Show that $MNPQ$ is a square
$begingroup$
Let $ ABCD $ a quadrilateral s.t. $AC=BD $ and $m (angle AOD)=30°$ where $O=ACcap BD $.
Let $triangle ABM, triangle DCN, triangle ADN, triangle CBQ $ equilateral triangles with $Int (triangle ABM)cap Int (ABCD)=emptyset$, $Int (triangle DCP)cap Int (ABCD)=emptyset$, $Int (triangle ADN)cap Int (ABCD) neq emptyset$, $Int (triangle CBQ)cap Int (ABCD)neqemptyset$.
Show that $MNPQ$ is a square.
I have no idea how to start.
geometry complex-numbers vectors euclidean-geometry geometric-transformation
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
$endgroup$
add a comment |
$begingroup$
Let $ ABCD $ a quadrilateral s.t. $AC=BD $ and $m (angle AOD)=30°$ where $O=ACcap BD $.
Let $triangle ABM, triangle DCN, triangle ADN, triangle CBQ $ equilateral triangles with $Int (triangle ABM)cap Int (ABCD)=emptyset$, $Int (triangle DCP)cap Int (ABCD)=emptyset$, $Int (triangle ADN)cap Int (ABCD) neq emptyset$, $Int (triangle CBQ)cap Int (ABCD)neqemptyset$.
Show that $MNPQ$ is a square.
I have no idea how to start.
geometry complex-numbers vectors euclidean-geometry geometric-transformation
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
$endgroup$
$begingroup$
Just drawn a picture and it didn't look like a square to me.
$endgroup$
– Quang Hoang
Jan 10 at 21:02
1
$begingroup$
Edit: I think I see the problem here. All equilateral triangles should be drawn in the opposite direction and the statement would be true. I still don't know how to approach though...
$endgroup$
– Edward H.
Jan 10 at 21:23
$begingroup$
Ok so here's an idea assuming that the equilateral triangles are meant to be drawn oppositely: Rotate ΔABC by 60° around B and you'll get ΔMBQ. Do this 4 times and use AC=BD, and you'll find that MNPQ is a rhombus. Moreover if you keep track of all the angles of rotation and use ∠AOD = 30°, you'll find that the angles of the rhombus are all right, and hence it is a square.
$endgroup$
– Edward H.
Jan 10 at 21:49
$begingroup$
As given it's clearly false. $mangle AMB =60$ and $mangle MBA = 60$ and $mangle QBC = 60$ so $Q$ is in the interior of $angle AMB$. As is $N$ so $mangle NMQ < mangle AMB = 60 < 90$. Can't be a square. But if all the intersections are empty so the equi triangles pop out.
$endgroup$
– fleablood
Jan 10 at 22:29
add a comment |
$begingroup$
Let $ ABCD $ a quadrilateral s.t. $AC=BD $ and $m (angle AOD)=30°$ where $O=ACcap BD $.
Let $triangle ABM, triangle DCN, triangle ADN, triangle CBQ $ equilateral triangles with $Int (triangle ABM)cap Int (ABCD)=emptyset$, $Int (triangle DCP)cap Int (ABCD)=emptyset$, $Int (triangle ADN)cap Int (ABCD) neq emptyset$, $Int (triangle CBQ)cap Int (ABCD)neqemptyset$.
Show that $MNPQ$ is a square.
I have no idea how to start.
geometry complex-numbers vectors euclidean-geometry geometric-transformation
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
$endgroup$
Let $ ABCD $ a quadrilateral s.t. $AC=BD $ and $m (angle AOD)=30°$ where $O=ACcap BD $.
Let $triangle ABM, triangle DCN, triangle ADN, triangle CBQ $ equilateral triangles with $Int (triangle ABM)cap Int (ABCD)=emptyset$, $Int (triangle DCP)cap Int (ABCD)=emptyset$, $Int (triangle ADN)cap Int (ABCD) neq emptyset$, $Int (triangle CBQ)cap Int (ABCD)neqemptyset$.
