Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view the complete problem 1042.
Saturday, September 6, 2014
Geometry Problem 1042: Scalene Triangle, Isosceles, Equal Angles, Parallelogram, Congruence
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Let BC=k BE, then AB=k AD and AC=k AF.
ReplyDeleteLet S(P, k, α) be the homothetic rotation transformation,
with center P, scale factor k, and angle α.
Consider first S(B, k, −α) and then S(A, 1/k, α).
B→B→D
E→C→F
On the other hand, the combined transformation
S(A, 1/k, α)。S(B, k, −α) = T(v)
which is a translation by vector v.
[It is because (k)(1/k)=1 and (α)+(−α)=0]
Hence, vector v = vector BD = vector EF.
i.e. BEFD is a parallelogram.
Let DF cut BC at M and EF cut AB at N.
ReplyDeleteNote that ABC is the image of ADF in the spiral similarity transformation with
Rotation angle = α => ∠(DMB)= α => DF//BE
In the similar way we also get EF//BD so DFEB is a parallelogram
Tr.s AFC and ADB are similar hence AF / b = AD / c = p say.
ReplyDeleteSo AF = pb and AD = pc and since < DAF = < BAC it follows that Tr.s ADF and ABC are similar the respective sides being in the ratio of p
So DF = pa = BE since tr.s AFC and BEC are similar.
Similarly EF = BD = pc since Tr.s AFC and BEC are similar and Tr.s BEC and ABC are similar
Hence in quadrilateral BDFE the opposite sides are equal and hence BDFE is a parallelogram
Sumith Peiris
Moratuwa
Sri Lanka
Problem 1042
ReplyDeleteThe three equilateral triangles ADB ,BEC and AFC are similar to each other ,so AD/AB=BE/BC=AF/AC, But AD/AB=AF/AC or AD/AF=AB/AC with <DAF=α+<BAE=<BAC.
Then the triangle DAF is similar with triangle BAC so AD/AB=DF/BC=BE/BC.Therefore
DF=BE. And triangle FCE is similar with triangle ACB so FE/AB=CE/BC=(BE/BC=AD/AB) so
FE=DB.Therefore the DBEF is parallelogram.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE