Synthetic solution to this geometry problem?












6














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.)










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    6














    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.)










    share|cite|improve this question

























      6












      6








      6


      0





      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.)










      share|cite|improve this question













      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






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      asked Dec 5 at 1:06









      YiFan

      2,2691421




      2,2691421






















          3 Answers
          3






          active

          oldest

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          5














          Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.



          enter image description here



          $$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$






          share|cite|improve this answer





















          • 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





















          4














          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.






          share|cite|improve this answer































            3














            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, $MNgeq PQ=1$.



            Diagram






            share|cite|improve this answer























            • 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











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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            5














            Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.



            enter image description here



            $$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$






            share|cite|improve this answer





















            • 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


















            5














            Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.



            enter image description here



            $$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$






            share|cite|improve this answer





















            • 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
















            5












            5








            5






            Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.



            enter image description here



            $$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$






            share|cite|improve this answer












            Let $overline{PQ}$ be the midsegment of $triangle ABC$ parallel to $overline{BC}$, and note that $overline{MP}congoverline{NQ}$.



            enter image description here



            $$frac12|BC| = |PQ| = |M^prime N^prime| leq |MN|$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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




















            • 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













            4














            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.






            share|cite|improve this answer




























              4














              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.






              share|cite|improve this answer


























                4












                4








                4






                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.






                share|cite|improve this answer














                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 12 at 0:55

























                answered Dec 5 at 7:39









                tobi_s

                1512




                1512























                    3














                    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, $MNgeq PQ=1$.



                    Diagram






                    share|cite|improve this answer























                    • 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
















                    3














                    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, $MNgeq PQ=1$.



                    Diagram






                    share|cite|improve this answer























                    • 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














                    3












                    3








                    3






                    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, $MNgeq PQ=1$.



                    Diagram






                    share|cite|improve this answer














                    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, $MNgeq PQ=1$.



                    Diagram







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 5 at 4:43

























                    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


















                    • 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


















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