Show that $MNPQ$ is a square.
I have no idea how to start.
geometry complex-numbers vectors euclidean-geometry geometric-transformation
geometry complex-numbers vectors euclidean-geometry geometric-transformation
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
edited Jan 15 at 17:30
Maria Mazur
49.9k1361125
49.9k1361125
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
asked Jan 10 at 20:51
ProblemsolvingProblemsolving
917412
917412
$begingroup$
Just drawn a picture and it didn't look like a square to me.
$endgroup$
– Quang Hoang
Jan 10 at 21:02
1
$begingroup$
Edit: I think I see the problem here. All equilateral triangles should be drawn in the opposite direction and the statement would be true. I still don't know how to approach though...
$endgroup$
– Edward H.
Jan 10 at 21:23
$begingroup$
Ok so here's an idea assuming that the equilateral triangles are meant to be drawn oppositely: Rotate ΔABC by 60° around B and you'll get ΔMBQ. Do this 4 times and use AC=BD, and you'll find that MNPQ is a rhombus. Moreover if you keep track of all the angles of rotation and use ∠AOD = 30°, you'll find that the angles of the rhombus are all right, and hence it is a square.
$endgroup$
– Edward H.
Jan 10 at 21:49
$begingroup$
As given it's clearly false. $mangle AMB =60$ and $mangle MBA = 60$ and $mangle QBC = 60$ so $Q$ is in the interior of $angle AMB$. As is $N$ so $mangle NMQ < mangle AMB = 60 < 90$. Can't be a square. But if all the intersections are empty so the equi triangles pop out.
$endgroup$
– fleablood
Jan 10 at 22:29
add a comment |
$begingroup$
Just drawn a picture and it didn't look like a square to me.
$endgroup$
– Quang Hoang
Jan 10 at 21:02
1
$begingroup$
Edit: I think I see the problem here. All equilateral triangles should be drawn in the opposite direction and the statement would be true. I still don't know how to approach though...
$endgroup$
– Edward H.
Jan 10 at 21:23
$begingroup$
Ok so here's an idea assuming that the equilateral triangles are meant to be drawn oppositely: Rotate ΔABC by 60° around B and you'll get ΔMBQ. Do this 4 times and use AC=BD, and you'll find that MNPQ is a rhombus. Moreover if you keep track of all the angles of rotation and use ∠AOD = 30°, you'll find that the angles of the rhombus are all right, and hence it is a square.
$endgroup$
– Edward H.
Jan 10 at 21:49
$begingroup$
As given it's clearly false. $mangle AMB =60$ and $mangle MBA = 60$ and $mangle QBC = 60$ so $Q$ is in the interior of $angle AMB$. As is $N$ so $mangle NMQ < mangle AMB = 60 < 90$. Can't be a square. But if all the intersections are empty so the equi triangles pop out.
$endgroup$
– fleablood
Jan 10 at 22:29
$begingroup$
Just drawn a picture and it didn't look like a square to me.
$endgroup$
– Quang Hoang
Jan 10 at 21:02
$begingroup$
Just drawn a picture and it didn't look like a square to me.
$endgroup$
– Quang Hoang
Jan 10 at 21:02
1
1
$begingroup$
Edit: I think I see the problem here. All equilateral triangles should be drawn in the opposite direction and the statement would be true. I still don't know how to approach though...
$endgroup$
– Edward H.
Jan 10 at 21:23
$begingroup$
Edit: I think I see the problem here. All equilateral triangles should be drawn in the opposite direction and the statement would be true. I still don't know how to approach though...
$endgroup$
– Edward H.
Jan 10 at 21:23
$begingroup$
Ok so here's an idea assuming that the equilateral triangles are meant to be drawn oppositely: Rotate ΔABC by 60° around B and you'll get ΔMBQ. Do this 4 times and use AC=BD, and you'll find that MNPQ is a rhombus. Moreover if you keep track of all the angles of rotation and use ∠AOD = 30°, you'll find that the angles of the rhombus are all right, and hence it is a square.
