Synthetic solution to this geometry problem?
up vote
6
down vote
favorite
Consider the following diagram. In the isosceles triangle $triangle ABC$ with $AB=AC$, it is given that $BC=2$. Two points $M,N$ lie on $AB,AC$ respectively so that $AM=NC$. Prove: $MN$ is at least $1$. (Source: 1990 High School Olympiad held in Xi'an, China)
I've already solved this problem by doing some coordinate geometry, setting $AM=NC=t$, finding $MN$ as a function of $t$, then minimising that function. But this is quite tedious, which led me to wonder what the synthetic geometry solution, which I couldn't find, is. (Btw, "synthetic" means without the use of coordinate geometry, and hopefully with as little algebra as possible as well.)
geometry contest-math euclidean-geometry
asked Dec 5 at 1:06
YiFan
2,2091421
add a comment |
up vote
6
down vote
favorite
Consider the following diagram. In the isosceles triangle $triangle ABC$ with $AB=AC$, it is given that $BC=2$. Two points $M,N$ lie on $AB,AC$ respectively so that $AM=NC$. Prove: $MN$ is at least $1$. (Source: 1990 High School Olympiad held in Xi'an, China)
I've already solved this problem by doing some coordinate geometry, setting $AM=NC=t$, finding $MN$ as a function of $t$, then minimising that function. But this is quite tedious, which led me to wonder what the synthetic geometry solution, which I couldn't find, is. (Btw, "synthetic" means without the use of coordinate geometry, and hopefully with as little algebra as possible as well.)
geometry contest-math euclidean-geometry
asked Dec 5 at 1:06
YiFan
2,2091421
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Consider the following diagram. In the isosceles triangle $triangle ABC$ with $AB=AC$, it is given that $BC=2$. Two points $M,N$ lie on $AB,AC$ respectively so that $AM=NC$. Prove: $MN$ is at least $1$. (Source: 1990 High School Olympiad held in Xi'an, China)
I've already solved this problem by doing some coordinate geometry, setting $AM=NC=t$, finding $MN$ as a function of $t$, then minimising that function. But this is quite tedious, which led me to wonder what the synthetic geometry solution, which I couldn't find, is. (Btw, "synthetic" means without the use of coordinate geometry, and hopefully with as little algebra as possible as well.)
geometry contest-math euclidean-geometry
asked Dec 5 at 1:06
YiFan
2,2091421
Consider the following diagram. In the isosceles triangle $triangle ABC$ with $AB=AC$, it is given that $BC=2$. Two points $M,N$ lie on $AB,AC$ respectively so that $AM=NC$. Prove: $MN$ is at least $1$. (Source: 1990 High School Olympiad held in Xi'an, China)
I've already solved this problem by doing some coordinate geometry, setting $AM=NC=t$, finding $MN$ as a function of $t$, then minimising that function. But this is quite tedious, which led me to wonder what the synthetic geometry solution, which I couldn't find, is. (Btw, "synthetic" means without the use of coordinate geometry, and hopefully with as little algebra as possible as well.)
geometry contest-math euclidean-geometry
geometry contest-math euclidean-geometry
asked Dec 5 at 1:06
YiFan
2,2091421
asked Dec 5 at 1:06
YiFan
2,2091421
asked Dec 5 at 1:06
YiFan
2,2091421
asked Dec 5 at 1:06
YiFan
2,2091421
asked Dec 5 at 1:06
YiFan
2,2091421
2,2091421
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.
$$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$
answered Dec 5 at 4:33
Blue
47.3k870149
How did you draw the diagram?
– Anubhab Ghosal
Dec 5 at 8:12
@AnubhabGhosal: I use GeoGebra.
– Blue
Dec 5 at 8:15
How do you mark the angles in geogebra?
– Anubhab Ghosal
Dec 5 at 8:16
1
@AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".)
– Blue
Dec 5 at 8:20
1
Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…).
