Esercizi sul Circocentro in Geometria

Esercizi sul Circocentro in Geometria

Esercizi sul Circocentro in Geometria

Versione italiana

Esercizi sul Circocentro in Geometria

Il circocentro è uno dei punti notevoli di un triangolo. È il punto di intersezione delle bisettrici dei lati del triangolo e rappresenta il centro della circonferenza circoscritta al triangolo.

Concetti Chiave

1. Definizione di Circocentro: Il circocentro di un triangolo è il punto in cui si incontrano le tre bisettrici dei lati del triangolo. È equidistante dai tre vertici del triangolo.

2. Equazione della circonferenza circoscritta: La circonferenza circoscritta ha come centro il circocentro e passa per i tre vertici del triangolo.

3. Coordinate del Circocentro: Se i vertici del triangolo sono A(x_1, y_1)$A(x_1, y_1)$, B(x_2, y_2)$B(x_2, y_2)$, e C(x_3, y_3)$C(x_3, y_3)$, le coordinate del circocentro O(x_O, y_O)$O(x_O, y_O)$ possono essere calcolate utilizzando le seguenti formule:
   x_O = \frac{(x_1^2 + y_1^2)(y_2 - y_3) + (x_2^2 + y_2^2)(y_3 - y_1) + (x_3^2 + y_3^2)(y_1 - y_2)}{2 \cdot \text{Area}} $x_O = \frac{(x_1^2 + y_1^2)(y_2 - y_3) + (x_2^2 + y_2^2)(y_3 - y_1) + (x_3^2 + y_3^2)(y_1 - y_2)}{2 \cdot \text{Area}}$
   y_O = \frac{(x_1^2 + y_1^2)(x_3 - x_2) + (x_2^2 + y_2^2)(x_1 - x_3) + (x_3^2 + y_3^2)(x_2 - x_1)}{2 \cdot \text{Area}} $y_O = \frac{(x_1^2 + y_1^2)(x_3 - x_2) + (x_2^2 + y_2^2)(x_1 - x_3) + (x_3^2 + y_3^2)(x_2 - x_1)}{2 \cdot \text{Area}}$
   dove l’area del triangolo può essere calcolata con la formula:
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Esercizi

Esercizio 1: Trovare il Circocentro di un Triangolo

Problema: Trova il circocentro del triangolo con vertici A(1, 2)$A(1, 2)$, B(4, 6)$B(4, 6)$, e C(5, 1)$C(5, 1)$.

Soluzione:

1. Calcola l’area del triangolo:
   \text{Area} = \frac{1}{2} \left| 1(6 - 1) + 4(1 - 2) + 5(2 - 6) \right| $\text{Area} = \frac{1}{2} \left| 1(6 - 1) + 4(1 - 2) + 5(2 - 6) \right|$
   = \frac{1}{2} \left| 1 \cdot 5 + 4 \cdot (-1) + 5 \cdot (-4) \right| = \frac{1}{2} \left| 5 - 4 - 20 \right| = \frac{1}{2} \left| -19 \right| = \frac{19}{2} $= \frac{1}{2} \left| 1 \cdot 5 + 4 \cdot (-1) + 5 \cdot (-4) \right| = \frac{1}{2} \left| 5 - 4 - 20 \right| = \frac{1}{2} \left| -19 \right| = \frac{19}{2}$

2. Calcola le coordinate del circocentro:

   • Calcola x_O$x_O$:
     x_O = \frac{(1^2 + 2^2)(6 - 1) + (4^2 + 6^2)(1 - 2) + (5^2 + 1^2)(2 - 6)}{2 \cdot \frac{19}{2}} $x_O = \frac{(1^2 + 2^2)(6 - 1) + (4^2 + 6^2)(1 - 2) + (5^2 + 1^2)(2 - 6)}{2 \cdot \frac{19}{2}}$
     = \frac{(1 + 4)(5) + (16 + 36)(-1) + (25 + 1)(-4)}{19} $= \frac{(1 + 4)(5) + (16 + 36)(-1) + (25 + 1)(-4)}{19}$
     = \frac{5 \cdot 5 - 52 - 104}{19} = \frac{25 - 52 - 104}{19} = \frac{-131}{19} \approx -6.89 $= \frac{5 \cdot 5 - 52 - 104}{19} = \frac{25 - 52 - 104}{19} = \frac{-131}{19} \approx -6.89$

