Synthetic solution to this geometry problem?
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
2,2691421
add a comment |
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
2,2691421
add a comment |
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
2,2691421
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
2,2691421
asked Dec 5 at 1:06
YiFan
2,2691421
asked Dec 5 at 1:06
YiFan
2,2691421
asked Dec 5 at 1:06
YiFan
2,2691421
asked Dec 5 at 1:06
YiFan
2,2691421
2,2691421
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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
47.4k870150
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 |
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 |
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
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
47.4k870150
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 |
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
47.4k870150
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 |
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
47.4k870150
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
47.4k870150
answered Dec 5 at 4:33
Blue
47.4k870150
answered Dec 5 at 4:33
Blue
47.4k870150
answered Dec 5 at 4:33
Blue
47.4k870150
47.4k870150
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 |
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 |
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 |
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 |
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 |
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 |
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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Post as a guest
Required, but never shown
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
