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10

$begingroup$

While working in GeoGebra I noticed something odd. I had a triangle with a point inside and the point was connected to each of the vertices. For each vertice I had drawn the circle passing through the vertice and the point, with the connection being the circle's diameter (see picture below).

What I noticed is that the overlapping circles completely covered the triangle. Further experimentation showed this was also the case if the point was outside the triangle (see below).

Still more experimentation appears to show this is the case for any polygon, simple or not:

Is this observation true or did GeoGebra lead me astray? I couldn't immediately find the result via Google.

geometry recreational-mathematics

share|cite|improve this question

asked yesterday

JensJens

4,02721032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is another remarkable detail that the intersection point of any pair of circles distinct from the red point lies on the line connecting two vertices. And this fact is quite easy to prove.
  $endgroup$
  – user
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user I noticed that too. Could be a starting point for proving the above is true (if it is true).
  $endgroup$
  – Jens
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes this can be the starting point. In the case of internal point of the triangle it is practically evident. In the case of polygons it suffices to consider its convex hull.
  $endgroup$
  – user
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Given any polygon $mathcalP = ABCcdots F$ and a point $X$. Consider triangles formed by $X$ and an edge of $mathcalP$. Let's say triangle $XAB$. Let $Y$ be the foot of the altitude through $X$ on $AB$. Split $XAB$ into two right angled triangles $XAY$ and $XYB$. These two triangles will be covered by the two circles with $XA$ and $XB$ as diameters....
  $endgroup$
  – achille hui
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For future reference, "vertice" is not a word. You can talk about each vertex of a triangle. Vertices is the plural of vertex. It's not the usual way of forming a plural in English, because it isn't English--it's Latin. If you were to speak of multiple vertexes, however, I wouldn't complain. That's a legitimate alternative plural formation in English without the Latin pretentiousness.
  $endgroup$
  – David K
  23 hours ago
  
  

  
  
  
  

add a comment | 

10

$begingroup$

While working in GeoGebra I noticed something odd. I had a triangle with a point inside and the point was connected to each of the vertices. For each vertice I had drawn the circle passing through the vertice and the point, with the connection being the circle's diameter (see picture below).

What I noticed is that the overlapping circles completely covered the triangle. Further experimentation showed this was also the case if the point was outside the triangle (see below).

Still more experimentation appears to show this is the case for any polygon, simple or not:

Is this observation true or did GeoGebra lead me astray? I couldn't immediately find the result via Google.

geometry recreational-mathematics

share|cite|improve this question

asked yesterday

JensJens

4,02721032

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is another remarkable detail that the intersection point of any pair of circles distinct from the red point lies on the line connecting two vertices. And this fact is quite easy to prove.
  $endgroup$
  – user
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user I noticed that too. Could be a starting point for proving the above is true (if it is true).
  $endgroup$
  – Jens
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes this can be the starting point. In the case of internal point of the triangle it is practically evident. In the case of polygons it suffices to consider its convex hull.
  $endgroup$
  – user
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Given any polygon $mathcalP = ABCcdots F$ and a point $X$. Consider triangles formed by $X$ and an edge of $mathcalP$. Let's say triangle $XAB$. Let $Y$ be the foot of the altitude through $X$ on $AB$. Split $XAB$ into two right angled triangles $XAY$ and $XYB$. These two triangles will be covered by the two circles with $XA$ and $XB$ as diameters....
  $endgroup$
  – achille hui
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For future reference, "vertice" is not a word. You can talk about each vertex of a triangle. Vertices is the plural of vertex. It's not the usual way of forming a plural in English, because it isn't English--it's Latin. If you were to speak of multiple vertexes, however, I wouldn't complain. That's a legitimate alternative plural formation in English without the Latin pretentiousness.
  $endgroup$
  – David K
  23 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

While working in GeoGebra I noticed something odd. I had a triangle with a point inside and the point was connected to each of the vertices. For each vertice I had drawn the circle passing through the vertice and the point, with the connection being the circle's diameter (see picture below).

What I noticed is that the overlapping circles completely covered the triangle. Further experimentation showed this was also the case if the point was outside the triangle (see below).

