recursal/Fanatic-Fandom huggingface dataset:
Tangent of Circle

Instruction: Construct a tangent to the circle from the external point.

Goal: 4L 5E 2V

Available Tools:

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

Pack: Kappa

Previous Level: 9.10

Next Level: 10.2

## 4L Solution

Let A be the center and B the given point.

1. Construct line AB.
2. Construct the perpendicular bisector of AB, intersecting AB at C.
3. Construct circle C with radius AC, intersecting circle A at D.
4. Construct line BD.

## 5E Solution

Let A be the center and B the given point.

1. Construct line AB, intersecting circle A at C.
2. Construct circle C with radius BC, intersecting AB at D.
3. Construct circle A with radius AB, intersecting AB at E.
4. Construct circle E with radius DE, intersecting the latest circle at F.
5. Construct line BF.

## 2V Solution

From the 4L Solution

1. Let the other intersection between circle A and circle C be E.
2. Construct line BE.

From the 5E Solution

1. Let the other intersection between circle E and circle A be G
2. Construct line BG.
Triangle Cleaver

## Level Information

### Instruction:

Construct a line through the midpoint of a side of the triangle that bisects its perimeter.

## 3L solution

Let A, B, and C be the vertices of the triangle

1. Construct the perpendicular bisector of AC intersecting line AC at midpoint D
2. Construct the line BE that bisects the angle ABC
3. Draw the line parallel to BE that intersects D

## 7E solution

Let A, B, and C be the vertices of the triangle.

1. Construct perpendicular bisector of side AC. It intersects two sides; AC at point D and the other side at point E.
2. Construct circle B, with center B and radius BE. B intersects the third side at F.
3. Construct circle E, with center E and radius ED.
4. Construct circle F, with center F and radius FD. It intersects circle E at point G.
5. Draw line DG.

## 3V solution

From 3L solution

1. Repeat the steps to construct the line AF where F is the midpoint between B and C
2. Repeat the steps to construct the line CG where G is the midpoint between A and B
Parallelogram

## Level Information

### Instruction:

Construct a parallelogram with the given side and the midpoint of the opposite side in the given point.

## 5L solution

Given the segment AB and the point C.

1. Construct the perpendicular bisector of AB, intersecting AB at D
2. Use the non-collapsing compass to construct the circle with radius AD and center C
3. Construct the parallel to AB through C, intersecting circle C at E and F
4. Construct line AE
5. Construct line BF

Alternative 5L solution:

1. Construct the perpendicular bisector of AB, intersecting AB at point D
2. Construct line CD
3. Construct the parallel to AB through C
4. Construct the parallel to CD through A
5. Construct the parallel to CD through B

## 8E solution

Given the segment AB and the point C.

1. Construct line AC
2. Construct the circle with center A and radius AC, intersecting AB at D and AC at E; if it doesn't intersect the segment AB without you needing to prolong it, either move B, or switch A and B including for the previous step
3. Construct the circle with center D and radius DE, intersecting circle A at F
4. Construct line CF
5. Construct the circle with center C and radius AC, intersecting AC at G
6. Construct line BG, intersecting CF at H
7. Construct the circle with center C and radius CH, intersecting CH at I
8. Construct line AI
Parallel Tool

Note that the Parallel tool costs 4 elementary (E) moves. — Euclidea Manual

## Explanation

The parallel tool works by clicking on a point then clicking on a line, by clicking on a line then on a point, or by click-and-hold on a line and dragging to a point. When starting on the line, the tool visualizes where the parallel line will land before selecting the point. The line (or line segment) must already be constructed; there is no way to draw a line parallel to the implied line between two arbitrary points (including when the implied line is a decoration such as the one drawn by the Perpendicular Bisector Tool).
Circle Tangent to Square Side

As long as algebra and geometry have been separated, their progress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched together towards perfection. — Joseph Louis Lagrange

## Level Information

### Instruction:

Construct a circle that is tangent to a side of the square and goes through the vertices of the opposite side.

## 3L solution

Given the square ABCD (AB on the bottom for the solution to match the hint mode orientation)

1. Construct the perpendicular bisector of AB, intersecting CD at E and AB at F
2. Construct the perpendicular bisector of AE, intersecting EF at G
3. Construct the circle with center G and radius AG (or EG)

## 6E solution

Given the square ABCD.

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at E and F
3. Construct line EF, intersecting AB at G
4. Construct line CG, intersecting circle A at H
5. Construct line AH, intersecting EF at I
6. Construct the circle with center I and radius AI

Alternate 6E solution:

Given the square ABCD.

1. Construct the circle with center A and radius AB
2. Construct the circle with center B and radius AB, intersecting circle A at E and F
3. Construct line EF, intersecting AB at G and CD at H
4. Construct the circle with center H and radius GH, intersecting circle A at I and J
5. Construct line IJ, intersecting EF at K
6. Construct the circle with center K and radius AK

## 4V solution

From the 3L solution:

1. Construct the perpendicular bisector of AD, intersecting AD at H, EF at I, and BC at J
2. Construct the circle with center I and radius GI, intersecting IJ at K, EI at L, and HI at M
3. Construct the circle with center K and radius HK
4. Construct the circle with center L and radius FL
5. Construct the circle with center M and radius JM

