## Thursday, May 20, 2010

### Problem 458: Square, Semicircle, Circular Sector, Internal Common Tangent, Measurement

Geometry Problem
Click the figure below to see the complete problem 458 about Square, Semicircle, Circular Sector, Internal Common Tangent, Measurement.

See also:
Complete Problem 458

Level: High School, SAT Prep, College geometry

#### 9 comments:

1. Let M be the midpt. of AB. Draw GN perpendicular to AD. AMEF is concyclic. Ang. AME = Ang. EFN or GN/FN=AD/AM or FN = a/2 and FG^2 = FN^2 + GN^2 = a^2 + a^2/4 = 5a^2/4 or FG = √5(a/2)
Ajit

2. draw GH//CD
▲FED ~ ▲AOD ( 90° & common ADO)
=> AOD = GFH
=>▲AOD = ▲FGH ( right tr, GH = AD, AOD = GFH )
=> FH = a/2
=> FG² = a² + a²/4

FG = a√5 /2
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3. Shift FG all the way such that G coincides with C. Since this is only translation distance is preserved.

Suppose the semicircle has center O. Draw OD.

I now claim that Triangle F'CD is congruent to triangle ODA.

Proof: Angle AOD is congruent to angle ODC since AB and CD are parallel. Angle A and angle FDC are right angles, therefore they are also congruent. AD is congruent to CD by properties of squares. Therefore ASA will show that the two triangles are congruent.

This then implies FG = OD.

OD = a (sqrt 5)/2 as can be easily shown with Pythagorean Theorem. Therefore FG = a (sqrt 5)/2

4. Let say the centre of cir.AEB as O.
AO=a/2, AD=a
∴ OD= a/2 √5 (Pythagoras theorem)
Construct GX ⊥AD.

OEF=OAF=90°(tangent of both circles)
∴OEAF is a cyclic quadrilateral.
In⊿AOD and ⊿GFX,
AD=GD,
AOD=GFX (OEAF is a cyclic quadrilateral.)
∠OAD=∠GXF=90°
∴⊿AOD ≡ ⊿GFX.

So, OD=GF (by congruency)
∴FG= a/2 √5

5. Let X be the mid point of AB
Drop a perpendicular from G to meet AD at Y
Tr.s FGY & AXD are congruent ASA and so FG = DX = (1/2)a.sqrt5

Sumith Peiris
Moratuwa
Sri Lanka

6. Let GE=GB=x and FE=y
Let O be midpoint of AB
Since m(OEG)=m(DEG)=90=> O,E,D are collinear
we know OD=OE+ED=a/2+ED=√5a/2 (apply pythogras to AOD)
=>ED=(√5-1)a/2 -----------(1)

Observe that triangles AOD and EFD are similar (AAA)
=>FE=y=ED/2=(√5-1)a/4 (from 1 ) --------(2)

Connect OG,GD and form the triangle OGD where OD (=√5a/2) is base and GE (x) is height
Area of square = AOD+OBG+GCD+DGO
=> a^2=a^2/4+ax/4+a(a-x)/2+√5ax/4
=> x=a/(√5-1)-----------(3)

GF=(2)+(3)=√5a/2 Q.E.D

7. rv.littleman@sciences.heptic.frMarch 18, 2021 at 1:53 AM

See the drawing

Define O middle of AB, r=ED
AD^2+AO^2=DO^2
a^2+(a/2)^2=(a/2+r)^2
a^2+a^2/4=a^2/4+r^2+ar
r^2+ar-a^2=0
∆=a^2+4a^2=5a^2
r>0 => r=(a.sqr(5)-a )/2
Draw a square symetryc of ABCD by O
=>D becomes D’, C becomes C’, E becomes E’
=> C’D’BA is a square, D, E, O, E’, D’ are colinear
=> EG=E’G’ and FE=F’E’
FG ꓕ DD’ => F’G’ ꓕ DD’
∆ADO is similar to ∆E’DG’ and EDF (aa)
=>AD/AO=2=E’D/E’G’=ED/EF
E’D=a+r = 2 EG and ED=r=2 EF
=>EF+EG=(r+a+r)/2=r+a/2=(a.sqr(5)-a )/2+a/2= a.sqr(5)/2
Therefore FG= a.sqr(5)/2

1. rv.littleman@sciences.heptic.frMarch 19, 2021 at 1:36 AM

With the help of my friend Greg, a synthetic solution :

Define O middle of AB, OD=a.sqr(5) /2
Draw a square symetric of ABCD by O
=>D becomes D’, C becomes C’, E becomes E’
D, E, O, E’, D’ are collinear, EG=E’G’, FE=F’E’ and FG ꓕDD’ => F’G’ ꓕDD’
∆ADO is similar to ∆E’DG’ and EDF (aa) =>AD/AO=2=E’D/E’G’=ED/EF
FG=EF+EG=(E’D+ED)/2=ED+a/2=OD = a.sqr(5)/2
Therefore FG=a.sqr(5)/2

8. See diagram here

Let O middle of AB
Rotate ∆ADO clockwise by π/2 around D to produce ∆A’D’O’ with A’=C and D’=D
OD ꓕ FG => by construction O’D // FG => DFGO’ is a parallelogram and FG = DO’ = DO = a.sqrt(5)/2 QED