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Geometry Difficulty 5.7 AIME, harder Prove it Romania

Let ABCABC be a triangle with AB>ACAB > AC. Point P(AB)P \in (AB) is such that ACP=ABC\angle ACP = \angle ABC. Let DD be the reflection of PP into the line ACAC and let EE be the point in which the circumcircle of BCDBCD meets again the line ACAC. Prove that AE=ACAE = AC.

Solution

Let QQ be the point in which the circumcircle of BCDBCD meets again the line ABAB. Then QEA=QBC=ECP\angle QEA = \angle QBC = \angle ECP, hence EQPCEQ \parallel PC. Moreover, ECP=ECD\angle ECP = \angle ECD implies QDQD is parallel to ECEC, hence EQ=CD=CPEQ = CD = CP. It follows that EQCPEQCP is a parallelogram, which leads to the conclusion.

Figure 1

Second solution. (Given in the contest.) If {O}=BDAC\{O\} = BD \cap AC, it is easy to prove that BABA and BOBO are isogonal in the angle EBC\angle EBC, therefore, in order to show that BABA is the median, one has to prove that BOBO is the symmedian. This follows readily by computation, using Steiner's Theorem.
Indeed, EBA=BACBEA=180ABCACBBDC=180ECDECBBDC=CBD\angle EBA = \angle BAC - \angle BEA = 180^\circ - \angle ABC - \angle ACB - \angle BDC = 180^\circ - \angle ECD - \angle ECB - \angle BDC = \angle CBD.
Triangles OEBOEB and ODCODC are similar, hence EOOD=EBCD\frac{EO}{OD} = \frac{EB}{CD}. (1)
Triangles OCBOCB and ODEODE are also similar, therefore OCOD=BCED\frac{OC}{OD} = \frac{BC}{ED}. (2)
We have ADC=APC=PBC+PCB=ACB=EDB\angle ADC = \angle APC = \angle PBC + \angle PCB = \angle ACB = \angle EDB, which shows that the triangles ACDACD and EBDEBD are similar. Hence ED=EBADACED = \frac{EB \cdot AD}{AC} and, as AD=CDACBCAD = \frac{CD \cdot AC}{BC} (from ΔACBΔAPCΔADC\Delta ACB \sim \Delta APC \sim \Delta ADC), it follows that ED=EBCDBCED = \frac{EB \cdot CD}{BC}.
From (1) and (2) we obtain EOOC=EBEDCDBC=(EBBC)2\frac{EO}{OC} = \frac{EB \cdot ED}{CD \cdot BC} = \left(\frac{EB}{BC}\right)^2, which shows that BOBO is indeed the symmedian.

Third solution. (Given in the contest by Andrei Mărginean.) Let a=EBAa = \angle EBA, x=ABCx = \angle ABC. We have that DCE=ACP=x\angle DCE = \angle ACP = x and EBC=a+x\angle EBC = a + x. As BB, CC, DD, EE are concyclic, it follows that CDE=180ax\angle CDE = 180^\circ - a - x, DEC=180EDCDCE=a=EBA\angle DEC = 180^\circ - \angle EDC - \angle DCE = a = \angle EBA. But AEAE is the perpendicular bisector of [PD][PD], hence PEA=DEC=EBA\angle PEA = \angle DEC = \angle EBA, which shows that the triangles APEAPE and AEBAEB are similar. It follows that AE2=ABAPAE^2 = AB \cdot AP. But triangles ACPACP and ABCABC are also similar, hence AC2=APAB=AE2AC^2 = AP \cdot AB = AE^2, and the conclusion follows immediately.

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