recursal/Fanatic-Fandom huggingface dataset:
Bricklin Group

The Malcolm Bricklin Group (d/b/a Bricklin) is a private conglomerate.

### Bricklin Products

* Subaru (construction pumps, motor scooter, portable generators, robot sweeper, small engines)

### Bricklin Retail

* **FasTrack** (Car Sales, RV Sales) - In 1971, Bricklin created a franchise named FasTrack, combining RV sales with shopping center parking lot race courses.
* **International Automobile Importers (IAI)** (Car Sales) - After Fiat left the U.S. market, Bricklin created International Automobile Importers (IAI) in 1982 to import various international car brands under one roof.

### Bricklin Visionary Vehicles

* Bricklin Motors
* Qoros
* ZAP!
Westfield Sarasota Square

**Westfield Sarasota Square**, formerly **Westfield Shoppingtown Sarasota Square** and **Sarasota Square Mall**, is a shopping mall in Sarasota, Florida. Its anchors are Costco, JCPenney, Macy's, Sears and AMC Sarasota 12.

## History

Arlen Realty and Development Corporation, a predecessor of CBL & Associates Properties, announced Sarasota Square Mall in 1976. Original tenants included Maas Brothers, JCPenney, and J. Byrons, with early tenants including Camelot Music, Kinney Shoes, and Morrison's Cafeteria. Sears was added in 1979. An expansion completed in 1988 added Parisian, which was the store's second location outside its home state of Alabama. Maas Brothers became Burdines in 1991, Burdines-Macy's in 2003, and Macy's in 2005.

In 1995, it was reported that Parisian had planned to sell its store to Dillard's. According to the plan, if Dillard's failed to acquire the Parisian store, it would instead build its own store adjacent to JCPenney, and the Parisian site would be sold to Jacobson's. Dillard's ultimately opened in the former Parisian building in 1996.

Westfield Group acquired the mall in 2003 from then-owners Coyote Group, and renamed it Westfield Shoppingtown Sarasota Square. The "Shoppingtown" was dropped in June 2005. In July 2019, Unibail-Rodamco-Westfield announced the shopping center would be renamed "Sarasota Square", dropping Westfield from its name and returning it to its original name before Westfield Group's acquisition of the property.

Dillard's closed its Westfield Sarasota Square store, along with a nearby store at DeSoto Square in Bradenton, in 2009. The Sarasota Square store was demolished and replaced with a Costco in 2012.
Triangle by Angle and Centroid

Instruction: Construct a segment connecting the sides of the angle to get a triangle whose centroid is in the point O.

Goal: 5L 7E

Available Tools:

* Move
* Point
* Line
* Circle
* Perpendicular Bisector
* Perpendicular
* Angle Bisector
* Parallel
* Non-collapsing Compass
* Intersect

Pack: Iota

Previous Level: 9.8

Next Level: 9.10

## 5L Solution

Let O be the vertex of the angle and A the given point.

1. Construct the line parallel to one line through A, intersecting the other line at B.
2. Construct the line parallel to BO through A, intersecting the other line at C.
3. Construct circle A with radius AB, intersecting AB at D.
4. Construct circle A with radius AC, intersecting AC at E.
5. Construct line DE.

## 7E Solution

Let O be the vertex of the angle and A the given point.

1. Construct line AO.
2. Construct circle A with radius AO, intersecting AO at B.
3. Construct circle B with radius BA, intersecting AO at C and circle A at D and E.
4. Construct line DE, intersecting one line at F and AO at G.
5. Construct circle G with radius FG, intersecting DE at H.
6. Construct line CH, intersecting the other line at I.
7. Construct line GI.
Outer Tangent

Instruction: Construct an outer tangent between two circles.

Goal: 6L 8E 2V

Available Tools:

* Move
* Point
* Line
* Circle
* Perpendicular Bisector
* Perpendicular
* Angle Bisector
* Parallel
* Non-collapsing Compass
* Intersect

Pack: Kappa

Previous Level: 10.1

Next Level: 10.3

## 6L Solution

Let A and B be the centers of the circles.

1. Construct line AB, intersecting the two circles at C and D, with C and D between A and B.
2. Construct the perpendicular bisector of AB, intersecting AB at E.
3. Construct the perpendicular bisector of CD, intersecting AB at F.
4. Construct circle E with radius AE, intersecting bisector F at G.
5. Construct circle G with radius CG, intersecting circle A at H and circle B at I.
6. Construct line HI.

## 8E Solution

Let A and B be the centers of the circles.

1. Construct line AB, intersecting circle A at C and circle B at D, with C on the other side of A from B and D between A and B.
2. Construct circle C with radius AC, intersecting circle A at E.
3. Construct circle D with radius BD, intersecting circle B at F, with F being the point closer to E.
4. Construct line EF, intersecting AB at G.
5. Construct circle D with radius DG, intersecting AB at H.
6. Construct circle B with radius BG, intersecting AB at I.
7. Construct circle I with radius HI, intersecting the latest circle at J.
8. Construct line GJ.

## 2V Solution

From the 4L Solution

1. Let the other intersection between circle E and bisector F be J.
2. Construct circle J with radius CJ, intersecting circle A at K and circle B at L.
3. Construct line KL.
Triangle by Three Sides

## Level Information

### Instruction:

Construct a triangle with the side AB and the two other sides equal to the given segments.

## 4L 12E solution

Given the other segments CD and EF.

1. Use the non-collapsing compass to construct the circle with radius CD and center A
2. Use the non-collapsing compass to construct the circle with radius EF and center B, intersecting circle A at G and H
3. Construct line AG
4. Construct line BG

## 4V solution

From the 4L and 12E solution.