   • Calcola y_O$y_O$:
     y_O = \frac{(1^2 + 2^2)(5 - 4) + (4^2 + 6^2)(1 - 5) + (5^2 + 1^2)(4 - 1)}{2 \cdot \frac{19}{2}} $y_O = \frac{(1^2 + 2^2)(5 - 4) + (4^2 + 6^2)(1 - 5) + (5^2 + 1^2)(4 - 1)}{2 \cdot \frac{19}{2}}$
     = \frac{(1 + 4)(1) + (16 + 36)(-4) + (25 + 1)(3)}{19} $= \frac{(1 + 4)(1) + (16 + 36)(-4) + (25 + 1)(3)}{19}$
     = \frac{5 - 208 + 78}{19} = \frac{-125}{19} \approx -6.58 $= \frac{5 - 208 + 78}{19} = \frac{-125}{19} \approx -6.58$

3. Coordinate del circocentro:
   Il circocentro O$O$ del triangolo è quindi:
   O\left(-\frac{131}{19}, -\frac{125}{19}\right) \approx O(-6.89, -6.58) $O\left(-\frac{131}{19}, -\frac{125}{19}\right) \approx O(-6.89, -6.58)$

Esercizio 2: Verificare la posizione del Circocentro

Problema: Considera un triangolo rettangolo con vertici A(0, 0)$A(0, 0)$, B(0, 4)$B(0, 4)$, e C(3, 0)$C(3, 0)$. Trova il circocentro e verifica se coincide con un vertice.

Soluzione:

1. Identifica il triangolo rettangolo: In questo caso, l’angolo retto è in A(0, 0)$A(0, 0)$.

2. Il circocentro di un triangolo rettangolo è il punto medio dell’ipotenusa. Calcoliamo le coordinate del circocentro.

   • L’ipotenusa è il segmento BC$BC$ con i punti B(0, 4)$B(0, 4)$ e C(3, 0)$C(3, 0)$.
   • Le coordinate del circocentro O$O$ sono date dalla media delle coordinate di B$B$ e C$C$:
     O\left(\frac{0 + 3}{2}, \frac{4 + 0}{2}\right) = O\left(\frac{3}{2}, 2\right) = O(1.5, 2) $O\left(\frac{0 + 3}{2}, \frac{4 + 0}{2}\right) = O\left(\frac{3}{2}, 2\right) = O(1.5, 2)$

3. Conclusione: Il circocentro O(1.5, 2)$O(1.5, 2)$ non coincide con nessuno dei vertici del triangolo.

Esercizio 3: Calcolare il raggio della circonferenza circoscritta

Problema: Utilizzando il triangolo con vertici A(1, 2)$A(1, 2)$, B(4, 6)$B(4, 6)$, e C(5, 1)$C(5, 1)$ trovato in precedenza, calcola il raggio della circonferenza circoscritta.

Soluzione:

1. Calcola la distanza dal circocentro a uno dei vertici. Utilizziamo il vertice A(1, 2)$A(1, 2)$ e il circocentro O\left(-\frac{131}{19}, -\frac{125}{19}\right)$O\left(-\frac{131}{19}, -\frac{125}{19}\right)$.

2. Formula per la distanza:
   r = \sqrt{\left(x_O - x_A\right)^2 + \left(y_O - y_A\right)^2} $r = \sqrt{\left(x_O - x_A\right)^2 + \left(y_O - y_A\right)^2}$
   Sostituendo i valori:
   r = \sqrt{\left(-\frac{131}{19} - 1\right)^2 + \left(-\frac{125}{19} - 2\right)^2} $r = \sqrt{\left(-\frac{131}{19} - 1\right)^2 + \left(-\frac{125}{19} - 2\right)^2}$
   = \sqrt{\left(-\frac{131 + 19}{19}\right)^2 + \left(-\frac{125 + 38}{19}\right)^2} $= \sqrt{\left(-\frac{131 + 19}{19}\right)^2 + \left(-\frac{125 + 38}{19}\right)^2}$
   = \sqrt{\left(-\frac{150}{19}\right)^2 + \left(-\frac{163}{19}\right)^2} $= \sqrt{\left(-\frac{150}{19}\right)^2 + \left(-\frac{163}{19}\right)^2}$
   = \frac{1}{19} \sqrt{150^2 + 163^2} $= \frac{1}{19} \sqrt{150^2 + 163^2}$
   = \frac{1}{19} \sqrt{22500 + 26569} = \frac{1}{19} \sqrt{49069} \approx \frac{221}{19} \approx 11.63 $= \frac{1}{19} \sqrt{22500 + 26569} = \frac{1}{19} \sqrt{49069} \approx \frac{221}{19} \approx 11.63$

Il raggio della circonferenza circoscritta è quindi circa 11.63$11.63$ unità.

