Epilogue - there is more than one way

I hope you had fun with the game, just like I do, the answer provided here are meant to be a sample solution. There are, of course, other ways to accomplish the same goal. I am sure the answers are correct, but in case you think my answer is inaccurate or you have better solutions, please leave me comments!

Level 20: Construct a line (segment) tangent to both circles. Construct the outer tangent line.

Hint: Think ratios.

Solution: First construct the ray AB, and let the outer tangent line intersect AB at O. Now OA:OB should be (radius of A: radius of B). To find such point O, we use the similar triangle NAO and MBO. Once we found the point O, we need to find a point C on circle A such that OCA is a right angle. Now we leverage another fact, angle subtended by the diameter on a circle is the right angle, so we construct the midpoint P of O and A, draw the circle with center P and radius PA, so the intersection between the circle would give us the point C and we are done.

Level 19: Construct two points, such that the segment is cut into three equal pieces.

Hint: Perhaps unrelated – study the centroid of a triangle.

Solution: The centroid of a triangle is found by intersecting the median, the median is defined by the intersection between the midpoint of an edge and the opposite vertex. The interesting thing about this is that it can be shown that the intersection point always cut the median into two-third on the vertex side and one third on the remaining side. It is this property that we will leverage for this level.

With that in mind, we make AB the median, which means A is the opposite vertex and B is the midpoint of some base line, construct the base line by creating a small circle at B and draw the diameter CD, find the midpoint E of AD and join CE for another median, now we find F which is the centroid of the triangle ACD. That cuts the line in two third/one third, and so we use another circle with center F and radius FB to find the remaining point G. 

Level 18: Construct two new circles of radius AB where each pair of the three circles is tangent.

Hint: Let’s play pool?

Solution: The pool configuration is exactly what we will need to have the pairwise tangent. The rest is just construct it, nothing special with this level.

Level 17: Given a point A, a line, and a point B. Construct a circle that passes through A and is tangent to the line at B.

Hint: The circle passes through A and B, what else?

Solution: The circle must also pass through the mirror image of A. The line is a tangent, which means the line through B perpendicular to it is the diameter. The point D, the mirror image of A, can be constructed by first constructing a parallel line through A parallel to the tangent, and intersect it with the circle with center B and radius AB. The rest? It is just level-15, the circumcircle of A, B and D!

For economy we could have just constructed the perpendicular bisector of AB and then the line through B perpendicular to the tangent and use that as the circumcenter, but it would be hard to explain why that’s true.

Level 16: Given a line, a line segment CD, and a point O. Construct a circle with center O that cuts off a segment congruent to CD on the given line.

Hint: The same old hint – perpendicular bisector of chord pass through center.

Solution: The segment cuts off by the required circle is a chord, and the perpendicular bisector of the chord must pass through center. So let construct the perpendicular bisector by constructing the line pass through O perpendicular to the given segment. That gives us the point J. The point J bisect the chord, and the chord must have length CD, so just find the midpoint of CD (name it E), and then draw a circle with center J with length CE, that gives us the point K, the circle with center O and radius OK give us the solution.

Level 15: Construct the circumcircle of a triangle

Hint: All three sides are chord of the circumcircle, and the perpendicular bisector of chord …

Solution: … pass through center! Just construct two perpendicular bisectors of two sides, their intersection must be the center of the circumcircle. The radius? Just use any of A,B or C!

Level 14: Construct the in-circle of a triangle:

Hint: The center of the in-circle must have same distance to the triangle lines. In particular, two of the three lines?

Solution: Consider the set of all points that is of the same distance to both AB and AC, what is it? It is the angle bisector of angle BAC! As the hint stated, the center must lie on the angle bisectors, so the angle bisectors intersects and that is the center of the in-circle. We only need two angle bisectors (of angle ABC and angle BAC) to find the center of the in-circle. To find the radius of the in-circle, note that the side must be perpendicular to the radius (as it is tangent), so we construct the perpendicular through the center to AB, that give us the intersection point F and that is the radius.

Level 13: Construct a line (segment) at B tangle to the circle

Hint: Tangent is perpendicular to radius

Solution: Construct the radius AB, and then construct the line at B perpendicular to AB, and we are done.

Level 12: Find the center of the circle

Hint: ALL Perpendicular bisector of line pass through center.

Solution: Construct two non-parallel perpendicular bisector, since they must both pass through center, which then must intersect at the center.

