Saturday, January 18, 2014

Geometry Problem 960. Quadrilateral, Trisection, Sides, Congruence, Similarity, Triangle

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 960.

1. Join PN, BD, EH.

In ΔABD, AP/AB=AN/AD=1/3, so PN/BD=1/3.
In ΔCBD, CE/CB=CH/CD=2/3, so EH/BD=2/3.
Thus ΔRPN~ΔRHE with PR/RH=1/2.

Similarly, UH/PU=1/2. Thus PR=RU=UH.
Similar results follow.

2. Let [AB] denote vector AB.

[ER]+[RH]
= [EC]+[CH]
= 2/3 ([BC]+[CD])
= 2 ([PA]+[AN])
= 2 ([PR]+[RN])

Since PRH and ERN are non-parallel line,
so [RH]=2 [PR], [ER]=2 [RN].

Similarly, [PU]=2 [UH]. Thus, PR=RU=UH.
Similar results follow.

1. Ah, you used vectors too! And I thought I was being original...

3. We use vector additions. Single capital letters will denote vectors. P = x A + y B + z C + w D will be written as P = (x, y, z, w). A, B, C, D are on a common plane (ie. the tetrahedron ABCD has volume of zero). AB means “length of AB” unless otherwise specified; most importantly, it does NOT mean “A times B”.

P = (1/3) (2A + B) because PA : PB = 1 : 2 and A, P, B are collinear. This is simply division of an interval in a given ratio, but with vectors rather than cartesian coordinates.

Hence P = (2/3, 1/3, 0, 0).
Similarly,
Q = (1/3, 2/3, 0, 0), E = (0, 2/3, 1/3, 0), F = (0, 1/3, 2/3, 0), G = (0, 0, 2/3, 1/3), H = (0, 0, 1/3, 2/3), N = (2/3, 0, 0, 1/3), M = (1/3, 0, 0, 2/3)

Define S’ as a point on line QG such that S’ = (1/3) (2Q + G), which means that S’Q : S’G = 1 : 2. Hence S’ = (2/9, 4/9, 2/9, 1/9).
Define S” as point on EN such that S” = (1/3) (2E + N). Hence S” = (2/9, 4/9, 2/9, 1/9).
This proves that S’ = S”. Hence, S, which is intersection of EN and QG and S must therefore divide EN and QG each in the ratios mentioned above.
Similarly, (1/3) (2G + Q) = T’, (1/3) (2F + M) = T”, both are equal to (1/9, 2/9, 4/9, 2/9), hence T’ = T”.
Similarly, (1/3) (2P + H) = R’, (1/3) (2N + E) = R”, both are equal to (4/9, 2/9, 1/9, 2/9), hence R’ = R”.
Similarly, (1/3) (2H + P) = U’, (1/3) (2M + F) = U”, both are equal to (2/9, 1/9, 2/9, 4/9), hence U’ = U”.

So we have QS : SG = 1 : 2, QT : TG = 2 : 1, which directly implies that QS : ST : TG = 1 : 1 : 1, which means QS = ST = TG.
Other conclusions rise similarly.

It’s interesting to note that the tetrahedron’s zero volume did not need to be invoked, which implies that ABCD need not be on a common plane.