# 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.)

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

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

• How did you draw the diagram? Commented Dec 5, 2018 at 8:12
• @AnubhabGhosal: I use GeoGebra.
– Blue
Commented Dec 5, 2018 at 8:15
• How do you mark the angles in geogebra? Commented Dec 5, 2018 at 8:16
• @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
Commented Dec 5, 2018 at 8:20
• Thanks. It is because of your help that I could draw this diagram(math.stackexchange.com/questions/3026660/…). Commented Dec 5, 2018 at 9:37

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.

• Is this assuming that length of MN as a function of the length of AM is quadratic (or at least convex)? That's true but proving it is a bit of work, isn't it? Commented Mar 28, 2021 at 17:13

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, $$MN\geq PQ=1$$.

• 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
Commented Dec 5, 2018 at 11:28
• @aml, I do not understand your concern. Commented Dec 5, 2018 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
Commented Dec 5, 2018 at 22:01
• @aml, that is right, but how do you prove that PQ bisects MN without any construction? Commented Dec 7, 2018 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
Commented Dec 8, 2018 at 12:43