12 multiple-choice questions, progressively harder.
Every vertex of a simple polygon moves aaa units horizontally, with its vertical coordinate unchanged. Why is the signed area unchanged?
Solution
Correct answer: C
Let did_idi be the original difference and di′d'_idi′ the new one. Replacing each horizontal coordinate xxx by x+ax+ax+a gives
Each vertical coordinate occurs once as a final coordinate and once as an initial coordinate. The added terms cancel around the closed walk.
A simple polygon in boundary order has signed area −9/4-9/4−9/4. Keep the vertex order, multiply each horizontal coordinate by −2-2−2, and multiply each vertical coordinate by 3. What is the new signed area?
Correct answer: A
Each diagonal product uses one horizontal and one vertical coordinate. Each product therefore gains a factor of (−2)(3)(-2)(3)(−2)(3), or −6-6−6. The signed area gains that same factor.
Use the four points A(0,0)A(0,0)A(0,0), B(8,0)B(8,0)B(8,0), C(1,6)C(1,6)C(1,6), and D(2,2)D(2,2)D(2,2). Both orders A,B,C,DA,B,C,DA,B,C,D and A,B,D,CA,B,D,CA,B,D,C make simple polygons. By how much do their ordinary areas differ?
Correct answer: D
For A,B,C,DA,B,C,DA,B,C,D, the differences are 000, 484848, −10-10−10, and 000. The ordinary area is 19 square units.
For A,B,D,CA,B,D,CA,B,D,C, the differences are 000, 161616, 101010, and 000. The ordinary area is 13 square units.
The two walks join the same points with different edges.
A triangle has boundary vertices P(1,1)P(1,1)P(1,1), Q(5,1)Q(5,1)Q(5,1), and R(3,t)R(3,t)R(3,t). Its ordinary area is 9. Which pair gives all possible values of ttt?
Correct answer: B
The differences are −4-4−4, 5t−35t-35t−3, and 3−t3-t3−t.
If the signed area is 999, then t=11/2t=11/2t=11/2.
If the signed area is −9-9−9, then t=−7/2t=-7/2t=−7/2. Both give a nondegenerate triangle.
The closed walk A,C,B,D,AA,C,B,D,AA,C,B,D,A uses A(0,0)A(0,0)A(0,0), B(9,0)B(9,0)B(9,0), C(9,6)C(9,6)C(9,6), and D(0,4)D(0,4)D(0,4). The slanted segments cross at (18/5,12/5)(18/5,12/5)(18/5,12/5). What are the absolute shoelace result and the total ordinary area of the two triangular regions, respectively?
The two nonzero product differences are −54-54−54 and 363636.
The absolute result is 9.
The left triangle has vertical base 4 and horizontal height 18/518/518/5. Its area is 36/536/536/5 square units.
The right triangle has vertical base 6 and horizontal height 27/527/527/5. Its area is 81/581/581/5 square units.
A simple triangle follows P(−2,1)P(-2,1)P(−2,1), Q(5,−1)Q(5,-1)Q(5,−1), R(6,4)R(6,4)R(6,4). Move QQQ to (5,2)(5,2)(5,2) while keeping the order. By how much does its signed area change?
Only the edges entering and leaving QQQ change. The first difference changes from −3-3−3 to −9-9−9, a change of −6-6−6. The second changes from 262626 to 888, a change of −18-18−18.
The signed area decreases by 12.
An internal diagonal splits a simple polygon into two pieces. The original boundary and both piece boundaries are followed clockwise. The original signed area is −41-41−41, and one piece has signed area −14-14−14. What is the signed area of the other piece?
The two clockwise piece boundaries use their shared diagonal in opposite directions. Its two product differences cancel, leaving the product differences for the original boundary.
Why does the vertical-strip proof use every distinct vertex horizontal coordinate as a cut?
Let aaa and bbb be two consecutive cut coordinates. A point lies in the open strip only if its horizontal coordinate satisfies
No vertex lies there. Each boundary segment that enters the open strip crosses its full width. The segments cannot meet there, so the upper and lower pairings stay the same across the strip.
Which value cannot be the ordinary area of a simple lattice polygon?
Both coordinates of every lattice vertex are integers, so the doubled signed area is an integer. Its magnitude 2K2K2K is an integer as well.
This is not an integer, so 19/419/419/4 cannot be the ordinary area of a simple lattice polygon.
A simple polygon has signed area −12-12−12. Reflect its points across the horizontal axis, then reverse the vertex order. What is the final signed area?
The reflection negates every vertical coordinate, so it negates both products for each edge. Reversing the order then swaps the products of every edge and negates the sum again.
The final signed area is −12-12−12.
A point QQQ is inserted strictly between PPP and RRR along the nonvertical boundary edge PRPRPR. Why do the two signed trapezoid contributions after insertion add to the old contribution?
Dropping a vertical line from QQQ to the reference line divides the old trapezoid into two. The boundary keeps moving in its original horizontal direction, so both parts carry its sign.
An edge runs from P(−6,3)P(-6,3)P(−6,3) to Q(4,−2)Q(4,-2)Q(4,−2). Use reference line y=−5y=-5y=−5. How do the original-coordinate shoelace difference and the signed trapezoid contribution compare?
Both original-coordinate products are 12.
The edge travels rightward, and its two endpoint heights above the reference line are positive. The signed trapezoid contribution is negative.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.