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Finding diameter of a circle using two chords and angle between them

Find the length of the chord and the distance between parallel chords, given their anglesFinding the angle between the $2$ radii of a circleCenter of a circle from two chords.How to calculate clock-wise and anti-clockwise arc lengths between two points on a circleCircle centre point from two angles and circle overalLength of chord and sector of a circleHow to prove angle bisector $angle A$, is also angle bisector between height $AH$ and diameter $AM$ in circumscribed circle of the triangleFinding the diameter of a circle from the length and position of a chordInternal point in circle and chordsMaximum number of common chords that are existent between two Conics

4

$begingroup$

Is it enough to find diameter of a circle using two arbitrary
crossover chords with known length of each partition and angle between this two chords? If it's possible, how?

circles

share|cite|improve this question

edited 8 hours ago

Mohamad Abasi

asked 8 hours ago

Mohamad AbasiMohamad Abasi

355

$endgroup$

add a comment | 

4

$begingroup$

Is it enough to find diameter of a circle using two arbitrary
crossover chords with known length of each partition and angle between this two chords? If it's possible, how?

circles

share|cite|improve this question

edited 8 hours ago

Mohamad Abasi

asked 8 hours ago

Mohamad AbasiMohamad Abasi

355

$endgroup$

add a comment | 

$begingroup$

Is it enough to find diameter of a circle using two arbitrary
crossover chords with known length of each partition and angle between this two chords? If it's possible, how?

circles

share|cite|improve this question

edited 8 hours ago

Mohamad Abasi

asked 8 hours ago

Mohamad AbasiMohamad Abasi

355

$endgroup$

share|cite|improve this question

edited 8 hours ago

Mohamad Abasi

asked 8 hours ago

Mohamad AbasiMohamad Abasi

355

share|cite|improve this question

edited 8 hours ago

Mohamad Abasi

asked 8 hours ago

Mohamad AbasiMohamad Abasi

355

asked 8 hours ago

Mohamad AbasiMohamad Abasi

355

asked 8 hours ago

Mohamad AbasiMohamad Abasi

355

3 Answers
3

active

oldest

votes

4

$begingroup$

If you do not know where they cross (i.e., in what proportions they partition each other), then no.

share|cite|improve this answer

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

289k23275510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if we have one chord partition lenghts is it possible then?with two chords partition lenght it's solvable?
  $endgroup$
  – Mohamad Abasi
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

If you know each of the partition sizes then it is solvable. I will first provide the outline for the proof, then make edits if you want to see the full proof.

Part I. consider first two chords only meet on the circle. You know length of both and the angle between them. calculating the radius is a straightforward exercise.

Part II. Prove that by one extra step of calculation, arbitrarily crossed chords with all four known partitions are equivalent to the previous case so it is possible to calculate it.

If you only know partitions on one of the two chords, you can't calculate the radius.

---

Proof:
I will use the following lemmas:

Lemma 1: If two angles and their common sides in a triangle is known, the triangle is completely known.

Lemma 2: If two sides and their common angle in a triangle is known, the triangle is completely known.

Part I:

By Lemma 2, the triangle ABC is completely known. We know $x = angle ACB$, $z = angle BAC$, $w = angle ABC$ as well as $a,b$ and $AB$. Since $OA=OB$, $y$ is denoted. By Lemma 1, if we know $y$ we know triangle $AOB$ completely. i.e. we know the radius.

for value of $y$ we joint $OC$. note $OB=OC=OA$. So
$$z+y = angle ACO = x-angle OCB = x- (w+y)$$

$z, x, w$ are all known so $y$ is solved. Done for part I.

---

Part II:

Using the result of Part I, we will be done if we know $AB$ and $angle ABD$ (equivalently, $angle ABE$).

These two things are obtained immediately, as we restrict our attention to triangle $ABE$. I claim that this triangle is completely known by Lemma 2, since we know $AE$, $BE$ as well as the angle $angle AEB$. Done for Part II.