1. Construct line AH
2. Construct line BH
3. Use the non-collapsing compass to construct the circle with radius EF and center A
4. Use the non-collapsing compass to construct the circle with radius CD and center B, intersecting the previous circle at I and J
5. Construct line AI
6. Construct line BI
7. Construct line AJ
8. Construct line BJ
Chord Trisection

Mathematics is the door and key to the sciences. — Roger Bacon

## Level Information

### Instruction:

Construct a chord of the larger circle through the given point that is divided into three equal segments by the smaller concentric circle.

## Solutions

### 3L Solution

Given the center A and point on the larger circle B.

1. Construct line AB, intersecting the smaller circle at C and D and the larger circle at E.
2. Construct circle E with radius CD, intersecting the smaller circle at F.
3. Construct line BF.

### 4E Solution

1. Construct line AB, intersecting the smaller circle at C (point closer to B) and the larger circle at D.
2. Construct circle C with radius CB, intersecting line AB in E.
3. Construct circle D with radius DE intersecting smaller circle in F.
4. Construct line BF.

### Alternative 4E Solution

Given the segment AB.

1. Construct line AB, intersecting the smaller circle at C and D.
2. Construct circle C with radius CD, intersecting the larger circle at E.
3. Construct line AE, intersecting the smaller circle at F.
4. Construct line BF.

### 2V Solution

From the 3L solution

1. Let the other intersection between circle E and the smaller circle be G.
2. Construct line BG.
Inscribed Circle

Circles to square and cubes to double would give a man exercise trouble. — Matthew Prior

## Solutions

### 4L solution

Given the triangle ABC

1. Construct the angle bisector of ∠CAB
2. Construct the angle bisector of ∠ABC, intersecting the angle bisector of ∠CAB at D
3. Construct the perpendicular to AB through D, intersecting AB at E
4. Construct the circle with center D and radius DE

### 8E solution

Given the triangle ABC

1. Construct the circle with center A and radius AB, intersecting AC at D
2. Construct the circle with center C and radius CD, intersecting BC at E
3. Construct the circle with center B and radius BE, intersecting AB at F
4. Construct the circle with center E and radius BE, intersecting circle B at G and H
5. Construct the circle with center F and radius BF, intersecting circle E at I
6. Construct line GH, intersecting BC at J
7. Construct line BI, intersecting GH at K
8. Construct the circle with center K and radius KJ

## Explanation

### 4L explanation

The incenter is the point where the internal angle bisectors, so in step 1-2 we construct 2 bisectors finding the center of incircle. Step 3 is perpendicular from any suitable side (doesn't work for a side that is adjacent to obtuse angle of triangle) to the center, finding a radius of incircle. Step 4 is constructing the circle itself.

### 8E explanation

First let's review steps 1-2. Let's construct these two circles and 2 angle bisectors for clarity (we don't need them in solution, only for explanation). We want to prove that BK = EK.

As AB=AD,

∠

{\displaystyle \angle}

BAK=

∠

{\displaystyle \angle}

DAK and side AK is common,

△

{\displaystyle \triangle}

BAK=

△

{\displaystyle \triangle}

DAK and BK=DK.

As CD=CE,

∠

{\displaystyle \angle}

DCK=

∠

{\displaystyle \angle}

ECK and side CK is common,

△

{\displaystyle \triangle}

CDK=

△

{\displaystyle \triangle}

ECK and DK=EK.

BK=DK=EK.

So, by building 2 circles in step 1-2, we know that for K - the center of incircle, BK=EK. K can be found by intersection of the perpendicular bisector of BE and any angle bisector. It's better than finding K as intersection of 2 angle bisectors as perpendicular bisector gives us also a radius of incircle.

In steps 3-7 we construct the perpendicular bisector of BE and the angle bisector of

∠

{\displaystyle \angle}

ABC. It would usually take 4 moves for an angle bisector and 3 moves for a perpendicular bisector, but here we are reusing 2 circles saving 2 moves.
Chord Midpoint

The study of the mathematics, like the Nile, begins in minuteness, but ends in magnificence. — Charles Colton

## Level Information

### Instruction:

Construct a chord whose midpoint is given.

## 2L 4E solution

Let A be the center and B the given midpoint.

1. Construct line AB
2. Construct the perpendicular to AB through B

## Explanation

The perpendicular bisector of any chord passes through the center of the circle. Line AB being that perpendicular bisector, the chord is simply the segment that's perpendicular to it and passes through the midpoint.
Cut Rectangle

Measure twice, cut once! — Proverb

## Level Information

### Instruction:

Construct a line through the given point that cuts the rectangle into two parts of equal area.

## 3L 3E solution

Given the rectangle ABCD and the point O.

1. Construct line AC, a diagonal of the rectangle
2. Construct line BD, the other diagonal, intersecting AC at E
3. Construct line OE

## Explanation

Any way you cut the rectangle with a line, if it passes through its centre, then it will cut the rectangle into two equal halves. So, you just need to find the centre of the rectangle by tracing its diagonals.
Six Flags Achievements (app)

**Six Flags Achievements** is a customer loyalty app where you can log on to rides in order to get points for free stuff.

