Which way did the bicycle go (Part 1)

I found the source of this classic smart question. It seems to have come from a book called which way did the bicycle go. This book is also a collection of awesome interesting questions, which contains many interesting questions that I have never seen before. I found an electronic version of this book and posted it to my website to share it with you. You can click here to download. The reader can be found here.

I sorted out some of my favorite questions. If you do not have time to repeat the entire book, it is also a good choice to look at these questions I have selected.

 

1. Given △abc, for any point of X on the plane, it belongs to the point set S. if and only if a point of D exists on the Line Segment BC, △adx is an equilateral triangle. What does point set S look like?

 

Answer: two line segments, which rotate 60 degrees clockwise or counterclockwise around a point by the BC line segment. This is because, given a and X points, the position of D points can be obtained by rotating 60 degrees around. Since the D point is on BC, it is clear that the X point should be on the line segment generated by the 60 degrees of rotation around a in BC.

 
 

2. Can I divide a square into seven isosceles right triangles, where any two triangles are incomplete?

 

Answer: Yes ..

 
 

3. Can I divide an equi-edge triangle into five equi-waist triangles so that
(1) are all five triangles not equilateral triangles?
(2) Is there an equi-edge triangle?
(3) are two triangles equal-edge triangles?

 

Answer: Yes ..

 
 

4. In △abc, AB = ac, then a = 20 °. P must be AP = BC on AB. Calculate the limit ACP.

 

Answer: Fold △abc twice to get △acd and △ade. Capture AQ on the edge of AE to make AQ = ap. Apparently △apq is an equi triangle, so AP = PQ = CD. In addition, SAS can obtain the full equality of △apc and △aqd. Therefore, Pc = qd, And the Quadrilateral pqdc is a parallelogram. From the full level, you can also obtain ∠ APC = ∠ Aqd. From this we can see that limit 1 = Limit 2. That is to say, the Quadrilateral pqdc is actually a rectangle. Therefore, ∠ ACP = ∠ ADQ = 90 °-∠ ADC = 10 °.

 
 

5. In △abc, M is the midpoint of the AB edge. Take AC as the side as the positive hexagonal, P as its center, BC as the side as the positive triangle, Q as its center. Proof: ∠ pmq is a right angle.

 

Answer: If the image is rotated 180 degrees around the M point, the Quadrilateral pqp 'q' is a parallelogram. Next we will prove the equality of the two shadow triangles. Apparently CP = BP 'and CQ = Bq. In addition, note that the three angles of △abc are α, β, and gamma, then ∠ pcq = 360 °-60 °-30 °-Gamma = 270 °-(180 °-α-β) = 90 ° + α + Beta = 60 ° + Alpha + BETA + 30 ° = ∠ P 'bq, so △pcq ≌ △p 'bq. Therefore, PQ = QP 'and quadrilateral pqp 'q' are diamond, and their two diagonal lines are perpendicular to each other.

 
 

6. In △abc, ad is the angular bisector, and m is the midpoint of BC. Parallel lines over M as ad, and AB are handed over to point N. Verify that Mn shares the perimeter of △abc.

 

Answer: I used C as the parallel line of Mn and handed over the extended line of Ba to E. E-certificate AC = AE, so the circumference of △abc is equal to BC + bE. You only need to note that Mn is the middle line of △bce, and the problem is justified.

 
 

7. The five circles are tangent in turn, and they are all tangent to two non-parallel straight lines. If the radius of the leftmost circle is 4 and the radius of the rightmost circle is 9, find the radius of the center circle.

 

Answer: 6. The following shows that the radius of the Five Circles is an proportional sequence. Write the Five Circles in ascending order as C1, C2, C3, C4, and C5, and mark the intersection of the two straight lines as p. The distance from the center of C1 and C2 to P is recorded as P1 and P2 respectively. Now, if the entire graph is reduced to the original P1/P2 with P as the center, the two straight lines are still in the original position, but C2 currently occupies the C1 position. In addition, since all the tangent relationships remain unchanged, the new C3 is the original C2, the new C4 is the original C3, and the new c5 is the original C4. This shows that each Ci is reduced to the original P1/P2 and the Ci-1 coincidence, that is, the ratio of the radius of each two adjacent circles is P1/P2.

 
 

8. Given a straight line and a point P outside the straight line, then given a point O on the straight line, and a Circle centered on O. How can we use only one ruler without a scale to create a vertical line with a known straight line crossing the P point?

 

Answer: connect the AP and the circle to Q. prolong the Pb and the circle to R. Then X of the extension line of AR and QB satisfies the PX limit L. This is because QX and PR are the height of triangles in △retrial, which means that point B is the center of a triangle, and there is a PX limit l naturally.

 
 

9. Proof: if a positive 400 edge is divided into parallelogram, there are at least 100 rectangles.

 

Answer: assume that the base side of the positive 400 edge is a horizontal line segment. Obviously, we can start from the top side, pass through a parallelogram, and walk all the way to the bottom side, so that the line passing through the road is a horizontal line segment; similarly, there is also such a path between the leftmost and rightmost ends of the 400 edge shape. Each side passing through the road is a vertical line segment. But the two paths obviously have an intersection. The quadrilateral where the intersection is located is obviously a rectangle. This operation can be performed 400 times in different directions of the positive 100 edge shape, so we can find 100 rectangles with different orientations.

 
 

10. As shown in the figure, a method is used to overwrite each vertex (including the vertex on the boundary) in a non-Intersecting Line Segment. The length of each line segment is not 0. Is it possible?
(1) how to overwrite each vertex in a triangle with an unmatched line segment whose length is not 0?
(2) How to overwrite each vertex in a circle with non-intersecting line segments whose lengths are not 0?

 

Answer: Yes ..

 
 

11. Given any quadrilateral ABCD and a point O outside the Quadrilateral. Translate AB to Oa, translate BC to OB, translate CD to OC, and then translate da to OD '. Returns the area ratio of the two quadrants.

 

Answer: BC to E, so the triangle (1) and (2) are the same area. Obviously, the triangle (2) and (3) are equal, so (1) and (3) are the same area. Likewise, we can see that the four triangles in the right quadrilateral are actually equal to △abc, △bcd, △cda, and △dab, so the area of the right quadrilateral is twice that of the left.

 
 

12. There are a total of six line segments between the four points in the figure. they satisfy the following requirements: one type of length occurs exactly once, one type of length occurs exactly twice, and one type of length occurs exactly three times. Whether there are five points on the plane. The 10 line segments between them meet the requirement that one length occurs exactly once, one length occurs exactly twice, and one length occurs exactly three times, is there a length that occurs exactly four times?

 

Answer: Yes. Is a simple structure: △abc is an equedge triangle, and O is its center. Take A as the center, AB as the radius as the arc, and the center vertical line of OB and this arc intersect at the Point D. These five points meet the requirements.

Due to the influence of many geometric propositions related to dimensions, many may think that the five points are already the most. Actually not. We have discovered some structures with n = 6, n = 7, or even n = 8. It shows a structure of N = 8. The person who constructs this stuff is really cool.

 
 

13. There is a light source in the Quadrilateral room, which illuminates most areas, with only two sides of the wall having shadow. Is there such a polygon room where the light source is placed so that each wall has a shadow?

 

Answer: Yes ..