Please help with Olympiad geometry question The 2019 Stack Overflow Developer Survey Results Are InSimilarity involving Miquel's TheoremMaximal area of triangle if angle and opposite side length is knownCompetition style problem circa 1992Given three concurrent lines $a,b$ and $c$, find the circunference tangent to $a$ and $b$ and with center at $c$Problem of axiomatic euclidean geometryConstruct a parallelogram subject to certain conditionsTwo Equilateral Triangles and the Golden Ratio: Simple Geomtric ProofAn interesting geometry problem with angle bisectors and tangentgeometry/combinatorics question: max # intersections of lines in a triangleRight triangle minimum area problem without calculus
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Please help with Olympiad geometry question
The 2019 Stack Overflow Developer Survey Results Are InSimilarity involving Miquel's TheoremMaximal area of triangle if angle and opposite side length is knownCompetition style problem circa 1992Given three concurrent lines $a,b$ and $c$, find the circunference tangent to $a$ and $b$ and with center at $c$Problem of axiomatic euclidean geometryConstruct a parallelogram subject to certain conditionsTwo Equilateral Triangles and the Golden Ratio: Simple Geomtric ProofAn interesting geometry problem with angle bisectors and tangentgeometry/combinatorics question: max # intersections of lines in a triangleRight triangle minimum area problem without calculus
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Please help me solve this problem.
Problem 2.27 (BAMO 2012/4). Given a segment AB in the plane, choose on it a point M different from A and B. Two equilateral triangles AMC and BMD in the plane are constructed on the same side of segment AB . The circumcircles of the two triangles intersect in point M and another point N.
(a) Prove that AD and BC pass through point N.
(b) Prove that no matter where one chooses the point M along segment AB, all lines MN
will pass through some fixed point K in the plane.
I have problem in part b. I don't want to use coordinates. Can someone give a hint?
euclidean-geometry
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
$endgroup$
add a comment |
$begingroup$
Please help me solve this problem.
Problem 2.27 (BAMO 2012/4). Given a segment AB in the plane, choose on it a point M different from A and B. Two equilateral triangles AMC and BMD in the plane are constructed on the same side of segment AB . The circumcircles of the two triangles intersect in point M and another point N.
(a) Prove that AD and BC pass through point N.
(b) Prove that no matter where one chooses the point M along segment AB, all lines MN
will pass through some fixed point K in the plane.
I have problem in part b. I don't want to use coordinates. Can someone give a hint?
euclidean-geometry
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
$endgroup$
$begingroup$
Well, typing "BAMO $2012/4$" in Google, open the first archive. Hint: Have a look at the sixth problem ;)
$endgroup$
– Dr. Mathva
Apr 1 at 20:32
$begingroup$
@Dr.Mathva thank you for your advice.
$endgroup$
– Shashwat Asthana
Apr 1 at 23:26
add a comment |
$begingroup$
Please help me solve this problem.
Problem 2.27 (BAMO 2012/4). Given a segment AB in the plane, choose on it a point M different from A and B. Two equilateral triangles AMC and BMD in the plane are constructed on the same side of segment AB . The circumcircles of the two triangles intersect in point M and another point N.
(a) Prove that AD and BC pass through point N.
(b) Prove that no matter where one chooses the point M along segment AB, all lines MN
will pass through some fixed point K in the plane.
I have problem in part b. I don't want to use coordinates. Can someone give a hint?
euclidean-geometry
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
$endgroup$
Please help me solve this problem.
Problem 2.27 (BAMO 2012/4). Given a segment AB in the plane, choose on it a point M different from A and B. Two equilateral triangles AMC and BMD in the plane are constructed on the same side of segment AB . The circumcircles of the two triangles intersect in point M and another point N.
(a) Prove that AD and BC pass through point N.
(b) Prove that no matter where one chooses the point M along segment AB, all lines MN
will pass through some fixed point K in the plane.
I have problem in part b. I don't want to use coordinates. Can someone give a hint?
euclidean-geometry
euclidean-geometry
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
asked Mar 30 at 17:33
Shashwat AsthanaShashwat Asthana
1098
1098
$begingroup$
Well, typing "BAMO $2012/4$" in Google, open the first archive. Hint: Have a look at the sixth problem ;)
$endgroup$
– Dr. Mathva
Apr 1 at 20:32
$begingroup$
@Dr.Mathva thank you for your advice.
