter}} = \ ?\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x_C}+ x_Ax_B(x-x_C) &=& \dfrac{1}{x_A}+ x_Bx_C(x-x_A) \\ \dfrac{1}{x_C}+ x_Ax_Bx-x_Ax_Bx_C &=& \dfrac{1}{x_A}+ x_Bx_Cx-x_Bx_Cx_A \\ \dfrac{1}{x_C}+ x_Ax_Bx &=& \dfrac{1}{x_A}+ x_Bx_Cx \\ x_Ax_Bx-x_Bx_Cx &=& \dfrac{1}{x_A}-\dfrac{1}{x_C} \\ x_Bx(x_A-x_C) &=& \dfrac{x_C-x_A}{x_Ax_C} \\ x_Bx(x_A-x_C) &=& -\dfrac{x_A-x_C}{x_Ax_C} \\ x_Bx &=& -\dfrac{1}{x_Ax_C} \\ \mathbf{x_{\text{orthocenter}}} & \mathbf{=} & \mathbf{ -\dfrac{1}{x_Ax_Bx_C} } \\ \hline \end{array}\)

 

4. \(y_{\text{orthocenter}} = \ ?\)

\(\begin{array}{|rcll|} \hline y_{\text{orthocenter}} &=& \dfrac{1}{x_C}+ x_Ax_B(x_{\text{orthocenter}}-x_C) \\ &=& \dfrac{1}{x_C}+ x_Ax_B(-\dfrac{1}{x_Ax_Bx_C}-x_C) \\ &=& \dfrac{1}{x_C}- \dfrac{x_Ax_B}{x_Ax_Bx_C}-x_Ax_Bx_C \\ &=& \dfrac{1}{x_C}- \dfrac{1}{ x_C}-x_Ax_Bx_C \\ \mathbf{y_{\text{orthocenter}}} & \mathbf{=} & \mathbf{ -x_Ax_Bx_C } \\ \hline \end{array}\)

 

\(\text{The orthocenter of triangle ABC lies on the curve $xy = 1$, if $x_{\text{orthocenter}}\cdot y_{\text{orthocenter}} = 1$ } \)

\(\begin{array}{|rcll|} \hline x_{\text{orthocenter}}\cdot y_{\text{orthocenter}} &=& -\dfrac{1}{x_Ax_Bx_C} \cdot (-x_Ax_Bx_C) \\ &=& \dfrac{x_Ax_Bx_C}{x_Ax_Bx_C} \\ &=& 1\checkmark \\ \hline \end{array}\)

 

laugh

heureka  Jun 8, 2018
 #2
avatar+90023 
+2

Very nice, heureka  !!!

 

 

cool cool cool

CPhill  Jun 8, 2018
 #3
avatar+20025 
+1

Thank you CPhill

 

laugh

heureka  Jun 11, 2018
 #4
avatar+1442 
+1

Thank you Heureka!

AnonymousConfusedGuy  Jun 11, 2018

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