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

enter image description here

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

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  • $\begingroup$ How did you draw the diagram? $\endgroup$ – Anubhab Ghosal Dec 5 '18 at 8:12
  • $\begingroup$ @AnubhabGhosal: I use GeoGebra. $\endgroup$ – Blue Dec 5 '18 at 8:15
  • $\begingroup$ How do you mark the angles in geogebra? $\endgroup$ – Anubhab Ghosal Dec 5 '18 at 8:16
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    $\begingroup$ @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".) $\endgroup$ – Blue Dec 5 '18 at 8:20
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    $\begingroup$ Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…). $\endgroup$ – Anubhab Ghosal Dec 5 '18 at 9:37
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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 $N\in AC$ and for $N=C$ we have $NM \ge 1$ from the triangle inequality.

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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 $MR+RN\geq PR+RQ$. Therefore, $MN\geq PQ=1$.

Diagram

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  • $\begingroup$ 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. $\endgroup$ – amI Dec 5 '18 at 11:28
  • $\begingroup$ @aml, I do not understand your concern. $\endgroup$ – Anubhab Ghosal Dec 5 '18 at 14:51
  • $\begingroup$ 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. $\endgroup$ – amI Dec 5 '18 at 22:01
  • $\begingroup$ @aml, that is right, but how do you prove that PQ bisects MN without any construction? $\endgroup$ – Anubhab Ghosal Dec 7 '18 at 15:02
  • $\begingroup$ 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). $\endgroup$ – amI Dec 8 '18 at 12:43

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