12 multiple-choice questions, progressively harder.
A convex parallelogram is placed A(0,0), B(6,0), D(t,4), C(t+6,4). For which real t are its diagonals congruent?
Solution
Correct answer: D
Equate their squared lengths.
At this value the figure is a 6 by 4 rectangle. The height is nonzero for every t, so no degenerate candidate was introduced.
A convex quadrilateral has A(-4,-2), B(2,1), C(6,6), D(0,3) in boundary order. Both diagonal midpoints are (1,2). Which additional computed fact rules out a rectangle?
Correct answer: B
Adjacent slopes are 1/2 and 5/4.
Both are defined and nonzero. Their product is not -1, so the adjacent angle is not right. Opposite-side equalities and diagonal bisection are already parallelogram properties.
A triangle has A(-5,2), B(7,-4), and C(t,11). Its centroid lies on the vertical line x=3. What is t?
Correct answer: C
Only the first-coordinate average is constrained.
B and C then have the same first coordinate but different heights, and A is off their line, so the triangle remains nondegenerate.
A triangle has A(0,0), B(18,0), C(-6,12). M and N are the midpoints of AC and BC. What is the area of triangle CMN?
Correct answer: A
The midpoints are M(-3,6) and N(6,6), giving base 9. C is 6 units above their horizontal line.
A simple pentagon has boundary vertices (-2,-1), (6,-1), (6,3), (1,7), and (-2,3). What is its exact perimeter?
The horizontal and vertical edges have lengths 8, 4, and 4. The slanted edges have gaps (5,4) and (3,4).
A simple quadrilateral has boundary vertices (0,0), (8,0), (6,5), and (0,5). What is its area?
Use an 8 by 5 containing rectangle and remove the upper-right triangle of horizontal leg 2 and vertical leg 5.
For positive a,b, a quadrilateral has A(-a,0), B(0,b), C(a,0), D(0,-b) in boundary order. Which condition exactly characterizes when it is a square?
All four side squares are the sum of a squared and b squared, so this is a rhombus. Its diagonals have lengths 2a and 2b.
For a rectangle and hence a square, these must agree.
A convex quadrilateral has congruent diagonals but their midpoints differ. What follows?
Every parallelogram has a common midpoint for its two diagonals. The failed necessary condition excludes that family.
Rectangles are parallelograms, so it excludes rectangles as well. Congruent diagonal lengths cannot repair the missing bisection.
A triangle has A(0,0), B(12,0), C(3,9). Its centroid is G. What is the area of triangle ABG?
The centroid is G(5,3), from the separate three-coordinate averages.
AB has length 12 and G has perpendicular height 3 above its line.
A simple polygon has boundary vertices (0,0), (10,0), (10,6), (u,6), (u,2), (0,2), where 0<u<10. Its area is 44 square units. What is u?
It contains a 10 by 2 lower rectangle and a right-hand upper rectangle of width 10-u and height 4.
The answer lies strictly between 0 and 10, preserving the stated simple boundary.
A coordinate proof shows that both pairs of opposite sides of a convex quadrilateral have equal squared lengths. Which conclusion is justified without further calculations?
Each side length is positive, so equality of squared lengths gives equality of the corresponding lengths.
Both opposite-side pairs are therefore congruent. The earlier opposite-side test proves a parallelogram, but supplies no extra right angle or adjacent-side equality.
For positive side lengths p and q this gives
Let a,c be positive and b real. A triangle has A(0,0), B(2a,0), C(2b,2c), and centroid G. When is the median from C perpendicular to AB?
The midpoint of AB is (a,0). Since AB is horizontal, its perpendicular median must be vertical. Thus C must have first coordinate a.
The positive height keeps C distinct from that midpoint.
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