## Explanation

Consider point A at coordinate (0, 0) and B at (8, 0), such that AB is length 8. The perpendicular bisector of AB then forms point E along CD at coordinate (4, 8) and F along AB at (0, 8). The length AF is thus 4; if we can find the point G along EF such that FG is 3 units away from line AB [that is, at coordinate (4, 3)], it will form a 3-4-5 Egyptian right triangle, where AG is then 5 units. And by subtraction from EF, the length of EG is 5 units, putting point G 5 units away from line CD. Thus, a circle of radius 5 from point G is tangent to CD and crosses through A and B. To find point G, consider that the slope of AE is 8:4, so its perpendicular bisector has a slope -4:8 or -1:2. Line AE crosses its perpendicular bisector at point (2, 4), and that bisector then crosses EF at point (4, 3), giving the desired point G at the center of the circle.
Move Tool

Our nature consists in motion; complete rest is death. — Blaise Pascal

## Level Information

### Instruction:

Pascal's theorem states that if a hexagon is inscribed in a circle, then three points of intersection of its opposite sides lie on a straight line. Use the Move tool to see how the theorem works: just drag the blue points.

## Solutions

Let A, B, C, D, E, and F be the 6 points of a hexagon inscribed in a circle. Points G, H, and I are then respectively the intersections of AB and DE, BC and EF, and CD and AF (provided that those pairs of sides are not parallel).

### 0L 0E Solution

1. With the move tool active, select any of the six points on the circle; as labeled here, F makes the effect easier to see. Observe that as the point is moved, G, H, and I remain collinear.

## Explanation

This level does not construct anything, but does demonstrate Pascal's theorem in use. As long as opposite sides of the hexagon are not parallel (which cannot be achieved in Euclidea using just the Move tool), they have an intersection; moving a hexagon vertex shows that the dependent intersections H, I, and J remain collinear.

When the move tool is active, one finger on an empty area scrolls, one finger on a point, line, or circle moves that item along with all other items whose construction depends on it, and double-tap-and-drag-up/down zooms.
Intersection of Perpendicular Bisectors

Knowledge exists to be imparted. — Ralph Emerson

## Level Information

### Instruction:

Construct a segment connecting the sides of the angle to get a triangle whose perpendicular bisectors are intersected in the point O.

## 2L 2E solution

Let A be the vertex of the angle.

1. Construct the circle with center O and radius OA, intersecting the sides of the angle at B and C
2. Construct line BC

## Explanation

The point O is the circumcenter of the triangle, which means that there's a circle with center O that passes through all three vertices of the triangle. You know the center and you know the radius.
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Rotation 60°

Instruction: Rotate the line segment 60° clockwise about the given point.

Goal: 4L 4E

Available Tools:

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

Pack: Kappa

Previous Level: 10.4

Next Level: 10.6

## 4L 4E Solution

Given the segment AB and O the point.

1. Construct circle A with radius AO.
2. Construct circle B with radius BO, intersecting circle A at C.
3. Construct circle O with radius AO, intersecting circle A at D, with D being clockwise around O from A.
4. Construct line CD, intersecting circle B at E.
5. Point at E.
Geometric Mean of Segments

Mathematics is the part of physics where experiments are cheap. — Vladimir Arnold

## Level Information

### Instruction:

Construct a point on the given ray to get a segment with length equal to the geometric mean of two adjacent segments.

## Solutions

### 5E Solution

1. Construct a circle with center O and radius |OA|, intersecting segment AB at C.
2. Construct a circle with center C and radius |CO|, intersecting circle O at D and E.
3. Construct line DE.
4. Construct a circle with center B and radius |BO|, intersecting line DE at F.
5. Construct a circle with center O and radius |OF|, intersecting ray O at G.
6. Mark the point G.

### 3L Solution

1. Construct the perpendicular bisector of |AB|.
2. Construct the perpendicular of ray O at point O, intersecting the perpendicular bisector at C.
3. Construct a circle with center C and radius |CB|, intersecting ray O at point D.
4. Mark the point D.

## Explanation

### 5E Explanation

The goal is to construct two isosceles triangles, one inside the other, with a common side. The interior triangle has base of length *a* and one of the two equal sides is the common side. The exterior triangle uses the common side as the base and the other side lengths are *b*. This means the base of the interior triangle sits on one of the sides of the exterior triangle. With this construction, it is guaranteed that the common side has length equal to the geometric mean of *a* and *b*.

In step 1, we are copying the line segment of length *a* onto the segment of length *b*, setting up the base of the interior triangle on one of the sides of the exterior triangle. Steps 2 and 3 construct a perpendicular bisector of that base, on which the third point of the interior triangle must lie. Note that usually we need 3 steps for a perpendicular bisector but here we are reusing the circle from step 1. Step 4 constructs the circle on which the third point of the exterior triangle must lie. Their intersection, F, must therefore be the common third point. Step 5 rotates that common side onto the ray, yielding the desired point G.

### 3L Explanation

Let's draw line OD for clarity (we don't need it in solution, only for explanation) intersecting circle O in points D and E.

As CD=CE (radii of the circle) and

∠

{\displaystyle \angle}

COD=90o,

O
# E

C

D
2

−
C

O
2

{\displaystyle OD=OE=\sqrt{CD^2-CO^2}}

.

By Intersecting chords theorem OD\*OE=OA\*OB=ab.

O
# E

a
b

{\displaystyle OD=OE=\sqrt{ab}}
Excircle

## 4L solution

Let A, B and C be the intersections of the lines.