$endgroup$
– Edward H.
Jan 10 at 21:49
$begingroup$
Ok so here's an idea assuming that the equilateral triangles are meant to be drawn oppositely: Rotate ΔABC by 60° around B and you'll get ΔMBQ. Do this 4 times and use AC=BD, and you'll find that MNPQ is a rhombus. Moreover if you keep track of all the angles of rotation and use ∠AOD = 30°, you'll find that the angles of the rhombus are all right, and hence it is a square.
$endgroup$
– Edward H.
Jan 10 at 21:49
$begingroup$
As given it's clearly false. $mangle AMB =60$ and $mangle MBA = 60$ and $mangle QBC = 60$ so $Q$ is in the interior of $angle AMB$. As is $N$ so $mangle NMQ < mangle AMB = 60 < 90$. Can't be a square. But if all the intersections are empty so the equi triangles pop out.
$endgroup$
– fleablood
Jan 10 at 22:29
$begingroup$
As given it's clearly false. $mangle AMB =60$ and $mangle MBA = 60$ and $mangle QBC = 60$ so $Q$ is in the interior of $angle AMB$. As is $N$ so $mangle NMQ < mangle AMB = 60 < 90$. Can't be a square. But if all the intersections are empty so the equi triangles pop out.
$endgroup$
– fleablood
Jan 10 at 22:29
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $R^{alpha}_O$ be a rotation in the plain by an angle $alpha$ around a point $O$.
Easy to see that to rotate a vector by an angle $alpha$ it's the same to rotate this vector around his tail.
Now, by using the beautiful Daniel Mathias's picture we obtain:
$$R^{90^{circ}}left(vec{NM}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{NA}+vec{AM}right)right)=R^{30^{circ}}left(vec{DA}+vec{AB}right)=$$
$$=R^{30^{circ}}left(vec{DB}right)=R^{60^{circ}}left(vec{AC}right)=R^{60^{circ}}left(vec{AB}+vec{BC}right)=vec{MB}+vec{BQ}=vec{MQ},$$
which says $NMperp MQ$ and $NM=MQ.$
Also,
$$R^{90^{circ}}left(vec{QP}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{QC}+vec{CP}right)right)=R^{30^{circ}}left(vec{BC}+vec{CD}right)=$$
$$=R^{30^{circ}}left(vec{BD}right)=R^{60^{circ}}left(vec{CA}right)=R^{60^{circ}}left(vec{CB}+vec{BA}right)=vec{QB}+vec{BM}=vec{QM},$$
which says $QMperp PQ$ and $QM=PQ.$
Can you end it now?
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
There is a synthetic proof? I try to prove it with congruences of triangles.
$endgroup$
– Problemsolving
Jan 11 at 5:44
1
$begingroup$
I like your solution +1, do you like mine?
$endgroup$
– Maria Mazur
Jan 14 at 20:42
$begingroup$
Do you have some more problems like this solved vith rotation?
$endgroup$
– Maria Mazur
Jan 19 at 11:14
$begingroup$
@greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152
$endgroup$
– Michael Rozenberg
Jan 19 at 11:17
$begingroup$
Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this?
$endgroup$
– Maria Mazur
Jan 19 at 11:19
|
show 1 more comment
$begingroup$
I believe this is the desired result. Note that the intersection/non-intersection is opposite from the description.
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
add a comment |
$begingroup$
Knowing complex numbers, this one is easy to solve. We need just this lemma:
Lemma: If $|vec{XY}| =|vec{ZT}|$ and $angle (vec{XY},vec{ZT}) = alpha$ then $$ZT = varepsiloncdot XY$$
where $varepsilon = cos alpha + i sin alpha$
So it is enought to prove $MN = icdot MQ;;; (*)$. We have
$$DB=varepsilon AC ;;;;;;;;{rm where};;;;varepsilon = cos {pi over 6} + i sin {pi over 6} $$
and if $delta = cos {pi over 3} + i sin {pi over 3} $ then $$MA = delta MB implies M = {A-delta Bover 1-delta}$$
$$QC = delta QB implies Q = {C-delta Bover 1-delta}$$
$$NA = delta ND implies N = {A-delta Dover 1-delta}$$
thus $$MN ={A-delta Dover 1-delta}-{A-delta Bover 1-delta} = {deltaover 1-delta}DB$$ $$ = {deltaover 1-delta}varepsilon cdot AC=i{ACover 1-delta} = iMQ$$
and we are done.
