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Problem 1565

National Olympiad, first round

Geometry Difficulty 6.1 Prove it Romanian Mathematical Olympiad · Romania

Consider a triangle $ABC$ , such that $\angle B = 90^\circ$ . Denote by $I$ the in-center and let $F$ , $D$ and $E$ be the points where the incircle touches sides $[AB]$ , $[BC]$ , and $[AC]$ respectively. If $CI \cap EF = \{M\}$ and $DM \cap AB = \{N\}$ , show that:

a) $AI = ND$ ;

b) $FM = \frac{EI \cdot EM}{EC}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

a. Triangle $AFE$ is isosceles with $AE = AF$ , and $AI \perp FE$ , hence $\angle AEF = 90^\circ - \angle A/2$ . In the same way from the isosceles triangle $CDE$ we get $\angle DEC = 90^\circ - \angle C/2$ . As a consequence $\overline{MED} = 180^\circ - \overline{AEF} - \overline{DEC} = 180^\circ - (180^\circ - (\hat{A} + \hat{C})/2) = 45^\circ$ . (*)

As $\triangle MDC \equiv \triangle MEC$ , we obtain $MD = ME$ (**). By () and (*), the triangle $\triangle MED$ is right angled and isosceles. As a consequence $DN \perp EF$ and, because $AI \perp EF$ , we obtain $DN \parallel AI$ . As $AN \parallel ID$ we conclude that the quadrilateral $ANDI$ is a parallelogram, so $AI = ND$ .

b. We have $\overline{EFD} = 180^\circ - \overline{AFE} - \overline{BFD} = (\hat{A} + \hat{B})/2 = 90^\circ - \hat{C}/2 = \overline{DIC}$ so $\triangle FMD \sim \triangle IDC$ . We conclude $\frac{FM}{ID} = \frac{MD}{DC}$ which implies
$FM = \frac{ID \cdot MD}{DC} = \frac{EI \cdot EM}{EC}.$

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