1. Construct the angle bisector A1 of ∠DAB, with D an arbitrary point on the line that intersects AB at A
2. Construct the angle bisector A2 of ∠EBA, with E an arbitrary point on the line that intersects AB at B, intersecting A1 at F
3. Construct the perpendicular to AB through F, intersecting AB at G
4. Construct the circle with center F and radius FG

## 8E solution

Let A, B and C be the intersections of the lines.

1. Construct the circle with center A and radius AB, intersecting AC at D, with D on the other side of A from C, if not, move C so C is inside circle A
2. Construct the circle with center C and radius CD, intersecting BC at E, E on the other side of C from B
3. Construct the circle with center B and radius BE, intersecting AB at F, F on the other side of A from B
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 B at I and J
6. Construct line GH
7. Construct line IJ, intersecting GH at K and AB at L
8. Construct the circle with center K and radius KL

## 3V solution

From the 4L solution:

1. Construct the angle bisector A3 of ∠HCA, with H an arbitrary point on the line that intersects AC at C, intersecting A1 at I and A2 at J
2. Construct the perpendicular to AC through I, intersecting AC at K
3. Construct the circle with center I and radius IK
4. Construct the perpendicular to BC through J, intersecting BC at L
5. Construct the circle with center J and radius JL
Octagon from Square

Math is not about following directions, it's about making new directions. — Paul Lockhart

## Level Information

### Instruction:

Construct a regular octagon by cutting the corners of the given square.

## Location Information

### Next Level:

8.9 Egyptian Triangle by Side of Length 4

## 7L solution

Let A, B, C and D be the corners of the square

1. Construct the perpendicular bisector of AB
2. Construct the perpendicular bisector of AD, intersecting the previous line at O the center of the square
3. Construct a circle with center O and radius OA, intersecting the first perpendicular bisector at points E and F and the second one at points G and H
4. Construct the line EG
5. Construct the line FG
6. Construct the line EH
7. Construct the line FH

## 8E solution

Let A, B, C and D be the corners of the square

1. Construct the line AC
2. Construct the line BD, intersecting AC at O the center of the square
3. Construct a circle with center A and radius AO, intersecting AD at E and AB at F
4. Construct a circle with center O and radius OE, intersecting the square at the 8 corners of the octagon E, F, G, H, I, J, K and L
5. Construct the line EF
6. Construct the line GH
7. Construct the line IJ
8. Construct the line KL
Angle Bisector

The description of right lines and circles, upon which geometry is founded, belongs to mechanics. Geometry does not teach us to draw these lines, but requires them to be drawn. — Isaac Newton

## Level Information

### Instruction:

Construct the line that bisects the given angle.

## Location Information

### 2L 4E Solution

1. Construct a circle with center A and an arbitrary radius, intersecting the sides of the angle at points B and C.
2. Construct the perpendicular bisector of BC.

### Alternative 4E solution

1. Construct a circle with center A and an arbitrary radius, intersecting the sides of the angle at points B and C.
2. Construct a circle with center B and radius AB.
3. Construct a circle with center C and radius AC, intersecting circle B at point D.
4. Construct line AD.

## Explanation

### Explanation of 2L 4E Solution

The circle allows you to construct an arbitrary isosceles triangle ABC with BC as its base. Then, constructing the perpendicular bisector of its base also allows you to find the required angle bisector as in an isosceles triangle, the angle bisector from the apex coincides with the perpendicular bisector of its base.

### Explanation of Alternative 4E Solution

The three circles allow you to construct a rhombus ABDC, which means its diagonal AD should bisect angle A.
Oakbrook Center/Tenant list

## Current

### Anchors

* Bonwit Teller
* I. Magnin
* Lord & Taylor
* Macy's
* Neiman Marcus
* Nordstrom
* Saks Fifth Avenue
* Sears

#### Minor anchors

* Barnes & Noble
* L.L.Bean

### Inline stores

* Artizia
* Ballard Designs
* Boss Hugo Boss
* The Container Store
* Crate & Barrel
* Eddie Bauer
* Forever 21/21 Kids
* Hollister/Gilly Hicks
* Hot Topic/ShockHound
* J.Crew
* Justice
* Louis Vuitton
* Lululemon Athletica
* Pandora
* Pirch
* Pottery Barn
* Pottery Barn Kids
* Spencer's
* Tommy Bahama
* Tiffany & Co.
* Urban Outfitters
* West Elm
* WTTW Store of Knowledge

### Dining

* The Cheesecake Factory
* Corner Bakery Cafe
* Maggiano's Little Italy

### Entertainment

* GCC Oakbrook Center 12
* Jeepers!
Reflection

## 5L 5E solution

Given the segment AB.

1. Construct a circle with an arbitrary center C on the line and radius AC
2. Construct a circle with another arbitrary center D on the line and radius AD, intersecting circle C at E
3. Construct a circle with an arbitrary center F on the line and radius BF
4. Construct a circle with another arbitrary center G on the line and radius BG, intersecting circle F at H
5. Construct line EH
Square by Two Vertices

The essential idea or method is always grasped intuitively long before any rational argument for the conclusion is devised. — Morris Kline

## Level Information

### Instruction:

Given two vertices of a square. Construct the two other vertices using only a compass.

## 7L 7E solution

Let A and B be the given vertices.