English version

Exercises on the Circumcenter in Geometry

The circumcenter is one of the notable points of a triangle. It is the point of intersection of the bisectors of the sides of the triangle and represents the center of the circumscribed circle of the triangle.

Key Concepts

1. Definition of Circumcenter: The circumcenter of a triangle is the point where the three bisectors of the sides of the triangle meet. It is equidistant from the three vertices of the triangle.

2. Equation of the circumscribed circle: The circumscribed circle has the circumcenter as its center and passes through the three vertices of the triangle.

3. Circumcenter Coordinates: If the vertices of the triangle are A(x_1, y_1)$A(x_1, y_1)$, B(x_2, y_2)$B(x_2, y_2)$, and C(x_3, y_3)$C(x_3, y_3)$, the circumcenter coordinates O(x_O, y_O)$O(x_O, y_O)$ can be calculated using the following formulas:
   x_O = \frac{(x_1^2 + y_1^2)(y_2 - y_3) + (x_2^2 + y_2^2)(y_3 - y_1) + (x_3^2 + y_3^2)(y_1 - y_2)}{2 \cdot \text{Area}} $x_O = \frac{(x_1^2 + y_1^2)(y_2 - y_3) + (x_2^2 + y_2^2)(y_3 - y_1) + (x_3^2 + y_3^2)(y_1 - y_2)}{2 \cdot \text{Area}}$
   y_O = \frac{(x_1^2 + y_1^2)(x_3 - x_2) + (x_2^2 + y_2^2)(x_1 - x_3) + (x_3^2 + y_3^2)(x_2 - x_1)}{2 \cdot \text{Area}} $y_O = \frac{(x_1^2 + y_1^2)(x_3 - x_2) + (x_2^2 + y_2^2)(x_1 - x_3) + (x_3^2 + y_3^2)(x_2 - x_1)}{2 \cdot \text{Area}}$
   where the area of ​​the triangle can be calculated with the formula:
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Exercises

Exercise 1: Finding the Circumcenter of a Triangle

Problem: Find the circumcenter of the triangle with vertices A(1, 2)$A(1, 2)$, B(4, 6)$B(4, 6)$, and C(5, 1)$C(5, 1)$.

Solution:

1. Calculate the area of ​​the triangle:
   \text{Area} = \frac{1}{2} \left| 1(6 - 1) + 4(1 - 2) + 5(2 - 6) \right| $\text{Area} = \frac{1}{2} \left| 1(6 - 1) + 4(1 - 2) + 5(2 - 6) \right|$
   = \frac{1}{2} \left| 1 \cdot 5 + 4 \cdot (-1) + 5 \cdot (-4) \right| = \frac{1}{2} \left| 5 - 4 - 20 \right| = \frac{1}{2} \left| -19 \right| = \frac{19}{2} $= \frac{1}{2} \left| 1 \cdot 5 + 4 \cdot (-1) + 5 \cdot (-4) \right| = \frac{1}{2} \left| 5 - 4 - 20 \right| = \frac{1}{2} \left| -19 \right| = \frac{19}{2}$

2. Calculate the coordinates of the circumcenter:

• Calculate x_O$x_O$:
  x_O = \frac{(1^2 + 2^2)(6 - 1) + (4^2 + 6^2)(1 - 2) + (5^2 + 1^2)(2 - 6)}{2 \cdot \frac{19}{2}} $x_O = \frac{(1^2 + 2^2)(6 - 1) + (4^2 + 6^2)(1 - 2) + (5^2 + 1^2)(2 - 6)}{2 \cdot \frac{19}{2}}$
  = \frac{(1 + 4)(5) + (16 + 36)(-1) + (25 + 1)(-4)}{19} $= \frac{(1 + 4)(5) + (16 + 36)(-1) + (25 + 1)(-4)}{19}$
  = \frac{5 \cdot 5 - 52 - 104}{19} = \frac{25 - 52 - 104}{19} = \frac{-131}{19} \approx -6.89 $= \frac{5 \cdot 5 - 52 - 104}{19} = \frac{25 - 52 - 104}{19} = \frac{-131}{19} \approx -6.89$