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $R^{alpha}_O$ be a rotation in the plain by an angle $alpha$ around a point $O$.
Easy to see that to rotate a vector by an angle $alpha$ it's the same to rotate this vector around his tail.
Now, by using the beautiful Daniel Mathias's picture we obtain:
$$R^{90^{circ}}left(vec{NM}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{NA}+vec{AM}right)right)=R^{30^{circ}}left(vec{DA}+vec{AB}right)=$$
$$=R^{30^{circ}}left(vec{DB}right)=R^{60^{circ}}left(vec{AC}right)=R^{60^{circ}}left(vec{AB}+vec{BC}right)=vec{MB}+vec{BQ}=vec{MQ},$$
which says $NMperp MQ$ and $NM=MQ.$
Also,
$$R^{90^{circ}}left(vec{QP}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{QC}+vec{CP}right)right)=R^{30^{circ}}left(vec{BC}+vec{CD}right)=$$
$$=R^{30^{circ}}left(vec{BD}right)=R^{60^{circ}}left(vec{CA}right)=R^{60^{circ}}left(vec{CB}+vec{BA}right)=vec{QB}+vec{BM}=vec{QM},$$
which says $QMperp PQ$ and $QM=PQ.$
Can you end it now?
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
There is a synthetic proof? I try to prove it with congruences of triangles.
$endgroup$
– Problemsolving
Jan 11 at 5:44
1
$begingroup$
I like your solution +1, do you like mine?
$endgroup$
– Maria Mazur
Jan 14 at 20:42
$begingroup$
Do you have some more problems like this solved vith rotation?
$endgroup$
– Maria Mazur
Jan 19 at 11:14
$begingroup$
@greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152
$endgroup$
– Michael Rozenberg
Jan 19 at 11:17
$begingroup$
Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this?
$endgroup$
– Maria Mazur
Jan 19 at 11:19
|
show 1 more comment
$begingroup$
Let $R^{alpha}_O$ be a rotation in the plain by an angle $alpha$ around a point $O$.
Easy to see that to rotate a vector by an angle $alpha$ it's the same to rotate this vector around his tail.
Now, by using the beautiful Daniel Mathias's picture we obtain:
$$R^{90^{circ}}left(vec{NM}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{NA}+vec{AM}right)right)=R^{30^{circ}}left(vec{DA}+vec{AB}right)=$$
$$=R^{30^{circ}}left(vec{DB}right)=R^{60^{circ}}left(vec{AC}right)=R^{60^{circ}}left(vec{AB}+vec{BC}right)=vec{MB}+vec{BQ}=vec{MQ},$$
which says $NMperp MQ$ and $NM=MQ.$
Also,
$$R^{90^{circ}}left(vec{QP}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{QC}+vec{CP}right)right)=R^{30^{circ}}left(vec{BC}+vec{CD}right)=$$
$$=R^{30^{circ}}left(vec{BD}right)=R^{60^{circ}}left(vec{CA}right)=R^{60^{circ}}left(vec{CB}+vec{BA}right)=vec{QB}+vec{BM}=vec{QM},$$
which says $QMperp PQ$ and $QM=PQ.$
Can you end it now?
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$begingroup$
There is a synthetic proof? I try to prove it with congruences of triangles.
$endgroup$
– Problemsolving
Jan 11 at 5:44
1
$begingroup$
I like your solution +1, do you like mine?
$endgroup$
– Maria Mazur
Jan 14 at 20:42
$begingroup$
Do you have some more problems like this solved vith rotation?