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
3. Construct the circle with center C and radius CD, intersecting circle A at E and circle B at F
4. Construct the circle with center E and radius BE
5. Construct the circle with center F and radius AF, intersecting circle E at G and H
6. Construct the circle with center A and radius AG, intersecting circle B at I and J, with I on the same side as G from AB
7. Construct the circle with center B and radius BG, intersecting circle A radius AB at K and L, with K on the same side as I from AB
8. Point at I
9. Point at K

## 3V solution

Starting from the 7L solution:

1. Point at J
2. Point at L
3. Construct the circle with center G and radius GI, intersecting circle A at M and circle B at N
4. Construct the circle with center K and radius KM
5. Construct the circle with center I and radius IN, intersecting circle K at O
6. Construct the circle with center A and radius AO
7. Construct the circle with center B and radius BO, intersecting circle A radius AO at P
8. Point at P

## Explanation

Since there are a large number of points involved, it may be easiest to map points to a coordinate system, by using the Pythagorean theorem on right triangles. Start with A at (0,0) and B at (1,0); the midpoint between A and B would be X at (1/2,0). The first two circles at A and B have radius 1, and construct C at (1/2,√3/2) and D at (1/2,-√3/2) because ABC and ABD are equilateral triangles. Then circle C has radius √3, and constructs E at (-1,0) and F at (2,0) (CEX is a 30-60-90 right triangle; we know leg CX is √3/2 and hypotenuse CE is √3, leaving leg EX at 3/2). Circles E and F each have radius 2, and construct G at (1/2,√7/2) and H at (1/2,-√7/2) (EGX is a right triangle with hypotenuse EG 2 and leg EX 3/2, leaving GX to be √7/2). At this point, we have triangle AGX with legs 1/2 and √7/2, which makes hypotenuse AG √2. So the second circles through A and B form 45-45-90 right triangles ABI, ABJ, ABK, and ABL, putting I at (1,1), J at (1,-1), K at (0,1), and L at (0,-1) giving the points needed for the first two squares (where A and B are adjacent vertices, side length 1).

The remaining variant for the 3V solution uses A and B as opposite rather than adjacent vertices (side length √2/2). With some effort, it can be shown that the triangles AGK and KIM are similar, so the ratio AG:AK = √2:1 = IK:KM = 1:KM, so the circle at K has radius KM = √2/2, likewise for the circle at I. From there, it follows that point O lies at (1/2,1/2) at the midpoint of AI and BK, and the final two circles through A and B reflect point O over AB to point P at (1/2,-1/2).
Circle in Rhombus

The mathematician's patterns, like the painter's or the poet's must be beautiful; the ideas like the colours or the words, must fit together in a harmonious way. — Godfrey Hardy

## 4L 6E solution

Given the rhombus ABCD.

1. Construct line AC, a diagonal of the rhombus
2. Construct line BD, the other diagonal of the rhombus, intersecting AC at O
3. Construct the perpendicular to AB through O, intersecting AB at E
4. Construct the circle with center O radius OE

## Explanation

In a rhombus, the inscribed circle is a circle with the same center as the rhombus and tangent to its sides. Tracing the diagonals gives you the center, then tracing the perpendicular to any of its sides through the center gives you the radius.
Annulus

## Level Information

### Instruction:

Construct a circle that is concentric with the given one and divides it into 2 parts of equal area.

## 4L solution

Let O be the center.

1. Construct an arbitrary line through O, intersecting the circle at A
2. Construct the perpendicular bisector of OA, intersecting at B
3. Construct the circle with center B and radius OB, intersecting the bisector at C
4. Construct the circle with center O and radius OC

## 5E solution

Let O be the center.

1. Construct an arbitrary line through O, intersecting the circle at A
2. Construct the circle with center A and radius OA, intersecting the circle at B and C
3. Construct line BC, intersecting OA at D
4. Construct the circle with center D and radius OD, intersecting BC at E
5. Construct the circle with center O and radius OE
Rhombus in Rectangle

Without mathematics there is no art. — Luca Pacioli

## 1.5 Rhombus in Rectangle

### 3L 5E Solution

1. Construct the perpendicular bisector of AC, intersecting AB at point E and CD at point F.
2. Construct line AF.
3. Construct line CE.

### 2V Solution

1. Construct the perpendicular bisector of BD, intersecting AB at G and CD at H.
2. Construct line BH.
3. Construct line DG.

## Explanation

In a rhombus, the diagonals are perpendicular bisectors to each other, so given one of the rectangle's diagonal (also one of the rhombus' diagonal), constructing its perpendicular bisector gives you the second diagonal. The intersections of that second diagonal with the rectangle give you the two remaining vertices of the rhombus.
Symmetry of Four Lines

## Level Information

### Instruction:

Three lines are intersected in a point. Construct a line so that the set of all 4 lines is mirror symmetric.

## 3L solution

Let O be the intersection of the three lines.

1. Construct a circle with center O and an arbitrary radius, intersecting the lines at six points; let A, B and C be a set of three consecutive points
2. Use the non-collapsing compass to construct the circle with radius AB and center C, intersecting circle O at D and E
3. Construct line OD

## 4E solution

Let O be the intersection of the three lines.

