## Sunday, December 27, 2009

### Problem 410: Two Regular Pentagons, Angle

Proposed Problem
Click the figure below to see the complete problem 410 about Two Regular Pentagons, Angle. Complete Problem 410
Level: High School, SAT Prep, College geometry

1. http://geometri-problemleri.blogspot.com/2009/12/problem-58-ve-cozumu.html

2. the two pentagons are similar ,by the similarity s,with center D,angle (108+m(CDF)),ratio GH/AB
s(B)=H;s(A)=G
x=108+m(CDF)
.-.

3. To Anonymous:

4. in the complex plane we can choose O center from the circumcircle of ABCDE as origin,affix of D
zD=1,then:
zC= exp(i2pi/5);zB= exp(i4pi/5);zA= exp(i6pi/5)
the similarity s can be written: z'=az+b
s(D)=D b=1-a
zG=azA+1-a;zH=azB+1-a
ang(HMG)=arg((zG-zA)/(zH-zB))
=arg[(exp(i4pi/5)-1)/(exp(i6pi/5)-1)]
=arg[exp(ipi/5).sin(2pi/5)/sin(3pi/5)]=pi/5
in degrees
x=180-36=144
.-.

5. The problem is simple if you assume that there is an answer to the problem. It is posed in a way that suggests that the answer is independent of the angle between the sides of the pentagon. If I assume that there is a unique answer to the problem, then I can orient the pentagons any way I choose without loss of generality. So I choose to orient them so that angle x occurs at D. Then x is simply the exterior angle of the well-know 36-72-72 triangle, and thus must measure 144 degrees.

6. http://img577.imageshack.us/img577/1528/problem410.png

Note that triangles ADG and BDH are congruence ( case SAS)
∠ (XMD)= ∠ (MHD) => quadrilateral MGHD is cyclic
And ∠ ( GMX)= ∠ (GDH) = 36 ----- ( both angles face same arc GH)
And x= 180-36=180

7. Problem 410

The circle described in both pentagons intersect at point Ν , then <ANB=36=<ANE=
=<END=<DNJ=<JNH ,but <BNG=36+36+36+36+36=180.Therefore section B, N, C and A, N, H is collinear so points M, N coincide.So <BMG=144.(<GMH=36).
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE