## Friday, July 13, 2018

### Geometry Problem 1362: Equilateral Triangle Inscribed in a Circle, Lunula, Ratio of Areas, Tangent, Sketch, iPad Apps

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Let be P the point Aq meet the circle Q.Join F to Q, F to P
The triangle PFQ eqilateral => Q midpoint of PQ
=> r = 2/3 (R)

2. Let the side of the equilateral triangle ABC be 3 units and AE=AF=x => FC=3-x
AEF is equilateral (since AE=AF) and AEQF and ABDC are similar cyclic quadrilaterals.
Also AOQD are collinear
Since m(EQF)=120=>m(EDF)=60 --------(1)
=> m(DFC)=60 -------(2)
and DF=AF=AE=EF and EDF is an equilateral triangle --------(3)
Since m(AFQ)=m(ACD)=90, the triangle DFC is a 30-60-90 triangle
=> DF/FC=2
=> x/(3-x)=2
=> x=2
So the ratios of the radius of the circles O and Q are 3/2
Areas = 9/4
=>Ar.of Blue lunula/Ar. of Yellow region=5/4

3. Let circle Q be of radius r and circle O be of radius R.

Since QA & OA both bisect <BAC it follows that A,O,Q,D are collinear points.

Since QAF is a 30-60-90 Tr.,
AQ = 2QF.
So R +(R-r) = 2r
Hence R/r = 3/2

So A(Circle O) / A(Circle Q) = 9/4

Therefore S/S1 = 5/4

Sumith Peiris
Moratuwa
Sri Lanka

4. Q and O are on AD; AE = AF;
QE = QF; thus ∠ EOA = 90°.
∠ EBO = 30° = ∠OAE. Thus ∠ AEO = 60°,
but ∠ AEQ = 90°, thus ∠OEQ = 30°.
BE = .5 * OB * sec(30°) = OE
QE = OE * sec(30°)
=.5 * OB * sec(30°) * sec(30°)
= OB * 2 / 3. Thus S / S_1 = 5 / 4.

5. See the drawing

- Define R radius of circle O and r radius of circle Q
- By construction and symmetry A, O, Q, D are collinear
=>DE=DF, AE=AF
- AD angle bisector of ∠BAC
- D symmetric of A by O => ED=EA=FD=FA : AEDF is a diamond
=>O middle of EF and EF ⊥ AD