1. Construct a circle with center O and an arbitrary radius, intersecting the lines at six points; let A, B, C, D and E be a set of five consecutive points
2. Construct line AC, intersecting OB at F
3. Construct the circle with center O and radius OF, intersecting AC at G
4. Construct line OG

## 3V solution

From the 3L solution:

1. Construct line OE
2. Use the non-collapsing compass to construct the circle with radius BC and center A, intersecting circle O at F
3. Construct line OF

From the 4E solution:

1. Construct line BD, intersecting OC at H
2. Construct the circle with center O and radius OH, intersecting BD at I
3. Construct line OI
4. Construct line CE, intersecting OD at J
5. Construct the circle with center O and radius OJ, intersecting CE at K
6. Construct line OK

## Explanation

### 3L Solution

* Let points A-F be the end points of the given lines, going clockwise from top right.
* Let point O be the point of intersection of the given lines.
* Let line GH be the line drawn such that the angle bisector of angle AOB is the same as the angle bisector of angle FOH.
* Let points A-H be equidistant from point O
* Angle GOC is congruent to angle FOH because they are opposite angles.
* Angle GOC therefore shares it's angle bisector with angle FOH.
* Through the transitive property Angle GOC shares it's angle bisector with angle AOB.
* Angle AOB is therefore centered in angle GOC.
* Angle GOA and angle BOC are therefore congruent.
* Triangle GOA and triangle BOC are therefore 2 congruent isocoles triangles.
* Line segment GA and BC are therefore congruent.
* Therefore line GH can also be defined as the line which passes through point O and point G, where point G is defined as the point which is the same distance from point D as the distance between points B and C, and is also the same distance from point O as the distance between points O and A.
Segment by Midpoint

## Level Information

### Instruction:

Construct a segment with the ends on the sides of the angle such that the given point is its midpoint.

## 3L solution

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

1. Construct parallel of one ray through B, intersecting the other ray at C
2. Construct the circle with center C and radius AC, intersecting AC at D
3. Construct line BD

## 5E solution

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

1. Construct a circle with center B and an arbitrary radius BC, with BC<BA, intersecting one ray at D and E
2. Construct line BD, intersecting circle B at F
3. Construct line BE, intersecting circle B at G
4. Construct line FG, intersecting the other ray at H
5. Construct line BH

## Explanation

Creating the parallel line sets up two 1:2 similar triangles
Compass Tool

Note that the Compass tool costs 5 elementary (E) moves. — Euclidea Manual

## Level Information

### Instruction:

You have discovered the Non-collapsing Compass tool. Click on two points to set the radius length, then on the center of the circle.

## Solutions

### 1L 5E Solution

1. With the compass tool active, click on one endpoint of the segment, then another endpoint of the segment, then on the destination point.

## Explanation

The compass tool works by selecting an initial radius (either click on the two points of the segment, or click-and-hold on one point of the segment and release on the second), followed by selecting a destination center with another click. The tool visualizes the size of the circle before it lands in its final location, then adds dashed line decorations indicating the two parallel and equivalent-length radii used between the original two points and the destination circle. Note that the decorations cannot be used as a target by other tools if they do not overlay an already-existing line. The two original points do not have to lie on a circle, and Euclidea does not allow the destination point to be either of the two source points (even if it did, using the Circle Tool would be a less expensive alternative). When copying an existing circle, it does not matter whether the first point is the center or on the circumference, as the selected radius will be the same either way for the destination center.
Angle of 15°

If we know anything we know it by mathematics. — Pierre Gassendi

## 3L solution

Let A be the initial point of the given ray.

1. Construct a circle with an arbitrary center B and radius AB, intersecting the ray at C
2. Construct the circle with center C and radius BC, intersecting circle B at D and E (D above ray to match orientation of explore mode)
3. Construct the angle bisector of ∠DAC

## 5E solution

Let A be the initial point of the given ray.

1. Construct a circle with an arbitrary center B and radius AB, intersecting the ray at C
2. Construct the circle with center C and radius BC, intersecting circle B at D and E (D above ray to match orientation of explore mode, E unused)
3. Construct the circle with center D and radius CD, intersecting circle C at B and F
4. Construct line BF, intersecting circle A at point G inside circles C and D
5. Construct line AG

## 2V solution

From the 3L solution:

1. Construct the angle bisector of ∠EAC

From the 5E solution:

1. Construct the circle with center C and radius CG, intersecting circle A at point H
2. Construct line AH

## Explanation

See Angle of 30° for the proof that points D and E form a 30° angle from A; the angle bisector thus gives the desired 15° angle. The 3L solution constructs the angle bisector directly; the 5E solution uses the fact that BCD is an equilateral triangle, so the perpendicular bisector of CD (line BG) divides arc CD of circle A into two equal parts. The 2V solution then reflects the first solution over ray A.
Nine Point Circle

## Level Information

### Instruction:

Construct a circle that passes through the midpoints of sides of the given acute triangle.

## 5L solution

Given the triangle ABC.

1. Construct the perpendicular to AB through C, intersecting AB at D
2. Construct the perpendicular to BC through A, intersecting BC at E
3. Construct the perpendicular bisector of DE, intersecting AC at F
4. Construct the perpendicular bisector of DF, intersecting the perpendicular bisector of DE at G
5. Construct the circle with center G and radius DG

## 9E solution

Given the triangle ABC.

1. Construct the perpendicular bisector of AB, intersecting AB at D
2. Construct the circle with center D and radius AD, intersecting AC at E and BC at F
3. Construct the circle with center E and radius DE, intersecting circle D at G and H
4. Construct line GH
5. Construct the circle with center F and radius DF, intersecting circle D at I and J
6. Construct line IJ, intersecting GH at K
7. Construct the circle with center K and radius DK
Intersection of Angle Bisectors

Have the courage to use your own reason. — Immanuel Kant

## 2.2 Intersection of Angle Bisectors

Given the triangle ABC.