$endgroup$
– Maria Mazur
Jan 19 at 11:14
$begingroup$
@greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152
$endgroup$
– Michael Rozenberg
Jan 19 at 11:17
$begingroup$
Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this?
$endgroup$
– Maria Mazur
Jan 19 at 11:19
|
show 1 more comment
$begingroup$
Let $R^{alpha}_O$ be a rotation in the plain by an angle $alpha$ around a point $O$.
Easy to see that to rotate a vector by an angle $alpha$ it's the same to rotate this vector around his tail.
Now, by using the beautiful Daniel Mathias's picture we obtain:
$$R^{90^{circ}}left(vec{NM}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{NA}+vec{AM}right)right)=R^{30^{circ}}left(vec{DA}+vec{AB}right)=$$
$$=R^{30^{circ}}left(vec{DB}right)=R^{60^{circ}}left(vec{AC}right)=R^{60^{circ}}left(vec{AB}+vec{BC}right)=vec{MB}+vec{BQ}=vec{MQ},$$
which says $NMperp MQ$ and $NM=MQ.$
Also,
$$R^{90^{circ}}left(vec{QP}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{QC}+vec{CP}right)right)=R^{30^{circ}}left(vec{BC}+vec{CD}right)=$$
$$=R^{30^{circ}}left(vec{BD}right)=R^{60^{circ}}left(vec{CA}right)=R^{60^{circ}}left(vec{CB}+vec{BA}right)=vec{QB}+vec{BM}=vec{QM},$$
which says $QMperp PQ$ and $QM=PQ.$
Can you end it now?
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
Let $R^{alpha}_O$ be a rotation in the plain by an angle $alpha$ around a point $O$.
Easy to see that to rotate a vector by an angle $alpha$ it's the same to rotate this vector around his tail.
Now, by using the beautiful Daniel Mathias's picture we obtain:
$$R^{90^{circ}}left(vec{NM}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{NA}+vec{AM}right)right)=R^{30^{circ}}left(vec{DA}+vec{AB}right)=$$
$$=R^{30^{circ}}left(vec{DB}right)=R^{60^{circ}}left(vec{AC}right)=R^{60^{circ}}left(vec{AB}+vec{BC}right)=vec{MB}+vec{BQ}=vec{MQ},$$
which says $NMperp MQ$ and $NM=MQ.$
Also,
$$R^{90^{circ}}left(vec{QP}right)=R^{30^{circ}}left(R^{60^{circ}}left(vec{QC}+vec{CP}right)right)=R^{30^{circ}}left(vec{BC}+vec{CD}right)=$$
$$=R^{30^{circ}}left(vec{BD}right)=R^{60^{circ}}left(vec{CA}right)=R^{60^{circ}}left(vec{CB}+vec{BA}right)=vec{QB}+vec{BM}=vec{QM},$$
which says $QMperp PQ$ and $QM=PQ.$
Can you end it now?
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
answered Jan 10 at 23:15
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
$begingroup$
There is a synthetic proof? I try to prove it with congruences of triangles.
$endgroup$
– Problemsolving
Jan 11 at 5:44
1
$begingroup$
I like your solution +1, do you like mine?
$endgroup$
– Maria Mazur
Jan 14 at 20:42
$begingroup$
Do you have some more problems like this solved vith rotation?
$endgroup$
– Maria Mazur
Jan 19 at 11:14
$begingroup$
@greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152
$endgroup$
– Michael Rozenberg
Jan 19 at 11:17
$begingroup$
Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this?
$endgroup$
– Maria Mazur
Jan 19 at 11:19
|
show 1 more comment
$begingroup$
There is a synthetic proof? I try to prove it with congruences of triangles.
$endgroup$
– Problemsolving
Jan 11 at 5:44
1
$begingroup$
I like your solution +1, do you like mine?
$endgroup$
– Maria Mazur
Jan 14 at 20:42
$begingroup$
Do you have some more problems like this solved vith rotation?
