Straightedge and Compass
The red lines and circles show the new constructions at each step. For more information about The Geometer's Sketchpad, see www.keypress.com. To learn about Cabri Geometry, head to education.ti.com.
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| 3. | A regular hexagon can be constructed in 10 steps (3+1+6).  | 4. | A regular octagon can be constructed in 20 steps (9+2+1+8). The first 9 steps make a square. |
1. Construct a pair of perpendicular lines using only one circular arc and as many straight lines as you wish.
2. Draw a circle. What is the minimum number of circles, all the same size as the original circle, needed to construct a square that circumscribes the circle (that is, the circle touches the square at just the four midpoints of the sides of the square)? Again, you may use as many straight lines as you wish.
Dynamic Geometry
1. The red point follows a curve called a cycloid that travels straight up and down at the point at which it touches the ground.
2. The red point follows a straight line, while the purple point follows an elliptical path.
3. The purple point swings back and forth along an arc that is slightly less than a semicircle. The red point traces a figure-eight curve called a lemniscate.
4. The red point follows a perfectly straight line, while the purple points trace arcs that are slightly less than semicircles. Discovered in 1864, the Peaucellier linkage was the first known hinged construction to trace a perfectly straight line. James Watt invented a similar mechanism for his steam engine, but the line was only approximately straight. Unfortunately, the Peaucellier linkage has too many parts to be of practical value in real machines.
Programs like Working Model simulate the full physics of a machine before it is built. Check it out at www.workingmodel.com.
Glad to Halve You
For the full story of the GLaD construction, see www.gfacademy.org/GlaD.
1. Here's why Q divides segment AB at the 1/3 point. The full red triangle PDM is proportional to the full blue triangle PBA because they have the same angles. So their side lengths are also proportional. M is the midpoint of segment CD, so BA is twice as long as DM. Because all the sides are in the same proportion, PB is twice as long as PD. But the dark red and dark blue triangles PDO and PBQ are also similar, so BQ is twice as long as DO. AQ is the same length as DO, so Q divides AB at the 1/3 point.
2. The same reasoning works for every step of the GLaD construction. The red and blue triangles are similar, so their edge lengths are proportional. We know that the top horizontal edge DM is 1/2 of our original segment AB, and the bottom horizontal edge QA is 1/3 of AB. So the ratio of DM to AB is (1/2) to (1/3), or 3/2, which means DN to RQ is also 3/2. DN is the same length as AR, so we get AR/RQ=3/2. Multiply both sides by RQ and you get AR=3/2(RQ). So AQ=5/2(RQ). Therefore, R divides AQ at the 3/5 point, which means it divides AB at the 1/5 point.
3. Points S, T, etc., divide segment AB at the 1/7, 1/9, 1/11, etc., points. In general, if the top and bottom points divide the top and bottom segments at the 1/x and 1/y points, then the point constructed by dropping a vertical line from the intersection of the two diagonals divides the bottom segment at the 1/(x+y) point. This is known as the harmonic mean of 1/x and 1/y.
4. The points Q, R, S, T, etc., divide AB at the even-numbered fractions 1/2, 1/4, 1/6, 1/8, etc.
Answers to Straightedge and Compass Bonus Question
Here are the answers to the additional constructions posed in the answers to "Straightedge and Compass," on page 86 of the January 2003 issue. The red lines and circles show the new constructions at each stage.
1. Challenge: Construct a pair of perpendicular lines using only one circular arc and as many straight lines as you wish.
Answer: The secret is that any angle inscribed in a semicircle is a right angle. (2+1+1=4 steps.)
2. Challenge: Draw a circle. What is the minimum number of circles, all the same size as the original circle, needed to construct a square that circumscribes the circle (that is, the circle touches the square at just the four midpoints of the sides of the square)? Again, you may use as many straight lines as you wish.
Answer: Just one. From two circles you can construct a triangular grid that extends as far as you like, which lets you construct perpendiculars to the original line. These perpendiculars form the vertical sides of the circumscribing square. Connect points at the tops and bottoms of the circles to construct the horizontal sides of the circumscribing square. (2+1+4+4+5=16 steps).
Several readers, including David B. Harper and Sonjia Stanford found a shorter 16-step solution to the construction of the octagon. (1+4+3+8).
This approach also leads to a different construction of the square that is more elegant but uses the same number of steps (1+4+4=9 steps).
Want to go back to the puzzle?
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