Level 11: Construct an angle on the given line equals to the given angle

Hint: Reproduce triangle with specified length and base

Solution: In the last level we already know how to reproduce a triangle with specified length and base. This is only another instance of it. Reproduce the triangle ABC with the given line D as base would solve the problem. Specifically, draw circle with center D with radius AB, mark the intersection point K. Draw circle with center D and radius AC, and draw circle with center K and radius BC. Draw the ray from D to the new intersection point L and we are done.

Level 10: Construct a triangle whose sides have the same length as the given segments. Use segment AB as base.

Hint: Use compass2 again.

Solution: Indeed, construct circle with center A and radius CD, and then construct circle with center B and radius EF, the intersection point is the vertex of the new triangle. 

Level 9: Construct a new point E on the line segment CD such that CE has the same length as AB

Hint: Use compass2.

Solution: Use the compass2 tool to draw a circle with center C with radius AB, now obviously the point E is the intersection between the circle and CD.

Level 8: Construct a circle with radius equals to line segment AB and center C.

Hint: Come on – we already know how to translate a line segment …

Solution: Translate the line segment AB to C, and then draw the circle as required.

Level 7: Construct a line segment with the same length and same direction as line segment AB but with begin point C.

Hint: Think parallelogram

Solution: Suppose we have the required line CD, join BD, then ABDC must be a parallelogram. Using the parallel line tool in level 6. We can easily construct this parallelogram by first constructing a line at B parallel to AC, and then a line at C parallel to AB, the new intersection point D is the parallelogram point. 

Level 6: Construct a parallel line through point C

Hint: A line perpendicular to a perpendicular line is parallel.

Solution: The hint spell out the answer! First, construct a line through C perpendicular to AB. Then construct a line through C perpendicular to the previous line, the line must pass through C, and it must perpendicular to the line perpendicular to AB, so it must be parallel to AB, so we are done. 

Level 5: Construct a line (segment) perpendicular to line segment AB going through point C.

Hint: Perpendicular bisector of a chord go through the center.

Solution: Without any particular reason, let construct two circles with center A and center B with radius AC and BC respectively. The two circle intersect at C and we name the other intersection E. Name the intersection between CE and AB as F. Now CF = EF because of symmetry. Triangle BCF is congruent to triangle BEF, so angle CFB = angle EFB, but they must also sum to 180 degree, so it is a right angle and we are done. 

Level 4: Construct a line (segment) that goes through point A perpendicular to the given line segment

Hint: Make some new point to unblock. Diagonal of rhombus are perpendicular to each other.

Solution: Given the hint, all we need to do is do redo the rhombus trick in level 2. But there is just one point? No worries, just create one using the point of the object tool. We just pick point D somewhere, to make A the midpoint, I used the circle tool to construct a circle with center A and radius AD. Mark the two intersections as E and F, and A is obviously the midpoint of EF. To complete the construction, note that we already know A is the intersection point between the diagonals, so we only need one equilateral triangle and join the vertex to A, and we are done.

Level 3: Construct a line (segment) that bisects the given angle

Hint: The angle bisector is an axis of symmetry.

Solution: It is natural to construct a circle with center A and radius AB, the circle intersect the other line at E. Now, imagine the angle bisector exists, it must be an axis of symmetry that maps B to E. Therefore, the midpoint of E must lie of the angle bisector, and we are done!

Level 2: Construct an equilateral triangle from the line segment AB.

Hint: The diagonals of a parallelogram bisect each other.

Solution: Given the hint, we want to make AB a diagonal of a parallelogram, we already know how to create equilateral triangle, so just stack two equilateral triangles together to become a parallelogram (actually a rhombus) and we are done. 

Level 1: Construct an equilateral triangle from the line segment AB.

Hint: You only need another point C so that it is equal distance from A and from B.

Solution: This is a very simple problem (after all this is just level 1) – the solution is as follow – you construct the other vertex C of the equilateral triangle. The point C must have AC = AB, so it must lie on the circle with center A and radius AB. The point C must also have BA = BC, so it must lie on the circle with center B and radius BA, therefore the solution is as follow

Euclid the game spoiler

This blog is written to show case the solutions for https://kasperpeulen.github.io/

Do NOT read this blog if you want to play the game for your own, it was fun for me so I don’t want to spoil your fun. But if you are stuck, or if you just want the solutions, here they are.