Maths Olympiad Prep

Library / /480 of 520

Geometry Difficulty 7.4 National olympiad, round 2 Prove it

Let ω1,ω2\omega_{1}, \omega_{2} be two circles tangent to each other at a point TT, such that ω1\omega_{1} is inside ω2\omega_{2}. Let MM and NN be two distinct points on ω1\omega_{1}, different from TT. Let [AB] and [CD] be two chords of the circle ω2\omega_{2} passing through MM and NN respectively. Suppose that the segments [BD], [AC][A C], and [MN][M N] intersect at a point KK.
Show that (TK) is the bisector of the angle MTN^\widehat{M T N}.

Solution

Let EE and FF be the points of intersection of (TM)(TM) and (TN)(TN) with ω2\omega_{2}, other than TT. Since ω1\omega_{1} and ω2\omega_{2} are tangent at TT, the homothety centered at TT that maps ω1\omega_{1} to ω2\omega_{2} also maps MM to EE and NN to FF. This implies that

TMTN=MENF \frac{\mathrm{TM}}{\mathrm{TN}}=\frac{\mathrm{ME}}{\mathrm{NF}}

Furthermore, the power of points MM and NN with respect to ω2\omega_{2} is respectively AM⋅MB=TM⋅MEA M \cdot M B=T M \cdot M E and CN⋅ND=TN⋅NFC N \cdot N D=T N \cdot N F. This shows that

TM2TN2=TM⋅METN⋅NF=AM⋅MBCN⋅ND \frac{\mathrm{TM}^{2}}{\mathrm{TN}^{2}}=\frac{\mathrm{TM} \cdot \mathrm{ME}}{\mathrm{TN} \cdot \mathrm{NF}}=\frac{\mathrm{AM} \cdot \mathrm{MB}}{\mathrm{CN} \cdot \mathrm{ND}}

On the other hand, in triangles AMKA M K and DNKD N K, the law of sines indicates that

AM⋅sin⁡(MAK^)=MK⋅sin⁡(AKM^) and that DN ⋅sin⁡(NDK^)=NK⋅sin⁡(DKN^). A M \cdot \sin (\widehat{M A K})=M K \cdot \sin (\widehat{A K M}) \text { and that DN } \cdot \sin (\widehat{N D K})=N K \cdot \sin (\widehat{D K N}) .

Since A,B,CA, B, C and DD are concyclic, we also have sin⁡(BAC^)=sin⁡(BDC^)\sin (\widehat{B A C})=\sin (\widehat{B D C}), so that

sin⁡(MAK^)=sin⁡(BAC^)=sin⁡(BDC^)=sin⁡(KDN^) \sin (\widehat{M A K})=\sin (\widehat{B A C})=\sin (\widehat{B D C})=\sin (\widehat{K D N})

It follows that AMDN=MK⋅sin⁡(AKM^)NK⋅sin⁡(DKN^)\frac{A M}{D N}=\frac{M K \cdot \sin (\widehat{A K M})}{N K \cdot \sin (\widehat{D K N})} and, similarly, that BMCN=MK⋅sin⁡(BKM^)NK⋅sin⁡(CKN^)\frac{B M}{C N}=\frac{M K \cdot \sin (\widehat{B K M})}{N K \cdot \sin (\widehat{C K N})}. Since the angles BKM^\widehat{B K M} and DKN^\widehat{D K N} on one hand, and AKM^\widehat{A K M} and CKM^\widehat{C K M} on the other, are vertically opposite, hence equal, we deduce that

AM⋅MBDN⋅NC=MK2NK2 \frac{A M \cdot M B}{D N \cdot N C}=\frac{M K^{2}}{N K^{2}}

We conclude that TM ⋅NK=TN⋅MK\cdot N K=T N \cdot M K.
The law of sines in triangles TKM and TKN also indicates that

TM⋅sin⁡(KTM^)=KM⋅sin⁡(MKT^) and that TN⋅sin⁡(KTN^)=KN⋅sin⁡(NKT^) \mathrm{TM} \cdot \sin (\widehat{\mathrm{KTM}})=\mathrm{KM} \cdot \sin (\widehat{\mathrm{MKT}}) \text { and that } \mathrm{TN} \cdot \sin (\widehat{\mathrm{KTN}})=\mathrm{KN} \cdot \sin (\widehat{\mathrm{NKT}})

Since MKT^\widehat{M K T} and NKT^\widehat{N K T} are supplementary, they have the same sine, so that

sin⁡(KTM^)sin⁡(KTN^)=KM⋅sin⁡(MKT^)⋅TNTM⋅KN⋅sin⁡(NKT^)=1 \frac{\sin (\widehat{K T M})}{\sin (\widehat{\mathrm{KTN}})}=\frac{\mathrm{KM} \cdot \sin (\widehat{\mathrm{MKT}}) \cdot \mathrm{TN}}{\mathrm{TM} \cdot \mathrm{KN} \cdot \sin (\widehat{\mathrm{NKT}})}=1

which proves that (TK) is the angle bisector of MTN^\widehat{M T N}.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.