Points D,E,F are on the sides BC,CA and AB, respectively which satisfy EF||BC, D1 is a point on BC, Make D1E1||DE,D1F1||DF which intersect AC and AB at E1 and F1, respectively. Make △PBC∼△DEF such that P and A are on the same side of BC. Prove that E,E1F1,PD1 are concurrent. [Edit by Darij: See my post #4 below for a possible correction of this problem. However, I am not sure that it is in fact the problem given at the TST... Does anyone have a reliable translation?]
Problem
Source: China Team Selection Test 2004, Day 2, Problem 1
Tags: geometry, geometric transformation, LaTeX, geometry solved
14.10.2005 18:22
Please post your solutions. Use LATEX please! Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in LATEX-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:
Happy Problem-Solving!
15.10.2005 13:05
orl wrote: Prove that E,E1F1,PD1 are concurrent. I don't understand.
18.11.2005 23:14
A hardy attempt to reconstruct the correct problem statement: Problem. Let D, E, F be points on the sides BC, CA, AB of a triangle ABC such that EF || BC. Let D1 be a point on the side BC, and let the parallels to the lines DE and DF through the point D1 meet the lines AC and AB at the points E1 and F1, respectively. Let the parallels to the lines DF and DE through the points B and C intersect at a point P. Prove that the lines EF, E1F1 and PD1 are concurrent if and only if the point D1 is the midpoint of the segment BC.
Solution. Since both of the lines BP and D1F1 were constructed as parallels to the line DF, we get a triple of parallel lines: BP∥D1F1∥DF. Similarly, CP∥D1E1∥DE. Let the lines BP and CP meet the line E1F1 at the points U and V, respectively. Let the lines PD1 and E1F1 intersect at R. Then, since BP∥D1F1, by Thales we have URF1R=PUD1F1=PRD1R, and since CP∥D1E1, Thales yields VRE1R=VPE1D1=PRD1R. Thus, URF1R=PUD1F1=VRE1R=VPE1D1. This equation splits into URF1R=VRE1R, what is equivalent to URF1R=RVRE1 or to URRV=F1RRE1, and into PUD1F1=VPE1D1, what becomes VPPU=E1D1D1F1. Since BC || FE, CP || ED and PB || DF (these are just the relations EF || BC, CP || DE and BP || DF, slightly rewritten), the triangles BCP and FED are homothetic. Thus, BPPC=FDDE. In other words, BPPC=DFED. Now we will use a fact which, if not known, is a nice little exercise for the reader (it should appear in good books, alas it doesn't...): Menelaos theorem for quadrilaterals. Let ABCD be a quadrilateral, and let X, Y, Z, W be four collinear points on its sidelines AB, BC, CD, DA, respectively. Then, AXXB⋅BYYC⋅CZZD⋅DWWA=1. Applying the Menelaos theorem for quadrilaterals to the quadrilateral UVCB with the collinear points R, P, D1, P on its sidelines UV, VC, CB, BU, respectively, we get URRV⋅VPPC⋅CD1D1B⋅BPPU=1. In other words: URRV⋅VPPU⋅BPPC⋅CD1D1B=1. Using the relations URRV=F1RRE1, VPPU=E1D1D1F1 and BPPC=DFED, this becomes F1RRE1⋅E1D1D1F1⋅DFED⋅CD1D1B=1. Now, again, this rewrites as F1RRE1⋅(E1D1ED:D1F1DF)⋅CD1D1B=1. Equivalently, (1) F1RRE1⋅(E1D1ED:F1D1FD)⋅CD1D1B=1. Now, since D1E1∥DE, Thales yields E1D1ED=CE1E1E. In other words, E1D1ED=CE1E1E⋅E1EAE⋅AECE. Another application of Thales using D1E1∥DE shows that CE1E1E=CD1D1D, so this becomes E1D1ED=CD1D1D⋅E1EAE⋅AECE. Similarly, F1D1FD=BD1D1D⋅F1FAF⋅AFBF. Thus, E1D1ED:F1D1FD=(CD1D1D⋅E1EAE⋅AECE):(BD1D1D⋅F1FAF⋅AFBF). Since EF || BC, by Thales we have AECE=AFBF, so that this simplifies to E1D1ED:F1D1FD=(CD1D1D⋅E1EAE):(BD1D1D⋅F1FAF) =CD1BD1⋅E1EAE⋅AFF1F=(−CD1D1B)⋅(−E1EEA)⋅(−AFFF1) =−CD1D1B⋅E1EEA⋅AFFF1. Hence, the equation (1) becomes F1RRE1⋅(−CD1D1B⋅E1EEA⋅AFFF1)⋅CD1D1B=1. In other words, −(F1RRE1⋅E1EEA⋅AFFF1)⋅(CD1D1B)2=1. Thus, (2) we have F1RRE1⋅E1EEA⋅AFFF1=−1 if and only if (CD1D1B)2=1. But according to the Menelaos theorem, applied to the triangle AE1F1 with the points R, E, F on its sidelines E1F1, AE1 and AF1, the equation