## Achievements

### SFFTAchievements

#### Conquer It Below The Track

Ride Goliath

* Day Pass: 10 points
* Season Pass: 20 points
* Gold Pass: 25 points
* Gold Plus Pass: 50 points
* Platinum Pass: 65 points
* Diamond Pass: 75 points
* Diamond Elite Pass: 100 points

### SFMMAchievements

#### A Giant Among Coasters

Ride Goliath

* Day Pass: 20 points
* Season Pass: 25 points
* Gold Pass: 30 points
* Gold Plus Pass: 100 points
* Platinum Pass: 110 points
* Diamond Pass: 120 points
* Diamond Elite Pass: 125 points

### SFOGAchievements

#### Bend Your Mind

Ride Mindbender

* Day Pass: 7 points
* Season Pass: 17 points
* Gold Pass: 22 points
* Gold Plus Pass: 25 points
* Platinum Pass: 50 points
* Diamond Pass: 55 points
* Diamond Elite Pass: 75 points

#### Conquer Georgia's Only Hypercoaster

Ride Goliath

* Day Pass: 20 points
* Season Pass: 25 points
* Gold Pass: 30 points
* Gold Plus Pass: 100 points
* Platinum Pass: 110 points
* Diamond Pass: 120 points
* Diamond Elite Pass: 125 points

#### Duh-na-na-na-na-na-na-na

Ride Batman: The Ride

* Day Pass: 10 points
* Season Pass: 20 points
* Gold Pass: 25 points
* Gold Plus Pass: 50 points
* Platinum Pass: 65 points
* Diamond Pass: 75 points
* Diamond Elite Pass: 100 points
Double Segment

A mathematical truth is neither simple nor complicated in itself, it is. — Emile Lemoine

## 3L 3E solution

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at D and E
3. Construct the circle with center D and radius DE, intersecting circle B at C
4. Point at C

## 2V solution

1. Construct the circle with center A and radius AC, intersecting circle D at F
2. Construct the circle with center F and radius AF, intersecting the previous circle at C', the diametrical opposite of C on this circle
3. Point at C'

## Explanation

ABD and ABE are both equilateral triangles by construction, with

∠

{\displaystyle \angle}

DAB = 60°. Although not traced, segment DE would be the perpendicular bisector of AB, intersecting AB at its midpoint M, and where

∠

{\displaystyle \angle}

For the 2V step construction, the second circle A has radius 2|AB|, and by construction, triangle FAC' is equilateral, so

∠

{\displaystyle \angle}

∠

{\displaystyle \angle}

FAD = 60°; and by adding the adjacent angles through A from C' to F to D to B, we have determined that C' is indeed on line AB.
Copy Segment

Let no man who is not a Mathematician read the elements of my work. — Leonardo da Vinci

## Level Information

### Instruction:

Construct a segment from the given point that is equal to the given segment and lies on the same line with it.

## 3L solution

Given the segment AB and the point C.

1. Construct line AC
2. Construct the perpendicular bisector of AC, intersecting AC at D
3. Construct the circle with center D and radius BD, intersecting AC at E
4. Point at E

## 4E solution

Given the segment AB and the point C.

1. Construct line AC
2. Construct the circle with center A and radius AC
3. Construct the circle with center C and radius AC, intersecting circle A at D
4. Construct the circle with center D and radius BD, intersecting AC at E
5. Point at E

## 2V solution

From the 3L solution.

1. Construct the circle with center C and radius CE, intersecting AC at F
2. Point at F
Drop a Perpendicular*

Instruction: Drop a perpendicular from the point to the line passing through the center of the circle using only a straightedge.

Goal: 5L 5E

Available Tools:

* Move
* Point
* Line
* Intersect

Pack: Iota

Previous Level: 9.3

Next Level: 9.5

## 5L 5E Solution

Let O be the center of the circle, A be the given point, and B and C be the intersections of the circle and line.

1. Construct line AB, intersecting the circle at D.
2. Construct line CD.
3. Construct line AC, intersecting the circle at E.
4. Construct line BE, intersecting CD at F.
5. Construct line AF.

## Explanation

Triangle BCD is inscribed in a circle of diameter BC, so

D
C
⊥
D
B

{\displaystyle DC \perp DB}

, or

D
C
⊥
A
B

{\displaystyle DC \perp AB}

. Hence DC is an altitude of triangle ABC.

Similarly, BCE is inscribed in a circle of diameter BC, so

E
B
⊥
E
C

{\displaystyle EB \perp EC}

, or

E
B
⊥
A
C

{\displaystyle EB \perp AC}

. Hence EB is an altitude of triangle ABC.

As the intersection of two altitudes of BCE, F is its orthocenter. Thus AF is the third altitude, and

A
F
⊥
B
C

{\displaystyle AF \perp BC}

.
Double Angle

Contradiction is not a sign of falsity, nor the lack of contradiction a sign of truth. — Blaise Pascal

## Level Information

### Instruction:

Construct an angle equal to the given one so that they share one side.

## Location Information

### 3L 3E Solution

1. Construct a circle with an arbitrary center A on one ray and an arbitrary radius so that it intersects the other ray at least once. Let B be one of those intersections.
2. Construct a circle with another arbitrary center C on the first ray and radius BC, intersecting circle A at D.
3. Construct line OD.

### Alternate 3L 3E solution:

1. Construct the circle with center O and an arbitrary radius, intersecting the first ray at A and the second ray at B.
2. Construct a circle with an arbitrary center C on the first ray and radius BC, intersecting circle O at D.
3. Construct line OD.

### 2V solution

#### From the 3L 3E solution (6E total):

1. Construct a circle with an arbitrary center E on the second ray and an arbitrary radius so that it intersects the first ray at least once. Let F be one of those intersections.
2. Construct a circle with another arbitrary center G on the second ray and radius FG, intersecting circle E at H.
3. Construct line OH.

#### From the alternate 3L 3E solution (5E total):

1. Construct a circle with an arbitrary center E on the second ray and radius AE, intersecting circle O at F.
2. Construct line OF.

## Explanation

The first circle allows you to pick any point on the other ray and the second circle allows you to create the reflection of that point through the first ray. Then by tracing the line OD, you're actually constructing the reflection of the second ray through the first ray, so it's angle to the first ray will be the same.
Segment Trisection*

Instruction: Construct points that divide the segment into three equal parts using only a straightedge. To help you, a line parallel to the segment is given.