Still more experimentation appears to show this is the case for any polygon, simple or not:

Is this observation true or did GeoGebra lead me astray? I couldn't immediately find the result via Google.

geometry recreational-mathematics

share|cite|improve this question

asked yesterday

JensJens

4,02721032

$endgroup$

share|cite|improve this question

asked yesterday

JensJens

4,02721032

share|cite|improve this question

asked yesterday

JensJens

4,02721032

asked yesterday

JensJens

4,02721032

asked yesterday

JensJens

4,02721032

2 Answers
2

active

oldest

votes

7

$begingroup$

We'll prove the case for $n=3$ and then generalize for an $n-$gon.

Let $triangle ABC$ be a triangle and $P$ an arbitraty point (inside or outside the triangle). Consider now the circumferences $omega$ and $tau$ with diameters $AP$ and $PB$ respectively. Consider furthermore the point $Din [AB]$, such that $PDbot AB$. In vitue of the converse of Thales' Theorem
$$angle ADP=90°implies Din omega qquadqquad angle PDB=90°implies Din tau$$ Analogously, we can prove that the intersections of the circles are $P$ and three points $D, Ein BC, Fin AC$ which lie on the sides of the triangles or on the extensions respectively.

Observe now that the triangles $triangle PBE, triangle PEC, triangle PCF, triangle PFA, triangle PDA$ and $triangle PBD$ are respectively inscribed in the circles $omega, tau$ and $rho$ (with the diameter $PC$). Thus, so is $triangle ABC$.

Now, once proven the case for a triangle, simply separate an $n-$gon into $(n-2)$ triangles, which will all be covered by the overlapping circles. Thus, the $n$-gon will also be covered $quadsquare$

---

For further reading: Episodes in nineteenth and twentieth century Euclidean Geometry (Honsberger), pages 79-86: Miquel's Theorem

share|cite|improve this answer

edited 19 hours ago

answered yesterday

Dr. MathvaDr. Mathva

2,857527

$endgroup$

add a comment | 

7

$begingroup$

The property you found boils down to the following: given a segment $AB$ and any point $P$ outside it, then the circles having $PB$ and $PA$ as diameter completely cover triangle $ABP$. And that is obvious, because those circles are the circumcircles of triangles $APH$ and $BPH$, where $H$ is the orthogonal projection of $P$ onto line $AB$.

share|cite|improve this answer

answered yesterday

AretinoAretino

25.4k21445

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add a comment | 

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7

$begingroup$

We'll prove the case for $n=3$ and then generalize for an $n-$gon.

Let $triangle ABC$ be a triangle and $P$ an arbitraty point (inside or outside the triangle). Consider now the circumferences $omega$ and $tau$ with diameters $AP$ and $PB$ respectively. Consider furthermore the point $Din [AB]$, such that $PDbot AB$. In vitue of the converse of Thales' Theorem
$$angle ADP=90°implies Din omega qquadqquad angle PDB=90°implies Din tau$$ Analogously, we can prove that the intersections of the circles are $P$ and three points $D, Ein BC, Fin AC$ which lie on the sides of the triangles or on the extensions respectively.

Observe now that the triangles $triangle PBE, triangle PEC, triangle PCF, triangle PFA, triangle PDA$ and $triangle PBD$ are respectively inscribed in the circles $omega, tau$ and $rho$ (with the diameter $PC$). Thus, so is $triangle ABC$.

Now, once proven the case for a triangle, simply separate an $n-$gon into $(n-2)$ triangles, which will all be covered by the overlapping circles. Thus, the $n$-gon will also be covered $quadsquare$

---

For further reading: Episodes in nineteenth and twentieth century Euclidean Geometry (Honsberger), pages 79-86: Miquel's Theorem

share|cite|improve this answer

edited 19 hours ago

answered yesterday

Dr. MathvaDr. Mathva

2,857527

$endgroup$

add a comment | 

7

$begingroup$

We'll prove the case for $n=3$ and then generalize for an $n-$gon.

Let $triangle ABC$ be a triangle and $P$ an arbitraty point (inside or outside the triangle). Consider now the circumferences $omega$ and $tau$ with diameters $AP$ and $PB$ respectively. Consider furthermore the point $Din [AB]$, such that $PDbot AB$. In vitue of the converse of Thales' Theorem
$$angle ADP=90°implies Din omega qquadqquad angle PDB=90°implies Din tau$$ Analogously, we can prove that the intersections of the circles are $P$ and three points $D, Ein BC, Fin AC$ which lie on the sides of the triangles or on the extensions respectively.