– Anubhab Ghosal
Dec 5 at 9:37
add a comment |
up vote
4
down vote
A very simple solution: $MN$ is invariant under exchange of $AM$ and $AN$ (this is mirror symmetry of right and left in your drawing). This implies that for $AM$ =$AN$ = $AB$/2 the length of $MN$ is extremal, i.e. it has either the largest or the smallest value it can take. In the symmetric case $AM = AN$ we have from similarity $MN = frac{1}{2} BC = 1$, which is the lower bound from the statement of the problem. So it remains to show that this is indeed a minimum, but the length cannot have more than one extremal value for $Nin AC$ and for $N=C$ we have $NM ge 1$ from the triangle inequality.
answered Dec 5 at 7:39
tobi_s
1512
add a comment |
up vote
3
down vote
Let $P$ and $Q$ be the mid-points of $AB$ and $AC$ respectively. Join $PQ$. Suppose $PQ$ meets $MN$ at $R$. Extend $PQ$ towards the side of $P$(if $M$ is nearer to $A$ as drawn in the diagram) to $R'$ such that $PR'=QR$. Now $MP=QN$ and $angle MPR'=angle NQR$. Therefore, $triangle MPR'cong triangle NQR$. Therefore, $MR+MR'ge RR'$ by triangle inequality, whence $MR+RNgeq PR+RQ$.
Therefore, $MNgeq PQ=1$.
answered Dec 5 at 4:02
Anubhab Ghosal
74212
Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A.
– amI
Dec 5 at 11:28
@aml, I do not understand your concern.
– Anubhab Ghosal
Dec 5 at 14:51
No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous.
– amI
Dec 5 at 22:01
@aml, that is right, but how do you prove that PQ bisects MN without any construction?
– Anubhab Ghosal
Dec 7 at 15:02
Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ).
– amI
Dec 8 at 12:43
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
up vote
5
down vote
accepted
Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.
$$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$
answered Dec 5 at 4:33
Blue
47.3k870149
How did you draw the diagram?
– Anubhab Ghosal
Dec 5 at 8:12
@AnubhabGhosal: I use GeoGebra.
– Blue
Dec 5 at 8:15
How do you mark the angles in geogebra?
– Anubhab Ghosal
Dec 5 at 8:16
1
@AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".)
– Blue
Dec 5 at 8:20
1
Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…).
– Anubhab Ghosal
Dec 5 at 9:37
add a comment |
up vote
5
down vote
accepted
Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.
$$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$
answered Dec 5 at 4:33
Blue
47.3k870149
How did you draw the diagram?
– Anubhab Ghosal
Dec 5 at 8:12
@AnubhabGhosal: I use GeoGebra.
– Blue
Dec 5 at 8:15
How do you mark the angles in geogebra?
– Anubhab Ghosal
Dec 5 at 8:16
1
@AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".)
– Blue
Dec 5 at 8:20
1
Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…).
– Anubhab Ghosal
Dec 5 at 9:37
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.
$$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$
answered Dec 5 at 4:33
Blue
47.3k870149
Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.
$$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$
answered Dec 5 at 4:33
Blue
47.3k870149
answered Dec 5 at 4:33
Blue
47.3k870149
answered Dec 5 at 4:33
Blue
47.3k870149
answered Dec 5 at 4:33
Blue
47.3k870149
47.3k870149
How did you draw the diagram?
– Anubhab Ghosal
Dec 5 at 8:12
@AnubhabGhosal: I use GeoGebra.
– Blue
Dec 5 at 8:15
How do you mark the angles in geogebra?
– Anubhab Ghosal
Dec 5 at 8:16
1
@AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".)
– Blue
Dec 5 at 8:20
1
Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…).
– Anubhab Ghosal
Dec 5 at 9:37
add a comment |
How did you draw the diagram?
– Anubhab Ghosal
Dec 5 at 8:12
@AnubhabGhosal: I use GeoGebra.
– Blue
Dec 5 at 8:15
How do you mark the angles in geogebra?
– Anubhab Ghosal
Dec 5 at 8:16
1
@AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".)
– Blue
Dec 5 at 8:20
1
Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…).
– Anubhab Ghosal
Dec 5 at 9:37
How did you draw the diagram?
– Anubhab Ghosal
Dec 5 at 8:12
How did you draw the diagram?
– Anubhab Ghosal
Dec 5 at 8:12
@AnubhabGhosal: I use GeoGebra.
– Blue
Dec 5 at 8:15
@AnubhabGhosal: I use GeoGebra.
– Blue
Dec 5 at 8:15
How do you mark the angles in geogebra?
– Anubhab Ghosal
Dec 5 at 8:16
How do you mark the angles in geogebra?
– Anubhab Ghosal
Dec 5 at 8:16
1
1
@AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".)
– Blue
Dec 5 at 8:20
@AnubhabGhosal: There's an angle-marking tool. :) If you use the online version of GeoGebra, the angle tool is the first tool in the "Measure" set; simply click on three points (in the proper order) to get an angle mark. (You can even use the Settings to set a "Decoration".)
– Blue
Dec 5 at 8:20
1
1
Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…).