Fun problem!

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Book Book BookBook Book Book

3417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can I calculate radius with two chords crossed on circle?
  $endgroup$
  – Mohamad Abasi
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Taking the bisectors of the two chords, its intersect in the center of the circle.
  $endgroup$
  – Piquito
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Not because a single chord and a segment of the other, determine a triangle which determines a unique circumcircle in which the $CD$ segment should be a chord in the attached figure.

share|cite|improve this answer

answered 7 hours ago

PiquitoPiquito

18.4k31640

$endgroup$

add a comment | 

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4

$begingroup$

If you do not know where they cross (i.e., in what proportions they partition each other), then no.

share|cite|improve this answer

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

289k23275510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if we have one chord partition lenghts is it possible then?with two chords partition lenght it's solvable?
  $endgroup$
  – Mohamad Abasi
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

If you do not know where they cross (i.e., in what proportions they partition each other), then no.

share|cite|improve this answer

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

289k23275510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if we have one chord partition lenghts is it possible then?with two chords partition lenght it's solvable?
  $endgroup$
  – Mohamad Abasi
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

If you do not know where they cross (i.e., in what proportions they partition each other), then no.

share|cite|improve this answer

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

289k23275510

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

289k23275510

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

289k23275510

answered 8 hours ago

Hagen von EitzenHagen von Eitzen

289k23275510

2

$begingroup$

If you know each of the partition sizes then it is solvable. I will first provide the outline for the proof, then make edits if you want to see the full proof.

Part I. consider first two chords only meet on the circle. You know length of both and the angle between them. calculating the radius is a straightforward exercise.

Part II. Prove that by one extra step of calculation, arbitrarily crossed chords with all four known partitions are equivalent to the previous case so it is possible to calculate it.

If you only know partitions on one of the two chords, you can't calculate the radius.

---

Proof:
I will use the following lemmas:

Lemma 1: If two angles and their common sides in a triangle is known, the triangle is completely known.

Lemma 2: If two sides and their common angle in a triangle is known, the triangle is completely known.

Part I:

By Lemma 2, the triangle ABC is completely known. We know $x = angle ACB$, $z = angle BAC$, $w = angle ABC$ as well as $a,b$ and $AB$. Since $OA=OB$, $y$ is denoted. By Lemma 1, if we know $y$ we know triangle $AOB$ completely. i.e. we know the radius.

for value of $y$ we joint $OC$. note $OB=OC=OA$. So
$$z+y = angle ACO = x-angle OCB = x- (w+y)$$

$z, x, w$ are all known so $y$ is solved. Done for part I.

---

Part II:

Using the result of Part I, we will be done if we know $AB$ and $angle ABD$ (equivalently, $angle ABE$).

These two things are obtained immediately, as we restrict our attention to triangle $ABE$. I claim that this triangle is completely known by Lemma 2, since we know $AE$, $BE$ as well as the angle $angle AEB$. Done for Part II.

Fun problem!

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Book Book BookBook Book Book

3417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can I calculate radius with two chords crossed on circle?
  $endgroup$
  – Mohamad Abasi
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Taking the bisectors of the two chords, its intersect in the center of the circle.
  $endgroup$
  – Piquito
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

If you know each of the partition sizes then it is solvable. I will first provide the outline for the proof, then make edits if you want to see the full proof.

Part I. consider first two chords only meet on the circle. You know length of both and the angle between them. calculating the radius is a straightforward exercise.

Part II. Prove that by one extra step of calculation, arbitrarily crossed chords with all four known partitions are equivalent to the previous case so it is possible to calculate it.

If you only know partitions on one of the two chords, you can't calculate the radius.

---

Proof:
I will use the following lemmas:

Lemma 1: If two angles and their common sides in a triangle is known, the triangle is completely known.

Lemma 2: If two sides and their common angle in a triangle is known, the triangle is completely known.