• Calculate y_O$y_O$:
  y_O = \frac{(1^2 + 2^2)(5 - 4) + (4^2 + 6^2)(1 - 5) + (5^2 + 1^2)(4 - 1)}{2 \cdot \frac{19}{2}} $y_O = \frac{(1^2 + 2^2)(5 - 4) + (4^2 + 6^2)(1 - 5) + (5^2 + 1^2)(4 - 1)}{2 \cdot \frac{19}{2}}$
  = \frac{(1 + 4)(1) + (16 + 36)(-4) + (25 + 1)(3)}{19} $= \frac{(1 + 4)(1) + (16 + 36)(-4) + (25 + 1)(3)}{19}$
  = \frac{5 - 208 + 78}{19} = \frac{-125}{19} \approx -6.58 $= \frac{5 - 208 + 78}{19} = \frac{-125}{19} \approx -6.58$

3. Coordinates of the circumcenter:
   The circumcenter O$O$ of the triangle is therefore:
   O\left(-\frac{131}{19}, -\frac{125}{19}\right) \approx O(-6.89, -6.58) $O\left(-\frac{131}{19}, -\frac{125}{19}\right) \approx O(-6.89, -6.58)$

Exercise 2: Verify the position of the Circumcenter

Problem: Consider a right-angled triangle with vertices A(0, 0)$A(0, 0)$, B(0, 4)$B(0, 4)$, and C(3, 0)$C(3, 0)$. Find the circumcenter and verify if it coincides with a vertex.

Solution:

1. Identify the right-angled triangle: In this case, the right angle is in A(0, 0)$A(0, 0)$.

2. The circumcenter of a right-angled triangle is the midpoint of the hypotenuse. Let’s calculate the coordinates of the circumcenter.

• The hypotenuse is the segment BC$BC$ with the points B(0, 4)$B(0, 4)$ and C(3, 0)$C(3, 0)$.
• The coordinates of the circumcenter O$O$ are given by the average of the coordinates of B$B$ and C$C$:
  O\left(\frac{0 + 3}{2}, \frac{4 + 0}{2}\right) = O\left(\frac{3}{2}, 2\right) = O(1.5, 2) $O\left(\frac{0 + 3}{2}, \frac{4 + 0}{2}\right) = O\left(\frac{3}{2}, 2\right) = O(1.5, 2)$

3. Conclusion: The circumcenter O(1.5, 2)$O(1.5, 2)$ does not coincide with any of the vertices of the triangle.

Exercise 3: Calculate the radius of the circumscribed circle

Problem: Using the triangle with vertices A(1, 2)$A(1, 2)$, B(4, 6)$B(4, 6)$, and C(5, 1)$C(5, 1)$ found previously, calculate the radius of the circumscribed circle.

Solution:

1. Calculate the distance from the circumcenter to one of the vertices. We use the vertex A(1, 2)$A(1, 2)$ and the circumcenter O\left(-\frac{131}{19}, -\frac{125}{19}\right)$O\left(-\frac{131}{19}, -\frac{125}{19}\right)$.

2. Formula for distance:
   r = \sqrt{\left(x_O - x_A\right)^2 + \left(y_O - y_A\right)^2} $r = \sqrt{\left(x_O - x_A\right)^2 + \left(y_O - y_A\right)^2}$
   Substituting the values:
   r = \sqrt{\left(-\frac{131}{19} - 1\right)^2 + \left(-\frac{125}{19} - 2\right)^2} $r = \sqrt{\left(-\frac{131}{19} - 1\right)^2 + \left(-\frac{125}{19} - 2\right)^2}$
   = \sqrt{\left(-\frac{131 + 19}{19}\right)^2 + \left(-\frac{125 + 38}{19}\right)^2} $= \sqrt{\left(-\frac{131 + 19}{19}\right)^2 + \left(-\frac{125 + 38}{19}\right)^2}$
   = \sqrt{\left(-\frac{150}{19}\right)^2 + \left(-\frac{163}{19}\right)^2} $= \sqrt{\left(-\frac{150}{19}\right)^2 + \left(-\frac{163}{19}\right)^2}$
   = \frac{1}{19} \sqrt{150^2 + 163^2} $= \frac{1}{19} \sqrt{150^2 + 163^2}$
   = \frac{1}{19} \sqrt{22500 + 26569} = \frac{1}{19} \sqrt{49069} \approx \frac{221}{19} \approx 11.63 $= \frac{1}{19} \sqrt{22500 + 26569} = \frac{1}{19} \sqrt{49069} \approx \frac{221}{19} \approx 11.63$

The radius of the circumscribed circle is therefore approximately 11.63$11.63$ units.