– Anubhab Ghosal
Dec 5 at 9:37
Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…).
– Anubhab Ghosal
Dec 5 at 9:37
add a comment |
up vote
4
down vote
A very simple solution: $MN$ is invariant under exchange of $AM$ and $AN$ (this is mirror symmetry of right and left in your drawing). This implies that for $AM$ =$AN$ = $AB$/2 the length of $MN$ is extremal, i.e. it has either the largest or the smallest value it can take. In the symmetric case $AM = AN$ we have from similarity $MN = frac{1}{2} BC = 1$, which is the lower bound from the statement of the problem. So it remains to show that this is indeed a minimum, but the length cannot have more than one extremal value for $Nin AC$ and for $N=C$ we have $NM ge 1$ from the triangle inequality.
answered Dec 5 at 7:39
tobi_s
1512
add a comment |
up vote
4
down vote
A very simple solution: $MN$ is invariant under exchange of $AM$ and $AN$ (this is mirror symmetry of right and left in your drawing). This implies that for $AM$ =$AN$ = $AB$/2 the length of $MN$ is extremal, i.e. it has either the largest or the smallest value it can take. In the symmetric case $AM = AN$ we have from similarity $MN = frac{1}{2} BC = 1$, which is the lower bound from the statement of the problem. So it remains to show that this is indeed a minimum, but the length cannot have more than one extremal value for $Nin AC$ and for $N=C$ we have $NM ge 1$ from the triangle inequality.
answered Dec 5 at 7:39
tobi_s
1512
add a comment |
up vote
4
down vote
up vote
4
down vote
A very simple solution: $MN$ is invariant under exchange of $AM$ and $AN$ (this is mirror symmetry of right and left in your drawing). This implies that for $AM$ =$AN$ = $AB$/2 the length of $MN$ is extremal, i.e. it has either the largest or the smallest value it can take. In the symmetric case $AM = AN$ we have from similarity $MN = frac{1}{2} BC = 1$, which is the lower bound from the statement of the problem. So it remains to show that this is indeed a minimum, but the length cannot have more than one extremal value for $Nin AC$ and for $N=C$ we have $NM ge 1$ from the triangle inequality.
answered Dec 5 at 7:39
tobi_s
1512
A very simple solution: $MN$ is invariant under exchange of $AM$ and $AN$ (this is mirror symmetry of right and left in your drawing). This implies that for $AM$ =$AN$ = $AB$/2 the length of $MN$ is extremal, i.e. it has either the largest or the smallest value it can take. In the symmetric case $AM = AN$ we have from similarity $MN = frac{1}{2} BC = 1$, which is the lower bound from the statement of the problem. So it remains to show that this is indeed a minimum, but the length cannot have more than one extremal value for $Nin AC$ and for $N=C$ we have $NM ge 1$ from the triangle inequality.
answered Dec 5 at 7:39
tobi_s
1512
edited Dec 12 at 0:55
answered Dec 5 at 7:39
tobi_s
1512
answered Dec 5 at 7:39
tobi_s
1512
answered Dec 5 at 7:39
tobi_s
1512
1512
add a comment |
add a comment |
up vote
3
down vote
Let $P$ and $Q$ be the mid-points of $AB$ and $AC$ respectively. Join $PQ$. Suppose $PQ$ meets $MN$ at $R$. Extend $PQ$ towards the side of $P$(if $M$ is nearer to $A$ as drawn in the diagram) to $R'$ such that $PR'=QR$. Now $MP=QN$ and $angle MPR'=angle NQR$. Therefore, $triangle MPR'cong triangle NQR$. Therefore, $MR+MR'ge RR'$ by triangle inequality, whence $MR+RNgeq PR+RQ$.
Therefore, $MNgeq PQ=1$.
answered Dec 5 at 4:02
Anubhab Ghosal
74212
Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A.
– amI
Dec 5 at 11:28
@aml, I do not understand your concern.
– Anubhab Ghosal
Dec 5 at 14:51
No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous.
– amI
Dec 5 at 22:01
@aml, that is right, but how do you prove that PQ bisects MN without any construction?
– Anubhab Ghosal
Dec 7 at 15:02
Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ).
– amI
Dec 8 at 12:43
add a comment |
up vote
3
down vote
Let $P$ and $Q$ be the mid-points of $AB$ and $AC$ respectively. Join $PQ$. Suppose $PQ$ meets $MN$ at $R$. Extend $PQ$ towards the side of $P$(if $M$ is nearer to $A$ as drawn in the diagram) to $R'$ such that $PR'=QR$. Now $MP=QN$ and $angle MPR'=angle NQR$. Therefore, $triangle MPR'cong triangle NQR$. Therefore, $MR+MR'ge RR'$ by triangle inequality, whence $MR+RNgeq PR+RQ$.