Point at D.

1. Construct the circle with center A and radius AB, intersecting AC at point D.
2. Construct the circle with center B and radius AB, intersecting BC at point E.
3. Construct the circle with center D and radius AD, intersecting circle B at point F.
4. Construct line AF.
5. Construct the circle with center E and radius BE, intersecting circle A at point G.
6. Construct line BG, intersecting AF at H.

Point at H.

## Explanation

Solution 6E has you manually construct the two angle bisectors. The first two circles has you construct two isosceles triangles ABD and ABE. Considering ABD with its base BD and the circle B, you can construct another circle with center D and the same radius and both circles would intersect along the perpendicular bisector of BD, which coincides with the angle bisector of ∠BAD since ABD is isosceles. Use the same reasoning by considering ABE with is base AE and the circle A and you can construct the perpendicular bisector of AE, which again coincides with the angle bisector of ∠ABE. Although the same point will be constructed for any two of the three angle bisectors of the triangle, the 6E solution only works when started from the shortest side, as otherwise you would also have to construct a line extension to the other side to locate the point forming the base of an isosceles triangle.
Circumscribed Square

A round man cannot be expected to fit in a square hole right away. He must have time to modify his shape. — Mark Twain

## Level Information

### Instruction:

Circumscribe a square about the circle. Two of its sides should be parallel to the given line.

## 6L solution

Let O be the center of the given circle.

1. Construct the perpendicular to the line through O, intersecting the circle at A and B
2. Construct the perpendicular to AB from A
3. Construct the perpendicular to AB from B
4. Construct the circle with center A and radius OA, intersecting the perpendicular to AB from A at C and D
5. Construct the perpendicular to CD from C
6. Construct the perpendicular to CD from D (total 16E)

## 11E solution

1. Construct the perpendicular to the line through O, intersecting the circle at A and B, and the given line at C
2. Construct the circle with center C and radius OC, intersecting the given line at D
3. Construct line OD
4. Construct the circle with center A and radius OA, intersecting OD at E
5. Construct the circle with center B and radius OB, intersecting OD at F
6. Construct line AE, intersecting circle A at G
7. Construct line BF, intersecting circle B at H
8. Construct line FG
9. Construct line EH

## Explanation

To have sides parallel with the given line, it is necessary to start with the diameter of the circle that is perpendicular to that line. From there, minimizing the number of E moves in the 6L solution prefers the use of further perpendiculars rather than the parallel tool, and circle A rather than a perpendicular through the center of the circle to find useful points for constructing the two sides.

The 11E solution uses circle C to find the diagonal of the square at a 45° angle. With that diagonal and the perpendicular midline, it is easy to create the two parallel lines AE and BF with the aid of circles A and B. That in turn gives enough points to trace the remaining sides FG and EH.
Inscribed Square

Where there is matter, there is geometry. — Johannes Kepler

## 1.7 Inscribed Square

Let O be the center of the circle and A the vertex.

2. Construct the perpendicular bisector of AB, intersecting the circle at points C and D.
3. Construct line AC.
4. Construct line BC.
5. Construct line BD.
6. Construct line AD.

### 7E Solution

1. Construct the circle with center A and radius OA, intersecting circle O at points B and C.
2. Construct the circle with center B and radius BC, intersecting circle Oat D.
3. Construct line OB, intersecting circle B at points E and F.
4. Construct line ED, intersecting circle O at point G.
5. Construct line AG.
6. Construct line DF, intersecting circle O at point H.
7. Construct line AH.

## Explanation

Solution 6L is easy as the inscribed square's center is the same as the circle's center, so you can find the vertex opposite to A as its diametrical opposite. That diameter is also a diagonal of the square, so by constructing its perpendicular bisector and second diagonal of the square, you can find the two remaining vertices and just need to trace the whole square.

Solution 7E is trickier. By construction, ABOC is a rhombus, and ABO and ACO are equilateral triangles as all their sides are equal. Therefore, AO and BC (the rhombus's diagonals) are perpendicular bisectors to each other, and angle BOC is 120°.

By construction as well, BCD is an isosceles triangle as BC=BD, and since angle ∠BDC is the inscribed angle that subtends the same arc as the central angle ∠BOC, ∠BDC=120°/2=60°. Therefore, BCD is also an equilateral triangle and AO is one of its bisectors and passes through D, making A, O and D collinear and A and D diametrical opposites on circle O.

Constructing line OB gives you EF, a diameter of circle B, and as such, EDF is a right triangle. And since AD is a diameter of circle O, AGD and AHD are right triangles as well. So, AGDH has three right angles and is therefore a rectangle. In fact, you can construct any diameter EF of circle B (that doesn't pass through point O) and similarly prove that the resulting AGDH is a rectangle.

Now, the fact that the diameter EF passes through point O allows you to determine several angles, for example angle ∠ADG.

∠ADG=∠ADB+∠BDG

∠ADB is known since the triangle BDC is equilateral and AD is one of its bisectors, so ∠ADB=60°/2=30°.