Part I:

By Lemma 2, the triangle ABC is completely known. We know $x = angle ACB$, $z = angle BAC$, $w = angle ABC$ as well as $a,b$ and $AB$. Since $OA=OB$, $y$ is denoted. By Lemma 1, if we know $y$ we know triangle $AOB$ completely. i.e. we know the radius.

for value of $y$ we joint $OC$. note $OB=OC=OA$. So
$$z+y = angle ACO = x-angle OCB = x- (w+y)$$

$z, x, w$ are all known so $y$ is solved. Done for part I.

---

Part II:

Using the result of Part I, we will be done if we know $AB$ and $angle ABD$ (equivalently, $angle ABE$).

These two things are obtained immediately, as we restrict our attention to triangle $ABE$. I claim that this triangle is completely known by Lemma 2, since we know $AE$, $BE$ as well as the angle $angle AEB$. Done for Part II.

Fun problem!

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Book Book BookBook Book Book

3417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How can I calculate radius with two chords crossed on circle?
  $endgroup$
  – Mohamad Abasi
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Taking the bisectors of the two chords, its intersect in the center of the circle.
  $endgroup$
  – Piquito
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If you know each of the partition sizes then it is solvable. I will first provide the outline for the proof, then make edits if you want to see the full proof.

Part I. consider first two chords only meet on the circle. You know length of both and the angle between them. calculating the radius is a straightforward exercise.

Part II. Prove that by one extra step of calculation, arbitrarily crossed chords with all four known partitions are equivalent to the previous case so it is possible to calculate it.

If you only know partitions on one of the two chords, you can't calculate the radius.

---

Proof:
I will use the following lemmas:

Lemma 1: If two angles and their common sides in a triangle is known, the triangle is completely known.

Lemma 2: If two sides and their common angle in a triangle is known, the triangle is completely known.

Part I:

By Lemma 2, the triangle ABC is completely known. We know $x = angle ACB$, $z = angle BAC$, $w = angle ABC$ as well as $a,b$ and $AB$. Since $OA=OB$, $y$ is denoted. By Lemma 1, if we know $y$ we know triangle $AOB$ completely. i.e. we know the radius.

for value of $y$ we joint $OC$. note $OB=OC=OA$. So
$$z+y = angle ACO = x-angle OCB = x- (w+y)$$

$z, x, w$ are all known so $y$ is solved. Done for part I.

---

Part II:

Using the result of Part I, we will be done if we know $AB$ and $angle ABD$ (equivalently, $angle ABE$).

These two things are obtained immediately, as we restrict our attention to triangle $ABE$. I claim that this triangle is completely known by Lemma 2, since we know $AE$, $BE$ as well as the angle $angle AEB$. Done for Part II.

Fun problem!

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Book Book BookBook Book Book

3417

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Book Book BookBook Book Book

3417

answered 8 hours ago

Book Book BookBook Book Book

3417

answered 8 hours ago

Book Book BookBook Book Book

3417

1

$begingroup$

Not because a single chord and a segment of the other, determine a triangle which determines a unique circumcircle in which the $CD$ segment should be a chord in the attached figure.

share|cite|improve this answer

answered 7 hours ago

PiquitoPiquito

18.4k31640

$endgroup$

add a comment | 

1

$begingroup$

Not because a single chord and a segment of the other, determine a triangle which determines a unique circumcircle in which the $CD$ segment should be a chord in the attached figure.

share|cite|improve this answer

answered 7 hours ago

PiquitoPiquito

18.4k31640

$endgroup$

add a comment | 

$begingroup$

Not because a single chord and a segment of the other, determine a triangle which determines a unique circumcircle in which the $CD$ segment should be a chord in the attached figure.

share|cite|improve this answer

answered 7 hours ago

PiquitoPiquito

18.4k31640

$endgroup$

share|cite|improve this answer

answered 7 hours ago

PiquitoPiquito

18.4k31640

answered 7 hours ago

PiquitoPiquito

18.4k31640

answered 7 hours ago

PiquitoPiquito

18.4k31640