Therefore, $MNgeq PQ=1$.
answered Dec 5 at 4:02
Anubhab Ghosal
74212
Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A.
– amI
Dec 5 at 11:28
@aml, I do not understand your concern.
– Anubhab Ghosal
Dec 5 at 14:51
No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous.
– amI
Dec 5 at 22:01
@aml, that is right, but how do you prove that PQ bisects MN without any construction?
– Anubhab Ghosal
Dec 7 at 15:02
Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ).
– amI
Dec 8 at 12:43
add a comment |
up vote
3
down vote
up vote
3
down vote
Let $P$ and $Q$ be the mid-points of $AB$ and $AC$ respectively. Join $PQ$. Suppose $PQ$ meets $MN$ at $R$. Extend $PQ$ towards the side of $P$(if $M$ is nearer to $A$ as drawn in the diagram) to $R'$ such that $PR'=QR$. Now $MP=QN$ and $angle MPR'=angle NQR$. Therefore, $triangle MPR'cong triangle NQR$. Therefore, $MR+MR'ge RR'$ by triangle inequality, whence $MR+RNgeq PR+RQ$.
Therefore, $MNgeq PQ=1$.
answered Dec 5 at 4:02
Anubhab Ghosal
74212
Let $P$ and $Q$ be the mid-points of $AB$ and $AC$ respectively. Join $PQ$. Suppose $PQ$ meets $MN$ at $R$. Extend $PQ$ towards the side of $P$(if $M$ is nearer to $A$ as drawn in the diagram) to $R'$ such that $PR'=QR$. Now $MP=QN$ and $angle MPR'=angle NQR$. Therefore, $triangle MPR'cong triangle NQR$. Therefore, $MR+MR'ge RR'$ by triangle inequality, whence $MR+RNgeq PR+RQ$.
Therefore, $MNgeq PQ=1$.
answered Dec 5 at 4:02
Anubhab Ghosal
74212
edited Dec 5 at 4:43
answered Dec 5 at 4:02
Anubhab Ghosal
74212
answered Dec 5 at 4:02
Anubhab Ghosal
74212
answered Dec 5 at 4:02
Anubhab Ghosal
74212
74212
Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A.
– amI
Dec 5 at 11:28
@aml, I do not understand your concern.
– Anubhab Ghosal
Dec 5 at 14:51
No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous.
– amI
Dec 5 at 22:01
@aml, that is right, but how do you prove that PQ bisects MN without any construction?
– Anubhab Ghosal
Dec 7 at 15:02
Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ).
– amI
Dec 8 at 12:43
add a comment |
Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A.
– amI
Dec 5 at 11:28
@aml, I do not understand your concern.
– Anubhab Ghosal
Dec 5 at 14:51
No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous.
– amI
Dec 5 at 22:01
@aml, that is right, but how do you prove that PQ bisects MN without any construction?
– Anubhab Ghosal
Dec 7 at 15:02
Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ).
– amI
Dec 8 at 12:43
Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A.
– amI
Dec 5 at 11:28
Note - The accepted figure, and this figure, and a third version with R'R = RQ, all prove that PQ bisects MN, so MN > PQ if M and N are on different sides of A.
– amI
Dec 5 at 11:28
@aml, I do not understand your concern.
– Anubhab Ghosal
Dec 5 at 14:51
@aml, I do not understand your concern.
– Anubhab Ghosal
Dec 5 at 14:51
No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous.
– amI
Dec 5 at 22:01
No concern - its just that PQ (bisecting AB and AC) must bisect MN, so it seems obvious that MN >= PQ, and adding new point[s] just makes the proof rigorous.
– amI
Dec 5 at 22:01
@aml, that is right, but how do you prove that PQ bisects MN without any construction?
– Anubhab Ghosal
Dec 7 at 15:02
@aml, that is right, but how do you prove that PQ bisects MN without any construction?
– Anubhab Ghosal
Dec 7 at 15:02
Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ).
– amI
Dec 8 at 12:43
Indeed - I think I just see it being folded (once horiz along PQ and twice vert) so MN becomes a parallelogram inside a small rectangle with diagonal AQ (=CQ). The perimeter of this parallelogram (MN) can't be any less than twice the base (=twice PQ/2 =PQ).
– amI
Dec 8 at 12:43
add a comment |
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