Geometry Problem. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Friday, July 13, 2018

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

Subscribe to:
Post Comments (Atom)

Let be P the point Aq meet the circle Q.Join F to Q, F to P

ReplyDeleteThe triangle PFQ eqilateral => Q midpoint of PQ

=> r = 2/3 (R)

Let the side of the equilateral triangle ABC be 3 units and AE=AF=x => FC=3-x

ReplyDeleteAEF 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)

Consider the triangle ADF, since m(FAD)=m(FDA)=30

=> 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

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

ReplyDeleteSince 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

Q and O are on AD; AE = AF;

ReplyDeleteQE = 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.