Goal: 8L 8E

Available Tools:

* Move
* Point
* Line
* Intersect

Pack: Kappa

Previous Level: 10.6

Next Level: 10.8

## 8L 8E Solution

Given the segment AB.

1. Construct a line connecting A with an arbitrary point C on the line.
2. Construct a line connecting B with another arbitrary point D on the line that intersects AC at E.
3. Construct line AD.
4. Construct line BC, intersecting AD at F.
5. Construct line EF, intersecting CD at G.
6. Construct line AG, intersecting BD at H.
7. Construct line FH, intersecting AB at I and CD at J.
8. Construct line EJ, intersecting AB at K.
9. Point at I.
10. Point at K.
Angle Isosceles

## Level Information

### Instruction:

Construct a line through the given point such that it cuts off equal segments on the sides of the angle.

## 2L solution

Let A be the vertex of the angle and B the point.

1. Construct the angle bisector A1 of the given angle
2. Construct the perpendicular to A1 through B

## 5E solution

Let A be the vertex of the angle and B the point.

1. Construct the circle with center A and radius AB, intersecting the sides of the angle at C and D
2. Construct the circle with center C and radius AC
3. Construct the circle with center D and radius AD, intersecting circle C at E
4. Construct the circle with center E and radius BE, intersecting circle A at F
5. Construct line BF

## Alternate 5E solution

Let A be the vertex of the angle and B the point.

1. Construct the circle with center A and radius AB, intersecting the sides of the angle at C and D
2. Construct the circle with center C and radius BC, intersecting the side of the angle at E and F
3. Construct the circle with center A and radius AE, intersecting side of the angle at G
4. Construct the circle with center D and radius DG, intersecting circle A at H and I
5. Construct line BH
Parallelogram on Four Lines

It is easier to perceive error than to find truth. — Johann Goethe

## Level Information

### Instruction:

Construct a parallelogram whose center is in the given point so that each of the given lines contains one of its vertices. Two of the lines are parallel.

## Location Information

### Previous Level:

14.3 Triangle by Tangent Point on Hypotenuse

## Solutions

### 8L solution

Initial trapezoid ABCD

1. Construct the line parallel to AB through O, intersecting AD in E, BC in F.
2. Construct the circle in E, radius EA, intersecting line AD at A and G.
3. Construct the circle in F, radius FC, intersecting line BC at C and H.
4. Construct line GO, intersecting AB at I.
5. Construct IH.
6. Construct GH.
7. Construct the line parallel to IH through G.
8. Construct the line parallel to GH through I.

### 12E solution

Initial trapezoid ABCD

1. Construct the line parallel to AB through O, intersecting AD in E, BC in F.
2. Construct the circle in E, radius EA, intersecting line AD at A and G.
3. Construct the circle in F, radius FC, intersecting line BC at C and H.
4. Construct line GO, intersecting AB at I.
5. Construct IH.
6. Construct GH.
7. Construct line HO, intersecting CD at J.
8. Construct JG.
9. Construct JI.

### 2V solution

The same but mirrored.

1. Construct the line parallel to AB through O, intersecting AD in E, BC in F.
2. Construct the circle in E, radius ED, intersecting line AD at D and G.
3. Construct the circle in F, radius FB, intersecting line BC at B and H.
4. Construct line HO, intersecting AB at I.
5. Construct IG.
6. Construct GH.
7. Construct line GO, intersecting CD at J.
8. Construct JI.
9. Construct JH.

## Explanation

First steps are common in 8L and 12E solutions. We want to build diagonal GI of parallelogram, where one vertex is on line AB and the opposite is on line AD. As O should be the center of this diagonal, we can reuse solution from Segment by Midpoint. Construct parallel line and a circle E making AE=EG. Then

△
G
E
O
∼
△
G
I
A

{\displaystyle \triangle GEO \sim \triangle GIA}

, so GO=OI and GI is diagonal.

Similar HJ is a diagonal of paralellogram, HO=OJ, but don't construct it, it's enough to have point H.

At this point we can add two sides of paralellogram - HI and HG. We have spent totally 6L 9E.

For 8L solution, construct two other sides as parallel to existing.

For 12E solution, construct 2nd diagonal and 2 other sides.
Point Tool

The Point Tool is introduced during the tutorial level Equilateral Triangle. When active, the tool tries to place a point (shown as a small blue circle) on the nearest intersection of two lines (whether straight lines or the circumference of a circle); highlighting a point (with this tool or with the Intersect Tool makes it easier to use that point in a subsequent construction. If there is not a nearby intersection point, the tool instead creates a free point (shown as a small red circle) on a nearby line, or on the blank canvas.

Some levels require the use of a free point to achieve a minimal L or E score. Several construction tools can create free points without the earlier use of the Point tool.
Egyptian Triangle by Side of Length 4

## Level Information

### Instruction:

Construct a triangle with sides whose lengths are equal to 3, 4, and 5. The side with length 4 is given.

## 5L solution

Let A and B be the given points.

1. Construct the perpendicular bisector of line AB, intersecting line AB at C.
2. Construct the perpendicular bisector of line AC, intersecting line AB at D.
3. Construct circle B with center B and radius BD.
4. Erect a perpendicular to line AB at point B, intersecting circle B at E.
5. Construct line AE.

## 7E solution

Let A and B be the given points.

1. Construct line AB.
2. Construct circle B centered at B with radius BA, intersecting line AB at A and C.
3. Construct circle A centered at A with radius AC, intersecting line AB at C and D.
4. Construct circle C centered at C with radius CA, intersecting circle A at E.
5. Construct line BE, intersecting circle B at F.
6. Construct line DF, intersecting circle A at D and G.
7. Construct line AG.