∠BDG=∠BEG since the triangle BED is isosceles. Also ∠BEG is the inscribed angle that subtends the same arc as the central angle ∠FBD in circle B, so ∠BEG=∠FBD/2. Also, in circle O, ∠FBD is the inscribed angle that subtends the same arc as the central angle ∠FOD, so ∠FBD=∠FOD/2. ∠FOD itself is known as it's the opposite angle of ∠AOB and is therefore 60° as AOB is equilateral. So ∠BDG=∠BEG=∠FBD/2=∠FOD/4=60°/4=15°

So ∠ADG=∠ADB+∠BDG=30°+15°=45°, which makes triangle AGD right and isosceles. Therefore, the two adjacent sides AG and GD of the rectangle AGDH are equal, which makes AGDH a square.
Sum of Areas of Squares

## Level Information

### Instruction:

Construct a square whose area equals the sum of the areas of the two given squares and all three have the common angle.

## 3L solution

Let O be the common vertex of the squares, A the second vertex of one square on one line and B the second vertex of the other square on the other line.

1. Use the non-collapsing compass to construct the circle with radius AB and center O, intersecting one line at C and the other line at D
2. Construct the perpendicular to OC from C
3. Construct the perpendicular to OD from D

## 6E solution

Given the smaller square ABCD with angle ∠ABC the common angle.

1. Construct line AD, intersecting the bigger square at E
2. Construct the circle with center B and radius BE, intersecting one line at F and the other line at G
3. Construct the circle with center F and radius BF
4. Construct line BD, intersecting circle F at H
5. Construct line FH
6. Construct line GH
Circumscribed Equilateral Triangle

Logic is the anatomy of thought. — John Locke

## Level Information

### Instruction:

Construct an equilateral triangle that is circumscribed about the circle and contains the given point.

## 5L solution

Let O be the center of the circle and A the point on it.

1. Construct line OA, intersecting the circle at B
2. Construct the perpendicular to OA from A
3. Construct the circle with center B and radius OB, intersecting the circle at C and D, and line OA at E
4. Construct line CE
5. Construct line DE

## 6E solution

Let O be the center of the circle and A the point on it.

1. Construct line OA, intersecting the circle at B
2. Construct the circle with center B and radius OB, intersecting the circle at C and D, and line OA at E
3. Construct the circle with center O and radius OE
4. Construct line CE, intersecting the latest circle at F
5. Construct line DE, intersecting the latest circle at G
6. Construct line FG

## Explanation

Line OA forms an altitude of the equilateral triangle, with A at the midpoint of the base. Both constructions use circle B to determine the two sides at a 30° angle from the apex; this can be seen since triangles BCO and BDO are equilateral by construction, and angles ECO and EDO are 90 by Thales' theorem. Having drawn tangents through C and D, point E is the apex of the desired triangle. The final leg of the triangle through A is built with the perpendicular tool for the 5L solution, and by inscribing an equilateral triangle into the second larger circle O in the 6E solution.
Angle of 45°

He is unworthy of the name of man who is ignorant of the fact that the diagonal of a square is incommensurable with its side. — Plato

## Level Information

### Instruction:

Construct an angle of 45° with the given side.

## 2L solution

Let A be the initial point of the given ray and B a distinct arbitrary point on that ray.

1. Construct the perpendicular to the ray from A; let C be a distinct arbitrary point on that perpendicular
2. Construct the angle bisector of ∠BAC

## 5E solution

Let A be the initial point of the given ray and B a distinct arbitrary point on that ray.

1. Construct the perpendicular to the ray from B
2. Construct the circle with center B and radius AB, intersecting the perpendicular at C and D
3. Construct line AC

Alternate 5E 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 C and D
3. Construct line BC, intersecting circle B at E
4. Construct the circle with center E and radius AE, intersecting line BC at F and G
5. Construct line AF

## 2V solution

From the 2L solution:

1. Construct the perpendicular of the first line at point A, for 3L total (either by the perpendicular tool for 10E total; or by the angle bisector on ∠BAD, where D is an arbitrary point on line AC on the side of A opposite from C, for 11E total)

From the 5E solution:

1. Construct line AD, for 6E total

From the alternate 5E solution:

1. Construct line AG, for 6E total

## Explanation

By tracing the perpendicular to the line, the angle bisector will divide the 90° angle into two 45° angles.

The first 5E solution has you construct a right isosceles triangle, which has two angles of 45°.
Square Root of 3

The square root of 3 is also known as Theodorus' constant. — Wikipedia

## Level Information

### Instruction:

Let |AB|=1. Construct a point C on the ray AB such that the length of AC is equal to √3.

## 3L 3E solution

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

## Explanation

* AB = 1 = BD,

∴

{\displaystyle \therefore}

 DE = 1
* AD = 2
* ∠

{\displaystyle \angle}

AED = 90o based on Thales Theorem

Using the 30-60-90 special triangle, we can deduce that side AE =

3

{\displaystyle \sqrt{3}}

. AE and AC are both radii of the circle centered at A.
Minimum Perimeter - 2

Instruction: Inscribe a triangle into the given one whose perimeter is the minimum possible.

Goal: 6L 8E

Available Tools:

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

Pack: Iota

Previous Level: 9.6

Next Level: 9.8

## 6L Solution

Let A, B, and C be the vertices of the triangle.

1. Construct the line perpendicular to AB through C, intersecting AB at D.
2. Construct the line perpendicular to AC through B, intersecting AC at E.
3. Construct the line perpendicular to BC through A, intersecting BC at F.
4. Construct line DE.
5. Construct line DF.
6. Construct line EF.

## 8E Solution

Let A, B, and C be the vertices of the triangle.

