## Saturday, November 28, 2009

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

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

Complete Problem 395

Level: High School, SAT Prep, College geometry

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

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

3. solution plz some1

4. in triangle ABE ang(ABE)=75;BE=AB/(2cos15)
with 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

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

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

ED = FC = a and EA = FB = a

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

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

8. See the drawing : Drawing

- 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

9. See diagram here.

- 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 QED

10. Let length of square = a & mid-pt of BC & AD be F & G respectively
tan15=EF/(a/2)
EF=atan15/2
EG=a(1-tan15/2)

11. Construct Equilateral Triangle BCF, F outside the square.

Triangles 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

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