$endgroup$
– Maria Mazur
Jan 19 at 11:14
$begingroup$
@greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152
$endgroup$
– Michael Rozenberg
Jan 19 at 11:17
$begingroup$
Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this?
$endgroup$
– Maria Mazur
Jan 19 at 11:19
$begingroup$
There is a synthetic proof? I try to prove it with congruences of triangles.
$endgroup$
– Problemsolving
Jan 11 at 5:44
$begingroup$
There is a synthetic proof? I try to prove it with congruences of triangles.
$endgroup$
– Problemsolving
Jan 11 at 5:44
1
1
$begingroup$
I like your solution +1, do you like mine?
$endgroup$
– Maria Mazur
Jan 14 at 20:42
$begingroup$
I like your solution +1, do you like mine?
$endgroup$
– Maria Mazur
Jan 14 at 20:42
$begingroup$
Do you have some more problems like this solved vith rotation?
$endgroup$
– Maria Mazur
Jan 19 at 11:14
$begingroup$
Do you have some more problems like this solved vith rotation?
$endgroup$
– Maria Mazur
Jan 19 at 11:14
$begingroup$
@greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152
$endgroup$
– Michael Rozenberg
Jan 19 at 11:17
$begingroup$
@greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152
$endgroup$
– Michael Rozenberg
Jan 19 at 11:17
$begingroup$
Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this?
$endgroup$
– Maria Mazur
Jan 19 at 11:19
$begingroup$
Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this?
$endgroup$
– Maria Mazur
Jan 19 at 11:19
|
show 1 more comment
$begingroup$
I believe this is the desired result. Note that the intersection/non-intersection is opposite from the description.
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
add a comment |
$begingroup$
I believe this is the desired result. Note that the intersection/non-intersection is opposite from the description.
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
add a comment |
$begingroup$
I believe this is the desired result. Note that the intersection/non-intersection is opposite from the description.
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
$endgroup$
I believe this is the desired result. Note that the intersection/non-intersection is opposite from the description.
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
answered Jan 10 at 22:08
Daniel MathiasDaniel Mathias
1,40518
1,40518
add a comment |
add a comment |
$begingroup$
Knowing complex numbers, this one is easy to solve. We need just this lemma:
Lemma: If $|vec{XY}| =|vec{ZT}|$ and $angle (vec{XY},vec{ZT}) = alpha$ then $$ZT = varepsiloncdot XY$$
where $varepsilon = cos alpha + i sin alpha$
So it is enought to prove $MN = icdot MQ;;; (*)$. We have
$$DB=varepsilon AC ;;;;;;;;{rm where};;;;varepsilon = cos {pi over 6} + i sin {pi over 6} $$
and if $delta = cos {pi over 3} + i sin {pi over 3} $ then $$MA = delta MB implies M = {A-delta Bover 1-delta}$$
$$QC = delta QB implies Q = {C-delta Bover 1-delta}$$
$$NA = delta ND implies N = {A-delta Dover 1-delta}$$
thus $$MN ={A-delta Dover 1-delta}-{A-delta Bover 1-delta} = {deltaover 1-delta}DB$$ $$ = {deltaover 1-delta}varepsilon cdot AC=i{ACover 1-delta} = iMQ$$
and we are done.
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
Knowing complex numbers, this one is easy to solve. We need just this lemma:
Lemma: If $|vec{XY}| =|vec{ZT}|$ and $angle (vec{XY},vec{ZT}) = alpha$ then $$ZT = varepsiloncdot XY$$
where $varepsilon = cos alpha + i sin alpha$
So it is enought to prove $MN = icdot MQ;;; (*)$. We have
$$DB=varepsilon AC ;;;;;;;;{rm where};;;;varepsilon = cos {pi over 6} + i sin {pi over 6} $$
and if $delta = cos {pi over 3} + i sin {pi over 3} $ then $$MA = delta MB implies M = {A-delta Bover 1-delta}$$
$$QC = delta QB implies Q = {C-delta Bover 1-delta}$$
$$NA = delta ND implies N = {A-delta Dover 1-delta}$$
thus $$MN ={A-delta Dover 1-delta}-{A-delta Bover 1-delta} = {deltaover 1-delta}DB$$ $$ = {deltaover 1-delta}varepsilon cdot AC=i{ACover 1-delta} = iMQ$$
and we are done.