## 4V solution

From the 5L solution

1. Let the other intersection between the perpendicular to AB at B and circle B be F.
2. Let the intersection between the perpendicular bisector of AB and line AE be G.
3. Construct line FA, intersecting the perpendicular bisector of AB at H.
4. Construct line BG.
5. Construct line BH.
6. Erect the perpendicular to line AB at point A
Line Tool

The Line Tool is introduced during the tutorial level Equilateral Triangle. When active, the tool creates lines by a straight-edge; either by two consecutive clicks to the two endpoints of a segment, or by clicking on the first point and dragging to the second. When selecting an endpoint, the tool tries to favor nearby intersections between lines (whether straight lines or the circumference of a circle) or points previously marked with the Point Tool; however, it is also possible for an endpoint to be placed on a free point.

The Line Tool costs one elementary move if it creates a new line or if it is used on an existing line segment or ray to extend it into a line. There is no cost for trying to re-create a line that has already been constructed.
Three equal segments - 1

All knowledge degenerates into probability. — David Hume

## Level Information

### Instruction:

Given an angle ABC and a point M inside it, find points D on BA and E on BC and construct segments DM and ME such that BD = DM = ME.

## Location Information

### Previous Level:

3.3 Intersection of Perpendicular Bisectors

## 4L 6E solution

1. Construct the perpendicular bisector of BM, intersecting AB at D
2. Construct the circle with center M and radius DM, intersecting BC at E and F
3. Construct line DM
4. Construct line ME

## Explanation

Point D is easy. Since it's equidistant to both B and M, then it must be on their perpendicular bisector.

Now that you know DM, since M is also equidistant to both D and E, then both D and E are on a circle with center M. You'll notice that there are two possibilities for E.
Perpendicular Bisector Tool

Note that the Perpendicular Bisector tool costs 3 elementary (E) moves. — Euclidea Manual

## Level Information

### Instruction:

You have discovered the Perpendicular Bisector tool. Click sequentially on two points to see how it works.

## Solutions

### 1L 3E Solution

1. With the bisector tool active, click on the two points to be bisected.

## Explanation

The bisector tool works even when the line between two points has not been drawn; however the dashed line shown by the bisector tool is just visual decoration, and cannot be used by the Point or Intersect tool for selecting the midpoint. It is also possible to tap-and-hold on one point, then drag to see where the perpendicular bisector will land before releasing at the second point.

In many levels, use of the perpendicular bisector tool can achieve a lower L score than constructing the perpendicular bisector by elementary tools; however, when minimizing the E score, there are often situations where replacing an application of the bisector tool by its three underlying elementary constructions can point out a common underlying circle that can be reused for two perpendicular bisectors.
Three Circles - 1

## Level Information

### Instruction:

Construct three circles with the given centers that are externally tangent each other.

## 7L Solution

Given the points A, B, and C.

1. Construct line AB.
2. Construct circle A with radius AC, intersecting AB at D, with D being closer to B.
3. Construct circle B with radius BC, intersecting AB at E, with E being closer to A.
4. Construct the perpendicular bisector of DE, intersecting AB at F.
5. Construct circle A with radius AF.
6. Construct circle B with radius BF.
7. Construct circle C with radius DF.

## 10E Solution

Given the points A, B, and C.

1. Follow the 7L solution steps 1-6.
2. Construct line BC, intersecting the latest circle at G.
3. Construct circle C with radius CG.
Hash

No other science than mathematics strengthens so much our faith in the power of the human mind. — Hugo Steinhaus

## Level Information

### Instruction:

Construct a line through the given point on which two pairs of parallel lines cut off equal line segments.

## 2L solution

Let O be the given point, and ABCD the parallelogram created by the two pairs of parallel lines (A on top to match the orientation of explore mode):

1. Construct line AC, a diagonal of the parallelogram
2. Construct the parallel to AC through O

## 4E solution

1. Construct a circle with center A and radius AO
2. Construct line AO, intersecting circle A at E
3. Construct the circle with center C and radius CE, intersecting circle A at F
4. Construct line OF

## 2V solution

Starting from the 2L solution:

1. Construct line BD, a diagonal of the parallelogram
2. Construct the parallel to BD through O

Starting from the 4E solution:

1. Construct a circle with center D and radius DO
2. Construct a circle with center B and radius BO, intersecting circle D at G
3. Construct line BG, intersecting circle B at H
4. Construct line OH

## Explanation

The diagonal of the parallelogram is a line which cuts off equal line segments from both the pairs of parallel lines, so all other solutions are parallel to a diagonal of the parallelogram.

The alternative 2V solution demonstrates both methods of constructing a line parallel to two points; OF by doing point then line reflection, OH by doing line then point reflection.
Secant Bisection

## Level Information

### Instruction:

Construct a secant through the given point that is divided into two equal segments by the circle.

## 4L Solution

Given the point A and centre B.

1. Construct the line AB intersecting the circle at C and D.
2. Construct the circle with centre B and radius AB, intersecting line AB at E.
3. Construct the circle with centre E and radius CD, intersecting circle B at F.
4. Construct the line AF.

## 5E Solution

Given the point A and centre B.

1. Construct the line AB intersecting the circle at C and D with D on the opposite side of B from A.
2. Construct the circle with centre B and radius AB.
3. Construct the circle with centre D and radius CD, intersecting the larger circle B at E.
4. Construct the line BE, intersecting the smaller circle B at F.
5. Construct the line AF.

## 2V Solution

From the 4L Solution

1. Let the other intersection between circle E the circle B be G.
2. Construct the line AG.
Golden Section

The Golden ratio has inspired thinkers of all disciplines like no other number in the history of mathematics. — Mario Livio

## Level Information

### Instruction:

Construct a point that divides the segment in the "golden section".

