Problem A. 702. (September 2017)
A. 702. Fix a triangle \(\displaystyle ABC\). We say that triangle \(\displaystyle XYZ\) is elegant if \(\displaystyle X\) lies on segment \(\displaystyle BC\), \(\displaystyle Y\) lies on segment \(\displaystyle CA\), \(\displaystyle Z\) lies on segment \(\displaystyle AB\), and \(\displaystyle XYZ\) is similar to \(\displaystyle ABC\) (i.e., \(\displaystyle \angle A=\angle X\), \(\displaystyle \angle B=\angle Y\), \(\displaystyle \angle C=\angle Z\)). Of all the elegant triangles, which one has the smallest perimeter?
(5 pont)
Deadline expired on October 10, 2017.
Solution 1. Denote the sides and angles of \(\displaystyle \triangle ABC\) by \(\displaystyle \alpha,\beta,\gamma\) and \(\displaystyle a,b,c\). Then the sides of \(\displaystyle \triangle XYZ\) are \(\displaystyle \lambda a,\lambda b,\lambda c\) for some \(\displaystyle \lambda\in\mathbb{R}\).
Let \(\displaystyle \angle BZX=u\), \(\displaystyle \angle CYX=v\). Then \(\displaystyle \angle BXZ+\angle CXY=180^\circ-\alpha\) implies that \(\displaystyle u+v=2\alpha\).
Applying the Law of Sines to triangles \(\displaystyle BXZ\) and \(\displaystyle CXY\), we have
\(\displaystyle BX=\frac{XZ}{\sin \beta}\cdot \sin u,\qquad CX=\frac{XY}{\sin\gamma}\cdot \sin v.\)
Substituting and adding:
\(\displaystyle BC=\frac{\lambda b}{\sin\beta}\cdot \sin u+\frac{\lambda c}{\sin \gamma}\cdot \sin v.\)
Applying the Law of Sines to \(\displaystyle \triangle ABC\), multiplying this equation by \(\displaystyle \frac{\sin\alpha}{a}=\frac{\sin\beta}{b}=\frac{\sin\gamma}{c}\), we find that
\(\displaystyle \sin \alpha=\lambda(\sin u+\sin v)=\lambda\cdot 2\sin\frac{u+v}{2}\cos\frac{u-v}{2}.\)
Now \(\displaystyle \cos\frac{u-v}{2}\le 1\) and \(\displaystyle \frac{u+v}{2}=\alpha\) implies \(\displaystyle \lambda\ge \frac12\). Here, \(\displaystyle \lambda\) attains its minimum iff \(\displaystyle u=v=\alpha\), i.e. iff \(\displaystyle XYZ\) is the medial triangle of \(\displaystyle ABC\). Therefore, the elegant triangle with the minimal perimeter is the medial triangle.
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Solution 2. The key idea is to use circles to make angles appear in different places. In particular, intersecting circles \(\displaystyle BXZ\) and \(\displaystyle CXY\) can make angles \(\displaystyle u,v\) in Solution 1 appear at the second intersection of these circles. We will use directed angles (for a quick exposition, see this handout).
In directed angles, we have
\(\displaystyle \angle BPC=\angle BPX+\angle XPC=\angle BZX+\angle XYC=\angle(BA,ZX)+\angle(XY,AC)+\angle(ZX,XY)-\angle(ZX,XY)=\angle(BA,AC)-\angle(ZX,XY)=2\angle BAC,\)
because triangles \(\displaystyle ABC\) and \(\displaystyle XYZ\) have the same orientation and are similar. Circle theorems imply that if \(\displaystyle O\) is the center of circle \(\displaystyle ABC\), then \(\displaystyle \angle BOC=2\angle BAC\) as well. Hence \(\displaystyle \angle BPC=\angle BOC\), i.e. \(\displaystyle P\) lies on circle \(\displaystyle BCO\). Analogously, \(\displaystyle P\) also lies on circle \(\displaystyle CAO\) and \(\displaystyle ABO\) and thus \(\displaystyle P=O\).
We can also prove that \(\displaystyle P\) is the orthocenter of \(\displaystyle \triangle XYZ\). Indeed,
\(\displaystyle \angle(XP,YZ)=\angle(XP,PB)+\angle(PB,BZ)+\angle(BZ,ZX)+\angle(ZX,YZ)=\angle XPB+\angle PBZ+\angle BZX+\angle XZY=\)
\(\displaystyle =\angle XYB+\angle OBA+\angle BYX+\angle ACB=\angle OBA+\angle ACB=90^\circ,\)
where in the last step we considered the right triangle formed by \(\displaystyle O\), \(\displaystyle B\) and the midpoint of \(\displaystyle AB\). This shows that \(\displaystyle XP\perp YZ\), and \(\displaystyle YP\perp ZX\), \(\displaystyle ZP\perp XY\) can be proven analogously.
This proves that all possible quadrangles \(\displaystyle XYZP\) are similar to each other, and so \(\displaystyle |PX|\) is directly proportionate to the perimeter of \(\displaystyle XYZ\). However, since \(\displaystyle X\) lies on \(\displaystyle \overline{BC}\), \(\displaystyle |PX|\) is minimized when \(\displaystyle X\) is the projection of \(\displaystyle P=O\) onto \(\displaystyle \overline{BC}\), i.e. the midpoint of \(\displaystyle \overline{BC}\).
This again proves that the elegant triangle with the minimal perimeter is the medial triangle.
Statistics:
23 students sent a solution. 5 points: Beke Csongor, Bukva Balázs, Csépányi István, Daróczi Sándor, Gáspár Attila, Győrffy Ágoston, Imolay András, Janzer Orsolya Lili, Kerekes Anna, Matolcsi Dávid, Németh 123 Balázs, Schrettner Jakab, Simon Dániel Gábor, Szabó Kristóf, Szemerédi Levente, Weisz Máté. 4 points: Egri Máté, Hervay Bence, Márton Dénes, Szabó 417 Dávid, Záhorský Ákos. 3 points: 1 student. 0 point: 1 student.
Problems in Mathematics of KöMaL, September 2017