F1RRE1⋅E1EEA⋅AFFF1=−1 is equivalent to the assertion that the points R, E, F are collinear, i. e. that the point R lies on the line EF; since the point R is defined as the point of intersection of the lines PD1 and E1F1, this assertion is thus equivalent to stating that the lines EF, E1F1 and PD1 are concurrent. On the other hand, the equation (CD1D1B)2=1 is equivalent to CD1D1B=1 (since CD1D1B=−1 is impossible, unless the point D1 is infinite, what we don't want to allow); and this is actually just the statement that the point D1 is the midpoint of the segment BC. Hence, the result (2) rewrites as follows: The lines EF, E1F1 and PD1 are concurrent if and only if the point D1 is the midpoint of the segment BC. And the problem is solved... Darij
19.11.2005 00:44
I do not see why D1 has to be the midpoint of BC. The configuration is projective. There should be no limitations on position of D1. The only changes needed is correcting typo to "line EF", and not "line E" and orientation of PBC as Darij did. ( Please, do not think that I translated from original Chineese text!) Thank you. M.T.
19.11.2005 05:05
Here is my solution(?) : Using central projection transform triangles ABC and PBC into triangles with a point D into middle of BC. This will put point P onto A angle median.Then move line EF until it becomes midline, P will be in the A angle vertex. Lines E1F1 and PD1 transform into diagonals of an inscribed parallelogram. They intersect on the mid-line. Thank you. M.T.
19.11.2005 12:16
armpist wrote: I do not see why D1 has to be the midpoint of BC. The configuration is projective. No, it isn't. There is a lot of parallels around in the problem statement. armpist wrote: Here is my solution(?) : Using central projection transform triangles ABC and PBC into triangles with a point D into middle of BC. This will put point P onto A angle median.Then move line EF until it becomes midline, P will be in the A angle vertex. Lines E1F1 and PD1 transform into diagonals of an inscribed parallelogram. They intersect on the mid-line. Not that I would understand it, but at least it is nice that you call it a "solution(?)", not a "solution". And as I said above, you can't just project everything expecting the parallel lines to remain parallel after the projection, and particularly you can't force the triangle DEF to be the medial triangle of the triangle ABC. Darij
19.11.2005 19:20
Dear Darij, In your wonderful post http://www.mathlinks.ro/Forum/topic-59919.html you wrote about a line through midpoint: "..... It is not undefined. The midpoint between an Euclidean point and an infinite point is the latter infinite point (the Euclidean point is "insignificant"). Only the midpoint between two infinite points is undefined. And yes, assertion (c) very well holds with the point X being infinite. ...... " So it is possible to send BC to infinity (preserving parallels) and EF (chasing BC in parallel fashion) only half way there? Some experts have found it being hysterically funny. But really, where are all the lines that are parallel to the line being on it's way to infinity? Well, I think differently: the more unexplained phenomena - the better. People tend to like it because of the amusement and entertainment components of it... Thank you. M.T.
19.11.2005 19:32
I have just read the official version of this problem. Here is my translation: Points D,E,F are on the sides BC,CA,AB of an acute-angled triangle ABC respectively, which satisfy EF//BC. D1 is a point on side BC(different from B,D,C), through D1 draw D1E1//DE,D1F1//DF, which intersect AC and AB at points E1,F1 respectively. Join E1F1. Construct,on the same side of A, triangle PBC, such that △PBC∼△DEF. Join PD1. Prove that EF,E1F1,PD1 are concurrent. I know very well this must be the worse translation in the world...