## Location Information

### 4L Solution

1. Construct a circle with center A and radius AB.
2. Construct the perpendicular bisector of AB, marking intersection with AB as point C, and marking intersection with circle A as point D.
3. Construct a circle with center C and radius AB, marking intersection with circle A at E. E should be on the opposite side of AB from D, or else the solution will not work.
4. Construct line DE, intersecting with AB at point F.
5. Mark the point F.

### 5E Solution

1. Construct line AB.
2. Construct a circle with center A and radius AB, intersecting line AB at C.
3. Construct a circle with center B and radius BA, intersecting the previous circle at D.
4. Construct a circle with center C and radius CB, intersecting the previous circle at E. E should be on the opposite side of AB from D, or else the solution will not work.
5. Construct line DE, intersecting line AB at F.
6. Mark the point F.

### 2V Solution

Starting from the 4L solution:

1. Construct a circle with center C and radius CF, intersecting line AB at G.
2. Mark the point G.

Starting from the 5E solution:

1. Construct a line passing through D and the other intersection of the same two circles, creating the perpendicular bisector of line segment AB, intersecting line AB at O.
2. Repeat the steps above, with point O here equal to point C above.
Lozenge

To think deeply of simple things. — Arnold Ross

## 5L solution

Given the side AB (A on the left to match the orientation in explore mode)

1. Construct the perpendicular to AB from A; let C be a distinct arbitrary point on that perpendicular (above A, to match the orientation in explore mode)
2. Construct the angle bisector of ∠BAC
3. Construct the circle with center A and radius AB, intersecting the angle bisector at D and line AC at E with E on the side of A opposite from C
4. Construct the perpendicular to AC through D
5. Construct line BE

## 7E solution

Given the side AB (A on the left to match the orientation in explore mode).

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at C and D (C above A to match the orientation in explore mode)
3. Construct line CD, intersecting AB at E
4. Construct the circle with center E and radius AE, intersecting AC at F and AD at G
5. Construct line AF, intersecting circle A at H on the same side as C relative to AB
6. Construct line BG, intersecting circle B at I on the same side as C relative to AB
7. Construct line HI

## 4V solution

From the 5L solution:
Let F be the point diametrically opposite E on line AC; G at the intersection of circle A and the perpendicular constructed through AC from D; and H diametrically opposite D on line AD

1. Construct line AG, intersecting circle A at point I
2. Construct line HI
3. Construct line BF (total 8L 15E)

From the 7E solution:
Let J be the other intersection of line AF with circle A and K the other intersection of line BG with circle B.

1. Construct line JK
2. Construct line AG
3. Construct line BF (total 10L 10E)

## Explanation

For the 5L solution, the angle bisector of AB and its perpendicular AC forms the required 45° angle, and circle A determines an equidistant point D for side AD. Any perpendicular through AC is parallel to AB, forming the next side through D. Finally, since triangle ABE is a right isosceles triangle, the extension of line BE forms the final side also at a 45° angle.

For the 7E solution, E is constructed at the midpoint of AB, such that triangles AEF and BEG are both right isosceles triangles, forming the required angles once the hypotenuses are extended. Then circle A gives the right length for leg AH, and circle B gives the right length for leg BI, leaving the construction of line HI to conclude the rhombus.

The 4V solutions are based on the symmetry of swapping where A and B are labeled, and on where point C is chosen in relation to line AB, although constructing the later variants can reuse points determined earlier in the process.
Center of Quadrilateral

There is geometry in the humming of the strings. — Pythagoras

## Level Information

### Instruction:

Construct the midpoint of the segment that connects the midpoints of the diagonals of the quadrilateral.

## 4L 10E solution

Given the quadrilateral ABCD (arbitrary which vertex is A)

1. Construct the perpendicular bisector of AB, intersecting AB at E
2. Construct the perpendicular bisector of CD, intersecting CD at F
3. Construct the perpendicular bisector of EF
4. Construct line EF
5. Point at the intersection of EF and its perpendicular bisector

## Explanation

Consider the triangle ABC created by drawing diagonal AC with its midpoint H. The smaller triangle AEH shares angle A, and its two sides AE and AH are both half the length of the larger AB and AC, so line EH is parallel to line BC and half its length. Simlarly, considering triangle CDB by drawing diagonal BD with its midpoint I, the smaller triangle DFI demonstrates that FI is parallel to line BC and half its length. With two sides parallel and the same length, the quadrilateral EHFI is now known to be a parallelogram. Since the diagonals of a parallelogram intersect at their respective midpoints, the problem of finding the midpoint of diagonal HI is equivalent to finding the midpoint of diagonal EF.
Line Equidistant from Two Lines

Most people would die sooner than think — in fact they do so. — Bertrand Russell

## Level Information

### Instruction:

Construct a straight line parallel to the given parallel lines that lies at equal distance from them.

## 2L solution

1. Construct the perpendicular to one line from an arbitrary point A on it, intersecting the other line at B
2. Construct the perpendicular bisector of AB

## 5E solution

1. Construct a circle with an arbitrary center A on one line and an arbitrary radius at least half of the gap between the two parallel lines, intersecting the line through A at B and C
2. Construct the circle with center B and radius BC, intersecting the other line at D and E
3. Construct line CD, intersecting circle A at F
4. Construct line CE, intersecting circle A at G
5. Construct line FG

## Explanation

The desired line is the parallel line through any midpoint of any other line that crosses the original two lines. For the 2L construction, use of a perpendicular means that the perpendicular bisector can easily find the final line.

For the 5E solution, a different approach is used. Circle B is twice as large as circle A and tangent at C, so any line from C to a point on circle B has a corresponding point on circle A as its midpoint. By constructing B large enough to cross the second parallel line, then connecting C to the two points of intersection on the second line, the corresponding midpoints on circle A are then sufficient to trace the desired parallel line.
Drop a Perpendicular

The knowledge of which geometry aims is the knowledge of the eternal. — Plato

## 2L solution

Let O be the point.