Observe now that the triangles $triangle PBE, triangle PEC, triangle PCF, triangle PFA, triangle PDA$ and $triangle PBD$ are respectively inscribed in the circles $omega, tau$ and $rho$ (with the diameter $PC$). Thus, so is $triangle ABC$.

Now, once proven the case for a triangle, simply separate an $n-$gon into $(n-2)$ triangles, which will all be covered by the overlapping circles. Thus, the $n$-gon will also be covered $quadsquare$

---

For further reading: Episodes in nineteenth and twentieth century Euclidean Geometry (Honsberger), pages 79-86: Miquel's Theorem

share|cite|improve this answer

edited 19 hours ago

answered yesterday

Dr. MathvaDr. Mathva

2,857527

$endgroup$

add a comment | 

$begingroup$

We'll prove the case for $n=3$ and then generalize for an $n-$gon.

Let $triangle ABC$ be a triangle and $P$ an arbitraty point (inside or outside the triangle). Consider now the circumferences $omega$ and $tau$ with diameters $AP$ and $PB$ respectively. Consider furthermore the point $Din [AB]$, such that $PDbot AB$. In vitue of the converse of Thales' Theorem
$$angle ADP=90°implies Din omega qquadqquad angle PDB=90°implies Din tau$$ Analogously, we can prove that the intersections of the circles are $P$ and three points $D, Ein BC, Fin AC$ which lie on the sides of the triangles or on the extensions respectively.

Observe now that the triangles $triangle PBE, triangle PEC, triangle PCF, triangle PFA, triangle PDA$ and $triangle PBD$ are respectively inscribed in the circles $omega, tau$ and $rho$ (with the diameter $PC$). Thus, so is $triangle ABC$.

Now, once proven the case for a triangle, simply separate an $n-$gon into $(n-2)$ triangles, which will all be covered by the overlapping circles. Thus, the $n$-gon will also be covered $quadsquare$

---

For further reading: Episodes in nineteenth and twentieth century Euclidean Geometry (Honsberger), pages 79-86: Miquel's Theorem

share|cite|improve this answer

edited 19 hours ago

answered yesterday

Dr. MathvaDr. Mathva

2,857527

$endgroup$

share|cite|improve this answer

edited 19 hours ago

answered yesterday

Dr. MathvaDr. Mathva

2,857527

answered yesterday

Dr. MathvaDr. Mathva

2,857527

answered yesterday

Dr. MathvaDr. Mathva

2,857527

7

$begingroup$

The property you found boils down to the following: given a segment $AB$ and any point $P$ outside it, then the circles having $PB$ and $PA$ as diameter completely cover triangle $ABP$. And that is obvious, because those circles are the circumcircles of triangles $APH$ and $BPH$, where $H$ is the orthogonal projection of $P$ onto line $AB$.

share|cite|improve this answer

answered yesterday

AretinoAretino

25.4k21445

$endgroup$

add a comment | 

7

$begingroup$

The property you found boils down to the following: given a segment $AB$ and any point $P$ outside it, then the circles having $PB$ and $PA$ as diameter completely cover triangle $ABP$. And that is obvious, because those circles are the circumcircles of triangles $APH$ and $BPH$, where $H$ is the orthogonal projection of $P$ onto line $AB$.

share|cite|improve this answer

answered yesterday

AretinoAretino

25.4k21445

$endgroup$

add a comment | 

$begingroup$

The property you found boils down to the following: given a segment $AB$ and any point $P$ outside it, then the circles having $PB$ and $PA$ as diameter completely cover triangle $ABP$. And that is obvious, because those circles are the circumcircles of triangles $APH$ and $BPH$, where $H$ is the orthogonal projection of $P$ onto line $AB$.

share|cite|improve this answer

answered yesterday

AretinoAretino

25.4k21445

$endgroup$

share|cite|improve this answer

answered yesterday

AretinoAretino

25.4k21445

answered yesterday

AretinoAretino

25.4k21445

answered yesterday

AretinoAretino

25.4k21445