19.11.2005 19:41
mecrazywong wrote: I have just read the official version of this problem. Here is my translation: Points D,E,F are on the sides BC,CA,AB of an acute-angled triangle ABC respectively, which satisfy EF//BC. D1 is a point on side BC(different from B,D,C), through D1 draw D1E1//DE,D1F1//DF, which intersect AC and AB at points E1,F1 respectively. Join E1F1. Construct,on the same side of A, triangle PBC, such that △PBC∼△DEF. Join PD1. Prove that EF,E1F1,PD1 are concurrent. That's funny, since this means the official version is wrong... darij
19.11.2005 20:54
Hello guys, I think I see what is wrong. Darij put us in thinking that the triangle PBC can be constructed only by lines through B and C parallel to DE and DF. There is another triangle with vertex P still on the same line. And it has proper orientation. The problem is correct with only typo line E should be line EF. Now it all should work out, I just reduced the number of parallel lines for Darij. Thank you. M.T.
19.11.2005 22:08
armpist wrote: Hello guys, I think I see what is wrong. Darij put us in thinking that the triangle PBC can be constructed only by lines through B and C parallel to DE and DF. There is another triangle with vertex P still on the same line. And it has proper orientation. The problem is correct with only typo line E should be line EF. You mean that triangle PBC should be similar to triangle DEF (with corresponding letters for corresponding vertices)? Try it, it won't hold. Darij
19.11.2005 22:37
Dear Darij, PBC corresponds to FDE for direct similarity. I guess the problem statement just states similarity to make it harder to find the proper triangle. In my sketch it works out fine. The 3 lines intersect . Line EF if rather close to BC for angle D to be obtuse. If D and D1 on the same half of BC then PBC ~ EDF. Thank you. M.T.
22.11.2005 01:25
A vectorial solution (orientated segments). The fixed points: D, E, F, P⟹ the constant values d, k. The mobile points: D1, E1, F1, R⟹ the variable value x. AEEC=AFFB=k⟹¯|E=A+kC1+k, F=A+kB1+k|_. BDDC=d⟹¯|D=B+dC1+d|_. BD1D1C=x⟹¯|(1+x)D1=B+xC|_ (1). △PBC∼△DFE (D∈AP)⟹BPFD=ABAF=k+1k⟹ P=B+k+1k(D−F)⟹¯|k(1+d)P=−(1+d)A+(1+k)B+d(1+k)C|_ (2). EE1E1C=DD1D1C=D1−DC−D1=B+xC1+x−B+dC1+dC−B+xC1+x=x−d1+d⟹ E1=E+x−d1+dC1+x−d1+d=(1+d)E+(x−d)C1+x⟹ (1+x)E1=(1+d)A+kC1+k+(x−d)C⟹ ¯|(1+x)(1+k)E1=(1+d)A+(k+x+kx−d)C|_. FF1F1B=DD1D1B=D1−DB−D1=B+xC1+x−B+dC1+dB−B+xC1+x=d−xx(1+d)⟹ F1=F+d−xx(1+d)B1+d−xx(1+d)=x(1+d)A+kB1+k+(d−x)Bd(1+x)⟹ ¯|d(1+k)(1+x)F1=x(1+d)A+(kdx+kd+d−x)B|_. R∈EF∩E1F1⟺R=uE+(1−u)F=vE1+(1−v)F1⟺ ukx+k+x−d=1−ukdx+kd+d−x=1k(1+d)(1+x), v1=1−vd=11+d⟺ ¯|(1+k)(1+d)(1+x)R=(1+d)(1+x)A+(kdx+kd+d−x)B+(kx+k+x−d)C|_ (3). ========================================================================== D1∈PR⟺|01x−(1+d)1+kd(1+k)(1+d)(1+x)kdx+kd+d−xkx+k+x−d |=0 ⟺k(1+d)(x2−1)=0⟺x=1, i.e. D1 is the midpoint of the side [BC]. Remark. If the points P, A are on the same side of the line BC and △PBC∼△DEF, (PB∥DE, PC∥DF) then the official version is correctly !
01.02.2009 19:11
mecrazywong wrote: I have just read the official version of this problem. Here is my translation: Points D, E, F are on the sides BC, CA, AB of an acute-angled △ABC respectively, which satisfy EF∥BC. D1 is a point on side BC(different from B, D, C), through D1 draw D1E1∥DE,D1F1∥DF, which intersect AC and AB at points E1, F1 respectively. Join E1F1. Construct,on the same side of A, triangle PBC, such that △PBC∼△DEF. Join PD1. Prove that EF,E1F1,PD1 are concurrent. Yes, this is the official version appeared in the published book by the Chinese coach teams.