$endgroup$
– Shashwat Asthana
Apr 1 at 23:26
add a comment |
$begingroup$
Well, typing "BAMO $2012/4$" in Google, open the first archive. Hint: Have a look at the sixth problem ;)
$endgroup$
– Dr. Mathva
Apr 1 at 20:32
$begingroup$
@Dr.Mathva thank you for your advice.
$endgroup$
– Shashwat Asthana
Apr 1 at 23:26
$begingroup$
Well, typing "BAMO $2012/4$" in Google, open the first archive. Hint: Have a look at the sixth problem ;)
$endgroup$
– Dr. Mathva
Apr 1 at 20:32
$begingroup$
Well, typing "BAMO $2012/4$" in Google, open the first archive. Hint: Have a look at the sixth problem ;)
$endgroup$
– Dr. Mathva
Apr 1 at 20:32
$begingroup$
@Dr.Mathva thank you for your advice.
$endgroup$
– Shashwat Asthana
Apr 1 at 23:26
$begingroup$
@Dr.Mathva thank you for your advice.
$endgroup$
– Shashwat Asthana
Apr 1 at 23:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Part b: Draw tangents to circumscribed circles at points $A$ and $B$. These tangents interesect at some point $K$.
It's a known fact that the angle between a tangent and a chord is equal to the corresponding inscribed angle. Because of that:
$$angle KAB=angle KBA=60^circtag1$$
It follows that triangle $KAB$ is equilateral so:
$$KA=KBtag2$$
Therefore point $K$ has the same power with respect to both circumscribed circles. And because of that, it has to be on the radical axis $MN$. So all lines $MN$ pass through fixed point $K$ defined by (1) and (2).
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Part b: Draw tangents to circumscribed circles at points $A$ and $B$. These tangents interesect at some point $K$.
It's a known fact that the angle between a tangent and a chord is equal to the corresponding inscribed angle. Because of that:
$$angle KAB=angle KBA=60^circtag1$$
It follows that triangle $KAB$ is equilateral so:
$$KA=KBtag2$$
Therefore point $K$ has the same power with respect to both circumscribed circles. And because of that, it has to be on the radical axis $MN$. So all lines $MN$ pass through fixed point $K$ defined by (1) and (2).
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
$endgroup$
add a comment |
$begingroup$
Part b: Draw tangents to circumscribed circles at points $A$ and $B$. These tangents interesect at some point $K$.
It's a known fact that the angle between a tangent and a chord is equal to the corresponding inscribed angle. Because of that:
$$angle KAB=angle KBA=60^circtag1$$
It follows that triangle $KAB$ is equilateral so:
$$KA=KBtag2$$
Therefore point $K$ has the same power with respect to both circumscribed circles. And because of that, it has to be on the radical axis $MN$. So all lines $MN$ pass through fixed point $K$ defined by (1) and (2).
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
$endgroup$
add a comment |
$begingroup$
Part b: Draw tangents to circumscribed circles at points $A$ and $B$. These tangents interesect at some point $K$.
It's a known fact that the angle between a tangent and a chord is equal to the corresponding inscribed angle. Because of that:
$$angle KAB=angle KBA=60^circtag1$$
It follows that triangle $KAB$ is equilateral so:
$$KA=KBtag2$$
Therefore point $K$ has the same power with respect to both circumscribed circles. And because of that, it has to be on the radical axis $MN$. So all lines $MN$ pass through fixed point $K$ defined by (1) and (2).
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
$endgroup$
Part b: Draw tangents to circumscribed circles at points $A$ and $B$. These tangents interesect at some point $K$.
It's a known fact that the angle between a tangent and a chord is equal to the corresponding inscribed angle. Because of that:
$$angle KAB=angle KBA=60^circtag1$$
It follows that triangle $KAB$ is equilateral so:
$$KA=KBtag2$$
Therefore point $K$ has the same power with respect to both circumscribed circles. And because of that, it has to be on the radical axis $MN$. So all lines $MN$ pass through fixed point $K$ defined by (1) and (2).
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
edited Mar 30 at 19:03
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
answered Mar 30 at 18:55
OldboyOldboy
9,40711138
9,40711138
add a comment |
add a comment |
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$begingroup$
Well, typing "BAMO $2012/4$" in Google, open the first archive. Hint: Have a look at the sixth problem ;)
$endgroup$
– Dr. Mathva
Apr 1 at 20:32
$begingroup$
@Dr.Mathva thank you for your advice.
$endgroup$
– Shashwat Asthana
Apr 1 at 23:26