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
Knowing complex numbers, this one is easy to solve. We need just this lemma:
Lemma: If $|vec{XY}| =|vec{ZT}|$ and $angle (vec{XY},vec{ZT}) = alpha$ then $$ZT = varepsiloncdot XY$$
where $varepsilon = cos alpha + i sin alpha$
So it is enought to prove $MN = icdot MQ;;; (*)$. We have
$$DB=varepsilon AC ;;;;;;;;{rm where};;;;varepsilon = cos {pi over 6} + i sin {pi over 6} $$
and if $delta = cos {pi over 3} + i sin {pi over 3} $ then $$MA = delta MB implies M = {A-delta Bover 1-delta}$$
$$QC = delta QB implies Q = {C-delta Bover 1-delta}$$
$$NA = delta ND implies N = {A-delta Dover 1-delta}$$
thus $$MN ={A-delta Dover 1-delta}-{A-delta Bover 1-delta} = {deltaover 1-delta}DB$$ $$ = {deltaover 1-delta}varepsilon cdot AC=i{ACover 1-delta} = iMQ$$
and we are done.
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
Knowing complex numbers, this one is easy to solve. We need just this lemma:
Lemma: If $|vec{XY}| =|vec{ZT}|$ and $angle (vec{XY},vec{ZT}) = alpha$ then $$ZT = varepsiloncdot XY$$
where $varepsilon = cos alpha + i sin alpha$
So it is enought to prove $MN = icdot MQ;;; (*)$. We have
$$DB=varepsilon AC ;;;;;;;;{rm where};;;;varepsilon = cos {pi over 6} + i sin {pi over 6} $$
and if $delta = cos {pi over 3} + i sin {pi over 3} $ then $$MA = delta MB implies M = {A-delta Bover 1-delta}$$
$$QC = delta QB implies Q = {C-delta Bover 1-delta}$$
$$NA = delta ND implies N = {A-delta Dover 1-delta}$$
thus $$MN ={A-delta Dover 1-delta}-{A-delta Bover 1-delta} = {deltaover 1-delta}DB$$ $$ = {deltaover 1-delta}varepsilon cdot AC=i{ACover 1-delta} = iMQ$$
and we are done.
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
edited Jan 14 at 17:44
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
answered Jan 14 at 16:32
Maria MazurMaria Mazur
49.9k1361125
49.9k1361125
add a comment |
add a comment |
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$begingroup$
Just drawn a picture and it didn't look like a square to me.
$endgroup$
– Quang Hoang
Jan 10 at 21:02
1
$begingroup$
Edit: I think I see the problem here. All equilateral triangles should be drawn in the opposite direction and the statement would be true. I still don't know how to approach though...
$endgroup$
– Edward H.
Jan 10 at 21:23
$begingroup$
Ok so here's an idea assuming that the equilateral triangles are meant to be drawn oppositely: Rotate ΔABC by 60° around B and you'll get ΔMBQ. Do this 4 times and use AC=BD, and you'll find that MNPQ is a rhombus. Moreover if you keep track of all the angles of rotation and use ∠AOD = 30°, you'll find that the angles of the rhombus are all right, and hence it is a square.
$endgroup$
– Edward H.
Jan 10 at 21:49
$begingroup$
As given it's clearly false. $mangle AMB =60$ and $mangle MBA = 60$ and $mangle QBC = 60$ so $Q$ is in the interior of $angle AMB$. As is $N$ so $mangle NMQ < mangle AMB = 60 < 90$. Can't be a square. But if all the intersections are empty so the equi triangles pop out.
$endgroup$
– fleablood
Jan 10 at 22:29