# 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 (E1, F1) and (E2, F2 ) are interesting. Prove that

$\frac{E_1.E_2}{AB}=\frac{F_1.F_2}{AC}$

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 $\angle KEF=\angle KFE=\angle A$, which also means that the circumcircle ω1 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 $\angle KNS=\angle MNS=\angle LNT$. 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

$\angle KGE=\angle KNS=\angle SNM=\angle KLA=180^0-\angle KSA$

(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 $\angle KEF=\angle KSG=180^0-\angle TKS=\angle A$. Since KE = KF, we also have $\angle KFE=\angle KEF=\angle A$.

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 E1 and E2 on the side AC, two points F1 and F2 on the side AB, and two points K1 and K2 on the side BC, in a way that the triangles E1F1K1 and E2F2K2 are similar. Then the six circumcircles of the triangles AEiFi, BFiKi and CEiKi (i = 1, 2) meet at a common point Z. Moreover, Z is the centre of the spiral similarity that takes the triangle E1F1K1, to the triangle E2F2K2.

Proof.

Firstly, notice that for each i = 1, 2, the circumcircles of the triangles AEiFi, BFiKi, and CKiEi have a common point Zi by Miquel’s theorem. Moreover, we have

$\measuredangle (Z_iF_i,Z_iE_i)=\measuredangle (AB,CA),\measuredangle (Z_iK_i,Z_iF_i)=\measuredangle (BC,AB),\measuredangle (Z_iE_i,Z_iK_i)=\measuredangle (CA,BC)$.

This yields that the points Z1 and Z2 correspond to each other in similar triangles E1F1K1 and E2F2K2. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.

Finally, in order to show that Z1 = Z2, one may notice that $\measuredangle (AB,AZ_1)=\measuredangle (E_1F_1,E_1Z_1)=\measuredangle (E_2F_2,E_2Z_2)=\measuredangle (AB,AZ_2)$ (see Figure 2). Similarly, one has $\measuredangle (BC,BZ_1)=\measuredangle (BC,BZ_2)$ and $\measuredangle (CA,CZ_1)=\measuredangle (CA,CZ_2)$. This yields Z1 = Z2.

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 $\angle APQ=\angle B$ and $\angle RPC=\angle C$, which yields $\angle QPR=\angle A$. Similarly, we show that $\angle PQR=\angle A$. 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 E1F1K1 and E2F2K2 be two interesting triangles. By the lemma, Z is the centre of any spiral similarity taking one of the triangles E1F1K1, E2F2K2, and PQR to some other of them. Therefore the triangles ZE1E2 and ZF1F2 are similar, as well as the triangles ZE1F1 and ZPQ. Hence

$\frac{E_1E_2}{F_1F_2}=\frac{ZE_1}{ZF_1}=\frac{ZP}{ZQ}$

Moreover, the equalities $\angle AZQ=\angle APQ=\angle ABC=180^0-\angle QZR$ 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

$\frac{ZP}{ZQ}=\frac{ZP}{RC}.\frac{RB}{ZQ}=\frac{AZ}{AC}.\frac{AB}{AZ}=\frac{AB}{AC}$

which completes the solution.

Also Read : Shortlist Problem of geometry in IMO 2014

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