1. Construct a circle with center O and an arbitrary radius so that it intersects the line at two points A and B
2. Construct the perpendicular bisector of AB

## 3E solution

Let O be the point.

1. Construct a circle with an arbitrary center A on the line and radius OA
2. Construct a circle with center B at intersection of circle A and the line and radius OB, intersecting circle A at C
3. Construct line OC

## Explanation

Solution 2L has you construct an arbitrary isosceles triangle ABO with O as its apex. The perpendicular bisector of its base will pass through its apex and therefore will answer the puzzle.

Solution 3E has you construct the reflection of point O through the line, making that line the perpendicular bisector of the segment joining point O to its reflection. Just trace that segment.
Circle Equidistant from Four Points

## Level Information

### Instruction:

Construct a circle equidistant from the given four points so that exactly two of them lie inside the circle.

## 6L solution

1. Construct the perpendicular bisector of AB
2. Construct the perpendicular bisector of CD intersecting the previous line at E
3. Construct the line AE
4. Construct a circle with center E and radius EC, intersecting AE at F
5. Construct the perpendicular bisector of AF, intersecting AE at G
6. Construct a circle with center E and radius EG

## 3V solution

Continuing from either solution:

1. Repeat the steps using AC & BD
2. Repeat the steps using AD & BC
Circle through Point Tangent to Line

Knowledge comes, but wisdom lingers. — Alfred Tennyson

## Level Information

### Instruction:

Construct a circle through the point A that is tangent to the given line at the point B.

## 3L solution

1. Construct the perpendicular bisector of AB
2. Construct the perpendicular to the line from B, intersecting the perpendicular bisector at O
3. Construct the circle with center O and radius OA (or OB)

## 6E solution

1. Construct the circle with center A and radius AB, intersecting the line at C
2. Construct the circle with center B and radius AB, intersecting circle A at D and E
3. Construct line DE
4. Construct line AC, intersecting circle A at F
5. Construct line BF, intersecting DE at O
6. Construct the circle with center O and radius OA

## Explanation

The center has to be on the perpendicular bisector of AB since OA=OB and on the perpendicular to the line from B since the line is tangent to it at B.

Solution 6E has you manually construct both the perpendicular bisector and the perpendicular using a common circle for the two of them. Circles A and B having the same radius help you get the perpendicular bisector. With both B and C on the same circle A, constructing the diameter CF means BCF is a right triangle in B, which leads to BC, and by extension the given line, being perpendicular to BF and so BF is the needed perpendicular.
Line Equidistant from Three Points

## Level Information

### Instruction:

Construct a line that is at equal distance from the given three points.

## 3L 7E solution

Let A, B and C be the given points.

1. Construct line AB
2. Construct the perpendicular to AB through C, intersecting AB at D
3. Construct the perpendicular bisector of CD

## 3V solution

1. Construct line AC
2. Construct the perpendicular to AC through B, intersecting AB at E
3. Construct the perpendicular bisector of BE
4. Construct line BC
5. Construct the perpendicular to BC through A, intersecting AB at F
6. Construct the perpendicular bisector of AF
Angle of 30°

Inspiration is needed in geometry, just as much as in poetry. — Alexander Pushkin

## Level Information

### Instruction:

Construct an angle of 30° with the given side.

## Solutions

Let A be the initial point of the given ray.

### 3L 3E Solution

1. Construct a circle with an arbitrary center B and radius AB, intersecting the ray at point C. Although the image shown here places B on the ray, this construction works even when point B is not on the ray, so long as B is on the same half of the plane as the ray.
2. Construct a circle with center C and radius BC, intersecting circle B at points D and E.
3. Construct line AD.

## Explanation

You want the angle of 30° to be the inscribed angle of circle B that will subtend the same arc as a central angle of 60°, which you can get by constructing an equilateral triangle. By constructing circle C, you get BCD, an equilateral triangle, so the arc CD is the arc you want.
Line Equidistant from Two Points - 1

Mathematics is a tool for reasoning. — Richard Feynman

## Level Information

### Instruction:

Construct a line through the point C and at equal distance from the points A and B but that does not pass between them.

## 2L solution

Given points A, B, and C (C already labeled, choice of A and B is arbitrary):

1. Construct line AB
2. Construct the parallel to AB through C

## 4E solution

1. Construct the circle with center A and radius AC
2. Construct line AC, intersecting circle A at D
3. Construct the circle with center B and radius BD, intersecting circle A at E
4. Construct line CE

Alternative solution:

1. Construct the circle with center A and radius AC
2. Construct the circle with center B and radius BC, intersecting circle A at D
3. Construct line BD, intersecting circle B at E
4. Construct line CE

## Explanation

Two points are equidistant from any line parallel to the two points; the 2L solution constructs this line directly.

Both 4E solutions perform the steps for creating a parallel line without needing the line AB drawn. The first does a point-mirroring of C across A (steps 1 and 2) then a line-mirroring of D across AB (step 3); the second does a line mirroring of C across AB (steps 1 and 2) then a point-mirroring of D across B (step 3). Either way, it puts point E in line with C to be parallel to AB.
Inner Tangent

Instruction: Construct an inner tangent between two circles.

Goal: 6L 8E 2V

Available Tools:

* Move
* Point
* Line
* Circle
* Perpendicular Bisector
* Perpendicular
* Angle Bisector
* Parallel
* Non-collapsing Compass
* Intersect

Pack: Kappa

Previous Level: 10.2

Next Level: 10.4

## 6L Solution

Let A and B be the centers of the circles.

