Bidirectionally of the “Tangent Criterion”
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I've recently been reviewing some basic geometry concepts when I saw this one in Evan Chen's fantastic "Euclidean Geometry in Mathematical Olympiads" (EGMO).
Proving $(i)Rightarrow (iii)$ is quite simple.
Hint: Move point $C$ in the circumcircle so that $angle BAC=90°$
Nevertheless, I've had some issues trying to prove that this proposition is biconditional, i.e. $(i)iff (iii)$.
Since one of the directions is already proven, I only need to show $(iii)Rightarrow (i)$
What would you suggest?
euclidean-geometry triangle circle angle tangent-line
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
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add a comment |
$begingroup$
I've recently been reviewing some basic geometry concepts when I saw this one in Evan Chen's fantastic "Euclidean Geometry in Mathematical Olympiads" (EGMO).
Proving $(i)Rightarrow (iii)$ is quite simple.
Hint: Move point $C$ in the circumcircle so that $angle BAC=90°$
Nevertheless, I've had some issues trying to prove that this proposition is biconditional, i.e. $(i)iff (iii)$.
Since one of the directions is already proven, I only need to show $(iii)Rightarrow (i)$
What would you suggest?
euclidean-geometry triangle circle angle tangent-line
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
$endgroup$
add a comment |
$begingroup$
I've recently been reviewing some basic geometry concepts when I saw this one in Evan Chen's fantastic "Euclidean Geometry in Mathematical Olympiads" (EGMO).
Proving $(i)Rightarrow (iii)$ is quite simple.
Hint: Move point $C$ in the circumcircle so that $angle BAC=90°$
Nevertheless, I've had some issues trying to prove that this proposition is biconditional, i.e. $(i)iff (iii)$.
Since one of the directions is already proven, I only need to show $(iii)Rightarrow (i)$
What would you suggest?
euclidean-geometry triangle circle angle tangent-line
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
$endgroup$
I've recently been reviewing some basic geometry concepts when I saw this one in Evan Chen's fantastic "Euclidean Geometry in Mathematical Olympiads" (EGMO).
Proving $(i)Rightarrow (iii)$ is quite simple.
Hint: Move point $C$ in the circumcircle so that $angle BAC=90°$
Nevertheless, I've had some issues trying to prove that this proposition is biconditional, i.e. $(i)iff (iii)$.
Since one of the directions is already proven, I only need to show $(iii)Rightarrow (i)$
What would you suggest?
euclidean-geometry triangle circle angle tangent-line
euclidean-geometry triangle circle angle tangent-line
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
asked Dec 6 '18 at 18:11
Dr. MathvaDr. Mathva
1,098317
1,098317
add a comment |
add a comment |
1 Answer
1
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Unsurprisingly, the solution is
Move point $C$ in the circumcircle so that $angle ABC=90°$.
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
$endgroup$
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I can't believe I was so stupid I didn't see it! +1
$endgroup$
– Dr. Mathva
Dec 6 '18 at 19:16
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Well, you did all the work :D
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– Federico
Dec 6 '18 at 19:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Unsurprisingly, the solution is
Move point $C$ in the circumcircle so that $angle ABC=90°$.
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
$endgroup$
$begingroup$
I can't believe I was so stupid I didn't see it! +1
$endgroup$
– Dr. Mathva
Dec 6 '18 at 19:16
$begingroup$
Well, you did all the work :D
$endgroup$
– Federico
Dec 6 '18 at 19:28
add a comment |
$begingroup$
Unsurprisingly, the solution is
Move point $C$ in the circumcircle so that $angle ABC=90°$.
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
$endgroup$
$begingroup$
I can't believe I was so stupid I didn't see it! +1
$endgroup$
– Dr. Mathva
Dec 6 '18 at 19:16
$begingroup$
Well, you did all the work :D
$endgroup$
– Federico
Dec 6 '18 at 19:28
add a comment |
$begingroup$
Unsurprisingly, the solution is
Move point $C$ in the circumcircle so that $angle ABC=90°$.
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
$endgroup$
Unsurprisingly, the solution is
Move point $C$ in the circumcircle so that $angle ABC=90°$.
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
answered Dec 6 '18 at 18:16
FedericoFederico
5,029514
5,029514
$begingroup$
I can't believe I was so stupid I didn't see it! +1
$endgroup$
– Dr. Mathva
Dec 6 '18 at 19:16
$begingroup$
Well, you did all the work :D
$endgroup$
– Federico
Dec 6 '18 at 19:28
add a comment |
$begingroup$
I can't believe I was so stupid I didn't see it! +1
$endgroup$
– Dr. Mathva
Dec 6 '18 at 19:16
$begingroup$
Well, you did all the work :D
$endgroup$
– Federico
Dec 6 '18 at 19:28
$begingroup$
I can't believe I was so stupid I didn't see it! +1
$endgroup$
– Dr. Mathva
Dec 6 '18 at 19:16
$begingroup$
I can't believe I was so stupid I didn't see it! +1
$endgroup$
– Dr. Mathva
Dec 6 '18 at 19:16
$begingroup$
Well, you did all the work :D
$endgroup$
– Federico
Dec 6 '18 at 19:28
$begingroup$
Well, you did all the work :D
$endgroup$
– Federico
Dec 6 '18 at 19:28
add a comment |
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