1. Construct the perpendicular bisector of AB, intersecting AB at D
2. Construct circle D with radius AD, intersecting AC at E and BC at F.
3. Construct line EF.
4. Construct circle A with radius AE, intersecting circle D at G.
5. Construct line FG, intersecting AB at H.
6. Construct line EH.
Interior Angles

## Level Information

### Instruction:

Construct a line through the point that crosses the two lines so that the interior angles are equal.

## 2L solution

Let A be the point of intersection between the two lines. Let B be an arbitrary point on one of the lines, and C an arbitrary point on the other. Let O be the given point.

1. Bisect the angle BAC. The new line contains an arbitrary point D.
2. Construct the line perpendicular to AD through O.

## 4E solution

Let A be the point of intersection between the two lines. Let O be the given point.

1. Construct a circle with center A and a radius of AO. Circle A intersects with the two lines at 4 distinct points. Let B be the point from among those 4 which is closest to O. Let C be one of the two points not on line AB.
2. Construct a circle with center C and radius CO. Circle C intersects with circle A at point D.
3. Construct line BD, which intersects with circle C at point E.
4. Construct line OE.

## 2V solution

Continuing from either solution:

1. Construct the line perpendicular to the solution line through point O.
Third Proportional

Everything around us is geometry. — Le Corbusier

## Level Information

### Instruction:

Construct a point D on the ray such that the segment AD is the third proportional to the given line segments.

## 3L 3E solution

1. Construct circle A with center A and radius AB.
2. Construct circle C with center C and radius CA, intersecting circle A at E.
3. Construct circle E with center E and radius EA, intersecting circle O at D.

## Explanation

We have

△
A
E
D
∼
△
A
C
E

{\displaystyle \triangle AED\sim \triangle ACE}

, then

A

E

2

=
A
D
⋅
A
C

{\displaystyle AE^{2}=AD\cdot AC}

, or

A

B

2

=
A
D
⋅
A
C

{\displaystyle AB^{2}=AD\cdot AC}

, or

A
C

A
B

=

A
B

A
D

{\displaystyle \frac{AC}{AB}=\frac{AB}{AD}}

.
Cut Two Rectangles

Mathematics is the most reliable form of prophecy. — Wilhelm Schwebel

## 5L 5E solution

Given the rectangles ABCD and EFGH.

1. Construct line AC, a diagonal of the first rectangle
2. Construct line BD, the other diagonal, intersecting the previous diagonal at I
3. Construct line EG, a diagonal of the second rectangle
4. Construct line FH, the other diagonal, intersecting the previous diagonal at J
5. Construct line IJ

## Explanation

Any *straight* line through the center of a rectangle divides the rectangle into equivalent halves (180° rotational symmetry around its center). Find each rectangle's center and draw one line through both points.
Parallel Line

It is ignorance alone that could lead anyone to try to prove the axiom. — Aristotle

## Level Information

### Instruction:

Construct a line parallel to the given line and through the given point.

## 2L solution

Let A be the given point.

1. Construct the perpendicular to the line through A
2. Construct the perpendicular to the previous perpendicular from A

## 4E solution

Let A be the given point.

1. Construct a circle with an arbitrary center B on the line and radius AB, intersecting the line at C
2. Construct the circle with center C and radius BC
3. Construct the circle with center A and radius AB, intersecting circle C at D
4. Construct line AD

Alternate 4E solution:
Let A be the given point.

1. Construct a circle with an arbitrary center B on the line and radius AB, intersecting the line at C
2. Construct line AB, intersecting circle B at D
3. Construct the circle with center C and radius CD, intersecting circle B at E
4. Construct line AE

Alternate 4E solution:
Let A be the given point.

1. Construct a circle with an arbitrary center B on the line and radius AB, intersecting the line at C
2. Construct the circle with center C and radius CA, intersecting circle B at D
3. Construct line CD, intersecting circle C at E
4. Construct line AE

## Explanation

The 2L solution works because any two lines both perpendicular to a third a parallel to one another.

The first 4E solution is easily explained by noticing that a well-placed rhombus ABCD is constructed. A rhombus is a parallelogram thus its sides are, unsurprisingly, parallel.

The second 4E solution reflects point D over the line to E; this in turn means that DE is perpendicular to the original line; and by Thales' theorem, ∠DEA is 90° since AD is a diameter of circle B.

The third 4E solution reflects point A over the line to D; this in turn means that AD is perpendicular to the original line; and by Thales' theorem, ∠DAE is 90° since DE is a diameter of circle C.
Center of Rotation

## Level Information

### Instruction:

Construct the center of rotation that transforms one of the given equal segments to the other one.

## 2L Solution

Given one line AB and the other CD.

1. Construct the perpendicular bisector of AC.
2. Construct the perpendicular bisector of BD, intersecting the other bisector at E.
3. Point at E

## 5E Solution

Given one line AB and the other CD.

1. Construct the circle with centre A and radius AD, intersecting AB at E.
2. Construct the circle with centre D and radius AD, intersecting CD at F and circle A at G and H.
3. Construct line GH.
4. Construct the circle with centre F and radius EF, intersecting circle A at I.
5. Construct line DI, intersecting GH at J.
6. Point at J.

## 2V Solution

From the 2L solution.

1. Construct the perpendicular bisector of AD.
2. Construct the perpendicular bisector of BC, intersecting the other bisector at F.
3. Point at F.