1. Construct line AB, intersecting circle A at C and circle B at D, with C on the other side of A from B and D between A and B.
2. Construct the perpendicular bisector of AB, intersecting AB at E.
3. Construct the perpendicular bisector of CD, intersecting AB at F.
4. Construct circle E with radius AE, intersecting bisector F at G.
5. Construct circle G with radius CG, intersecting circle A at H and circle B at I.
6. Construct line HI.

## 8E Solution

Let A and B be the centers of the circles.

1. Construct line AB, intersecting circle A at C and circle B at D, with C and D between A and B.
2. Construct circle C with radius AC, intersecting circle A at E.
3. Construct circle D with radius BD, intersecting circle B at F, with F being the point further from E.
4. Construct line EF, intersecting AB at G.
5. Construct circle D with radius DG, intersecting AB at H.
6. Construct circle B with radius BG, intersecting AB at I.
7. Construct circle I with radius HI, intersecting the latest circle at J.
8. Construct line GJ.

## 2V Solution

From the 4L Solution

1. Let the other intersection between circle E and bisector F be J.
2. Construct circle J with radius CJ, intersecting circle A at K and circle B at L.
3. Construct line KL.
Triangle Mid-Segment

Neither thirty years, nor thirty centuries, affect the clearness, or the charm, of Geometrical truths. — Charles Dodgson

## Level Information

### Instruction:

Construct a mid-segment of the given acute triangle.

## 2L Solution

Let A, B, and C be the vertices of the triangle. To match the orientation of the solution visible in explore mode, AC must be the bottom leg; but the construction works for all 6 labeling choices.

1. Construct the perpendicular bisector of AB, intersecting AB at D.
2. Construct the line parallel to AC through D.

## 5E Solution

Let A, B, and C be the vertices of the triangle. To match the orientation of the solution visible in explore mode, A must be the top vertex; but the construction works for all 6 labeling choices.

1. Construct a circle with an arbitrary center D on AB and radius AD, intersecting AB at E.
2. Construct circle E with radius AE, intersecting BC at F and G, if the circle doesn't intersect BC at two points move D so that it does.
3. Construct line AF, intersecting circle D at H.
4. Construct line AG, intersecting circle D at I.
5. Construct line HI.

## 3V Solution

From the 2L Solution

1. Let the intersection between the parallel line through D and BC be E.
2. Construct the line parallel to AB through E, intersecting AC at F.
3. Construct line DF.

## Explanation

Every triangle has three mid-segments, each parallel to one of the legs. The 2L solution is fairly straight-forward: once a midpoint is found for one of the three legs, creating a line parallel to either remaining leg through that point completes the construction.

For the 5E solution, note that circle E has twice the radius of circle D and shares a common diameter along line AD; therefore, any chord from A to a point on circle E will intersect circle D at the chord's midpoint. By choosing D (and thus E) so that there are two intersections of circle E along leg BC, then tracing chords to those two points, the resulting intersections with circle D both lie on the mid-segment of ABC.
Minimum Perimeter - 1

Neither thirty years, nor thirty centuries, affect the clearness, or the charm, of Geometrical truths. — Charles Dodgson

## Level Information

### Instruction:

Construct a triangle whose perimeter is the minimum possible whose vertices lie on the two sides of the angle and the third vertex is in the point A.

## 6L 6E solution

Let O be the vertex of the angle and A the given point. Let B, C be abitary points on each ray, such that

A
B
>
A
O
,
A
C
>
A
O

{\displaystyle AB>AO,AC>AO}

.

1. Construct circle O with center O and radius OA.
2. Construct circle B with center B and radius BA, intersecting circle O at F.
3. Construct circle C with center C and radius CA, intersecting circle O at G.
4. Construct line FG, intersecting line OB at H, intersecting line OC at I.
5. Construct line AH.
6. Construct line AI.

## Explanation

For any points

P

{\displaystyle P}

 on

O
B

{\displaystyle OB}

 and

Q

{\displaystyle Q}

 on

O
C

{\displaystyle OC}

, we know:

>

A
P

=
F
P

Perpendicular Bisector Theorem

(
1
)

A
Q

=
Q
G

“ ”

(
2
)

A
H

=
F
H

“ ”

(
3
)

I
A

=
I
G

“ ”

(
4
)

A
P
A
Q
P
Q

=
F
P
Q
G
P
Q

Follows from steps 1

&

 2

(
5
)

A
H
H
I
I
A

=
F
H
H
I
I
G

Follows from steps 3

&

 4

(
6
)

A
H
H
I
I
A

=
F
G

(
7
)

F
P
P
Q
Q
G

≥
F
G

Triangle inequality

(
8
)

A
P
A
Q
P
Q

≥
F
H
H
I
I
G

(
9
)

∴

Perimeter of

△
A
P
Q

≥

Perimeter of

△
A
H
I

(
10
)

{\displaystyle >{\begin{aligned}AP&=FP&&{\text{Perpendicular Bisector Theorem}}&&(1)&\\AQ&=QG&&{\text{“ ”}}&&(2)&\\AH&=FH&&{\text{“ ”}}&&(3)&\\IA&=IG&&{\text{“ ”}}&&(4)&\\AP+AQ+PQ&=FP+QG+PQ&&{\text{Follows from steps 1 }}{\&}{\text{ 2}}&&(5)&\\AH+HI+IA&=FH+HI+IG&&{\text{Follows from steps 3 }}{\&}{\text{ 4}}&&(6)&\\AH+HI+IA&=FG&&&&(7)&\\FP+PQ+QG&\geq FG&&{\text{Triangle inequality}}&&(8)&\\AP+AQ+PQ&\geq FH+HI+IG&&&&(9)&\\\therefore {\text{Perimeter of }}\triangle APQ&\geq {\text{Perimeter of }}\triangle AHI&&&&(10)&\\\end{aligned}}}