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?
3 Answers
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What if we have one chord partition lenghts is it possible then?with two chords partition lenght it's solvable? – Mohamadreza Abasian Jun 10 '19 at 21:11
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!
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How can I calculate radius with two chords crossed on circle? – Mohamadreza Abasian Jun 10 '19 at 21:37
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Taking the bisectors of the two chords, its intersect in the center of the circle. – Piquito Jun 10 '19 at 22:46
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.
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