# Solution Shortlist problem about geometry in IMO 2014 ( Problem 6)

Problem 6:

Let ABC be a fixed acute-angled triangle. Consider some points E and F lying on the sides AC and AB, respectively, and let M be the midpoint of EF. Let the perpendicular bisector of EF intersect the line BC at K, and let the perpendicular bisector of MK intersect the lines AC and AB at S and T, respectively. We call the pair (E, F) interesting, if the quadrilateral KSAT is cyclic.

Suppose that the pairs (E_{1}, F_{1}) and (E_{2}, F_{2} ) are interesting. Prove that

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Solution 1.

For any interesting pair (E,F), we will say that the corresponding triangle EFK is also interesting.

Let EFK be an interesting triangle. Firstly, we prove that , which also means that the circumcircle ω_{1} of the triangle AEF is tangent to the lines KE and KF.

Denote by ω the circle passing through the points K, S, A, and T. Let the line AM intersect the line ST and the circle ω (for the second time) at N and L, respectively (see Figure 1).

Since EF || TS and M is the midpoint of EF, N is the midpoint of ST. Moreover, since K and M are symmetric to each other with respect to the line ST, we have . Thus the points K and L are symmetric to each other with respect to the perpendicular bisector of ST. Therefore KL || ST.

Let G be the point symmetric to K with respect to N. Then G lies on the line EF, and we may assume that it lies on the ray MF. One has

(if K “ L, then the angle KLA is understood to be the angle between AL and the tangent to ω at L). This means that the points K, G, E, and S are concyclic. Now, since KSGT is a parallelogram, we obtain . Since KE = KF, we also have .

After having proved this fact, one may finish the solution by different methods.

First method. We have just proved that all interesting triangles are similar to each other. This allows us to use the following lemma.

Lemma.

Let ABC be an arbitrary triangle. Choose two points E_{1} and E_{2} on the side AC, two points F_{1} and F_{2} on the side AB, and two points K_{1} and K_{2} on the side BC, in a way that the triangles E_{1}F_{1}K_{1 }and E_{2}F_{2}K_{2} are similar. Then the six circumcircles of the triangles AE_{i}F_{i}, BF_{i}K_{i} and CE_{i}K_{i} (i = 1, 2) meet at a common point Z. Moreover, Z is the centre of the spiral similarity that takes the triangle E_{1}F_{1}K_{1}, to the triangle E_{2}F_{2}K_{2}.

Proof.

Firstly, notice that for each i = 1, 2, the circumcircles of the triangles AE_{i}F_{i}, BF_{i}K_{i}, and CK_{i}E_{i} have a common point Z_{i} by Miquel’s theorem. Moreover, we have

.

This yields that the points Z_{1} and Z_{2} correspond to each other in similar triangles E_{1}F_{1}K_{1} and E_{2}F_{2}K_{2}. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.

Finally, in order to show that Z_{1} = Z_{2}, one may notice that (see Figure 2). Similarly, one has and . This yields Z_{1} = Z_{2}.

Now, let P and Q be the feet of the perpendiculars from B and C onto AC and AB, respectively, and let R be the midpoint of BC (see Figure 3). Then R is the circumcentre of the cyclic quadrilateral BCPQ. Thus we obtain and , which yields . Similarly, we show that . Thus, all interesting triangles are similar to the triangle PQR.

Denote now by Z the common point of the circumcircles of APQ, BQR, and CPR. Let E_{1}F_{1}K_{1} and E_{2}F_{2}K_{2} be two interesting triangles. By the lemma, Z is the centre of any spiral similarity taking one of the triangles E_{1}F_{1}K_{1}, E_{2}F_{2}K_{2}, and PQR to some other of them. Therefore the triangles ZE_{1}E_{2} and ZF_{1}F_{2} are similar, as well as the triangles ZE_{1}F_{1} and ZPQ. Hence

Moreover, the equalities show that the point Z lies on the line AR (see Figure 4). Therefore the triangles AZP and ACR are similar, as well as the triangles AZQ and ABR. This yields

which completes the solution.

**Also Read : Shortlist Problem of geometry in IMO 2014**