Let ABC be a triangle, and let D be a point on the internal angle bisector of BAC. Let x be the ellipse with foci B and C passing through D, y be the ellipse with foci A and C passing through D, and z be the ellipse with foci A and B passing through D. Ellipses x and z intersect at distinct points D and E, and ellipses x and y intersect at distinct points D and F. Prove that AD bisects angle EAF. Andrew Carratu
Problem
Source: ELMO Shortlist 2024 G8
Tags: geometry, ellipse, conics, ELMO Shortlist
23.06.2024 02:07
Let A=(−a,0), B=(b,d), C=(c,e), and D=(0,0). Let p=−a=AD, q=√b2+d2=BD, and r=√c2+e2=CD. We have AE+BE=p+q and BE+CE=q+r, so AE=(p+q−BE) and CE=(q+r−BE). Therefore, (q+r)AE2+(p−r)BE2−(p+q)CE2 is constant and describes the equation of line DE. Therefore, if E=(x,y), this is (q+r)(x+a)2+(p−r)(x−b)2−(p+q)(x−c)2+(q+r)y2+(p−r)(y−d)2−(p+q)(y−e)2so the equation of DE is x(aq+ar+br+cq+ac−ab)+y(ae−ad+dr+eq)=0.The equation of the ellipse with foci B and C passing through D and E is BE=q+r−CEBE2=(q+r)2+CE2−2(q+r)CE(BE2−CE2−(q+r)2)2=4(q+r)2CE2(BE2−CE2)2−2(q+r)2(BE2+CE2)+(q+r)4=0(−2bx+2cx−2dy+2ey+b2−c2+d2−e2)2−2(q+r)2(2x2−2(b+c)x+2y2−2(d+e)y+b2+c2+d2+e2)+(q+r)4=04x2((c−b)2−(q+r)2)+4y2((e−d)2−(q+r)2)+8xy(c−b)(e−d)+4x((c−b)(q2−r2)+(b+c)(q+r)2)+4y((e−d)(q2−r2)+(d+e)(q+r)2)=0(x(c−b)+y(e−d))2−(q+r)2(x2+y2)+2(q+r)(x(qc+rb)+y(qe+rd))=0. Since D lies on the angle bisector, (a+b)e+(a+c)d=0, so a=−be+cdd+e. Then, let m=aq+ar+br+cq+ac−ab and n=ae−ad+dr+eq. If x=Cn and y=−Cm, then we get C((n(c−b)−m(e−d))2−(q+r)2(m2+n2))=−2(q+r)(n(qc+rb)−m(qe+rd))C((r(cd−be)+q(cd−be)−a(q+r)(e−d))2−(q+r)2(m2+n2))=−2a(q+r)((e−d)(qc+rb)−(q+r+c−b)(qe+rd))C(q+r)2((cd−be−a(e−d))2−m2−n2)=−2a(q+r)(q(be−cd)+r(be−cd)−(q+r)(qe+rd))C((cd−be−a(e−d))2−m2−n2)=2a(−be+cd+qe+rd). Therefore, if s=(cd−be−a(e−d))2−m2−n2 and t=−be+cd+qe+rd, then C=2ats, so x=2atns and y=−2atms. The slope of AE is −2atm2atn+as=−2tm2tn+s. Then, we have 2tn+s=2tn+(cd−be−a(e−d))2−m2−n2=2tn+(t−n)2−m2−n2=t2−m2. Therefore, the slope of AE is −2tmt2−m2. Now, we have m=q(a+c)+r(a+b)+a(c−b)=qce−bed+e+rbd−cdd+e+(b−c)(be+cd)d+e=b−cd+e(−qe+rd+be+cd).Then, if t′=be−cd+qe+rd and m′=c−bd+e(qe−rd+be+cd), the slope of AF is −2t′m′t′2−m′2, so we need to show −2tmt2−m2=−−2t′m′t′2−m′2or tm−mt=−t′m′+m′t′.We claim tm=−t′m′, which implies the equality. This is equivalent to (−be+cd+qe+rd)c−bd+e(qe−rd+be+cd)=−(be−cd+qe+rd)b−cd+e(−qe+rd+be+cd)(−be+cd+qe+rd)(qe−rd+be+cd)=(be−cd+qe+rd)(−qe+rd+be+cd)(qe+cd)2−(be−rd)2=(rd+be)2−(qe−cd)2(rd+be)2+(rd−be)2=(qe+cd)2+(qe−cd)2r2d2+b2e2=q2e2+c2d2d2(c2+e2)+b2e2=(b2+d2)e2+c2d2. This is true, which implies that the slopes of AE and AF have quotient −1. Therefore, AE and AF are reflections over AD, so AD bisects EAF.
23.06.2024 02:38
how long did that take @above
23.06.2024 02:48
Rename D as I and let D be the intersection of y and z other than I. Let γA=(A,AI). Define γB and γC similarly. Claim. D is the center of the circle (D) which is internally tangent to γA and externally tangent to γB and γC. Proof. It suffices to show that AI−AD=BD−BI but is given since {I,D}∈z. Define (E) and (F) similarly. Let P,Q,R be reflections of I over BC, CA, AB. Now invert the diagram in I. The circles γA, γB, and γC form a triangle P′Q′R′ and the circles (D)′, (E)′, and (F)′ are the excircles of this triangle. Let γA touch (E) and (F) at Y and Z. then clearly A, Y, E are colinear and so are A, Z, F. Under inversion, Y and Z map to the tangency points Q′R′ with the excircles. Since I is on the A-angle bisector, AI is the perpendicular bisector of QR, and thus is the perpendicular bisector of Q′R′. Thus IY′=IZ′, so IY=IZ. Hence ∠YAI=∠ZAI. Extra: If I is incenter, then AD,BE,CF concur.
23.06.2024 04:59
bachkieu wrote: how long did that take @above around 2 hours