Proposed Problem

Click the figure below to see the complete problem 395 about Square, 15 Degree, Equilateral triangle, Congruence.

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Complete Problem 395

Level: High School, SAT Prep, College geometry

## Saturday, November 28, 2009

### Problem 395: Square, 15 Degree, Equilateral triangle

Labels:
15 degree,
congruence,
equilateral,
square,
triangle

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http://www.wolframalpha.com/input/?i=%281-1%2F2*tg%2815%29%29%5E2%2B1%2F4

ReplyDeletehttp://geometri-problemleri.blogspot.com/2009/11/problem-52-ve-cozumu.html

ReplyDeletesolution plz some1

ReplyDeletein triangle ABE ang(ABE)=75;BE=AB/(2cos15)

ReplyDeletewith the law of cosine

AE²=AB²+BE²-2AB.BE cos75=

AB²(1+(1/(2cos15)²-cos75/cos15)= AB² then

cos(75)=cos(45+30)=(sqr6-sqr2)/4

cos(15)=cos(45-30)=(sqr6+sqr2)/4

same result for CE

CE=AE=AD

ADE is equilateral

build on each side of the square (from the inside) isosceles triangle with 15 degrees..from there its easy..

ReplyDeleteConsider the equilateral triangle BCF at the exterior of the square ABCD say of side a. The quandrilateral FEDC and FEAB are rhombuses, therefore

ReplyDeleteED = FC = a and EA = FB = a

(This holds because ang(FEC) = 75 = ang(ECF), so FE = FC = a and FE parallel to CD = a)

See

ReplyDeletehttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=472125

See the drawing : Drawing

ReplyDelete- ABCD is a square

- Define J as the rotation of E by 90° from the center of the square

- E becomes J => BE=CE=CJ=DJ and ΔECJ is equilateral

- Define F the intersection of DJ et EC

- EDA=60°

- By symmetry, <EAD=60°

Therefore ΔEAD is equilateral

See diagram

ReplyDeletehere.- E can be uniquely defined as the intersection of the perpendicular bisector to BC and the ray from B forming a 15° angle with BC.

- Make F the intersection of circular arcs BD with center A and AC with center D.

- AD = AF = DF ⇒ ∆ADF is equilateral ⇒ F is on the perpendicular bisector to BC (same as that to AD) and ∠BAF = 30°.

- But on circular arc BD, ∠BAF = 30° ⇒ ∠CBF = 15° since CB is tangent in B to arc BD.

- So F = E and

∆AED is equilateral QEDLet length of square = a & mid-pt of BC & AD be F & G respectively

ReplyDeletetan15=EF/(a/2)

EF=atan15/2

EG=a(1-tan15/2)

tan<EAD=(1-tan15/2)/(1/2)

tan<EAD=2-tan15

tan<EAD=2-(2-sqrt3)

tan<EAD=sqrt3

<EAD=60

Construct Equilateral Triangle BCF, F outside the square.

ReplyDeleteTriangles CEF, BEF and ABE are all equilateral SAS (CF = BF = BA, CE = BE, included angle is 75 for all 3 triangles)

So Triangles BFE and EFC are isosceles 30-75-75) and so AEB is also isosceles and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

I should have said ,"Triangles CEF, BEF and ABE are all all congruent......." not equilateral

Delete