Geometry Problem

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

Click the figure below to view the complete problem 969.

## Tuesday, January 28, 2014

### Geometry Problem 969: Equilateral Triangle, Rectangle, Common Vertex, Right Triangles Areas

Labels:
area,
common vertex,
equilateral,
rectangle,
right triangle,
triangle

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Let z(P) be the complex number representing P.

ReplyDeleteLet z(C)=0, z(B)=−a, z(E)=−a−bi.

Then

z(F) = (−a−bi)(1/2 + √3/2 i) = (−1/2 a + √3/2 b) + (−√3/2 a − 1/2 b)i

z(D) = (−√3/2 a − 1/2 b)i

z(A) = −a + (−√3/2 a − 1/2 b)i

Thus

BC = a

BE = b

AE = √3/2 a − 1/2 b

DF = 1/2 a − √3/2 b

AF = 1/2 a + √3/2 b

CD = √3/2 a + 1/2 b

S₁ = 1/2 × AE × AF = √3/8 (a^2 − b^2) + 1/4 ab

S₂ = 1/2 × BC × BE = 1/2 ab

S₃ = 1/2 × CD × DF = √3/8 (a^2 − b^2) − 1/4 ab

Hence,

S₁ = S₂ + S₃

Use the area formula of triangle:

ReplyDeleteA=c^2/(2[cotA+cotB])

In this case, c^2/2 cancel out of the equation S2+S3=S1. What we are left with is a simple trigonometric identity which is always true.

Please if you want explain it more.

ReplyDeletethan you.

using trigonometry should always be the last resort. There must be another, direct, proof, using geometric reasoning.

ReplyDeleteFie E(0,e),F(f,o)atunci C(f/2+e√3/2; e/2+f√3/2)pentru a arata ca S₁ = S₂ + S₃ trebuie sa aratam ca

ReplyDeleteAE.AF=FD.CD+BE.BC adica ef= (f/2+e√3/2-f)(e/2+f√3/2)+(e/2+f√3/2-e)(f/2+e√3/2) care este adevarata

Think about pythagoras triangle BEC and DFC

ReplyDeletei think their might be wrong in the question'

ReplyDeleteTriangles EBC and FDC are congruent (RHS congruency)

ReplyDelete=> BE = FD = x let, AE = AF = a − x where a is length of the side of square ABCD.

By Pythagoras Theorem, S2 + S3 = ax/2 + ax/2 = ax

S1 = (1/2)(a − x)^2

Now notice that EF = CF = CE

so (a − x)^2 + (a − x)^2 = a^2 + x^2

so (a − x)^2 = 2ax

so (1/2)(a − x)^2 = ax

=> S1 = S2 + S3

To Abdul Problem 969:

DeletePlease more details to your first statement:

Triangles EBC and FDC are congruent (RHS congruency)

Thanks

BC = CD, CE = CF, m(EBC) = 90° = m(FDC)

ReplyDeleteTo Abdul: In your statement BC = CD

DeleteABCD is a rectangle (not necessary a square) therefore BC is not equal CD

Thanks.

Is there a simple geometric proof for this Antonio?

ReplyDeleteLet M be midpoint of EC so that AEMF and CDFM are cyclic quadrilaterals from which we can easily show that ADM is equilateral

ReplyDeleteLet BE = a, AE = b, AF = c, FD = d and FE = x

Apply Ptolemy to the cyclic quad AEMF we have

Sqrt3/2 xb + x/2 c = x(c+d) from which

b = (c+2d)/sqrt3.....(1)

Similarly from cyclic quad CDFM we have upon using Ptolemy and simplifying as before

a+b = (2c+d)/sqrt3 from which using (1) we have

a =(c-d)/sqrt3......(2)

So S 2 + S3 = 1/2a(c+d) + 1/2(a+b)d which upon substituting for a and b from (1) and (2)

= (1/2sqrt3){(c-d)(c+d) + (2c+d)d} = (1/2sqrt3)(c+2d)c = bc/2 = S1

Sumith Peiris

Moratuwa

Sri Lanka

https://goo.gl/photos/VRMyqEUhTyU6DB8e6

ReplyDeleteLet CD=AB=a and BC=AD=b

Draw equilateral triangle PDC

Let PE cut BC at F ( see sketch)

Triangle CDF congruent to CPE ( case SAS)

Triangles CPF and EBF are 30-60-90 triangles

With simple algebra calculation we have

EB= 2a-b.sqrt(3)

FD=EP=2b-ạsqrt(3)

And S2= ½.b(2a-b.sqrt(3))

S3= 1/2a(2b-ạsqrt(3))

S1= ½(b. sqrt(3)-a).(a .sqrt(3)-b)

Verify that S1=S2+S3

Angle chase and you have (x,90-x,90) (30+x,60-x,90) and (30-x, 60+x, 90) triangles.

ReplyDeleteTake them and line them all up to share the same hypotenuse. such that the that

C is the left corner of BCE, F is the left corner of CDF and E is the left corner of AEF,

Angle chase again and you'll now find six 30-60-90 triangles!

That's enough to get formulas for all three just in terms of BE and DF working through the sides.

See: https://pbs.twimg.com/media/DGXjr2YUQAAZPcv.jpg

From there its fairly simple to verify the equality.