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

Labels:
angle,
congruence,
isosceles,
parallelogram,
scalene,
triangle

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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