## Tuesday, August 15, 2023

### (x, why?) Mini: Box Plot

(Click on the comic if you can't see the full image.)
(C)Copyright 2023, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

That'd be a gallon of data.

This could be a valid chart, if someone had a use for it. You'd need a Six-Number Summary based on quintiles, instead of quartiles. If someone wanted just the 10% above and below the median, that would be the middle box. If for some reason, you wanted the central 60% instead of 50%, that would be the entire box containing the three smaller boxes.

And if anyone is looking for outliers, have a dram at the bar.

As for as the whiskey bottles go, the labels apper to be three flowers or butterflies or possibly an ancient family crest. That's my story and I'm sticking to it.

Cheers!

### I also write Fiction!

You can now order Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below.
Order the softcover or ebook at Amazon.

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Come back often for more funny math and geeky comics.

## Sunday, August 06, 2023

### Geometry Problems of the Day (Geometry Regents, June 2023)

This exam was adminstered in June 2023.

More Regents problems.

### Part I

17. The measure of one of the base angles of an isosceles triangle is 42°. The measure of an exterior angle at the vertex of the triangle is

(1) 42°
(2) 84°
(3) 96°
(4) 138°

If one base angle is 42° then the second base angle is also 42°. According to the Remote Angle Theorem, the measure of the exterior angle is equal to the sum of the two remote angles, which in the case of the vertex angle, the remote angles would be the two base angles. That sum is 84°, so Choice (2) is the correct answer.

If you forgot about the Remote Angle Teheorem, you could have used the sum of 84° to find the size of the vertex angle, which is 180 - 84 = 96°. Then the exterior angle would be supplementary to the vertex angle, so 180 - 96 = 84°. You not only got the same answer, but you just demonstrated the Remote Angle Theorem.

Two supplementary angles, A and B, add up to 180 degrees. The three angles of a triangle. B, C, and D, add up to 180 degrees. This means that A + B = B + C + D, so if you remove B (which in this problem is the vertex angle), you are left with A = C + D. This is Remote Angle Theorem.

18. In the diagram below, AFKB || CHLM, FH ≅ LH, FL ≅ KL, and LF bisects ∠HFK.

Which statement is always true?

(1) 2(m∠HLF) = m∠CHE
(2) 2(m∠FLK) = m∠LKB
(3) m∠AFD = m∠BKL
(4) m∠DFK = m∠KLF

Look at the choices one at a time.

Angle HLF is one of two base angles and is congruent to angle HFL. Angle FHL (and therefore angle CHE) will equal 180 - 2(m∠HLF), not 2(m∠HLF). Eliminate Choice (1).

Angle FLK is the vertex angle of isosceles triangle FLK. LKB is an exterior angle. That means that, according to the Remote Angle Theorem that the sum of KFL and FLK equals the measure of LKB. For Choice (2) to be true, angle KFL must be congruent to angle FLK and there is no reason to suggest that they always will be. Eliminate Choice (2).

Angles AFD and BKL have no relationship between them DE and KL are not given to be parallel. Eliminate Choice (3).

Angles DFK and KLF look like they have no relationship, but they do. You may want to mark the illustration with this information. Angles HFL and LFK are congruent becuase LF bisects HFK. Angles LFP and LKF are congruent because triangle FKL is isosceles (with FL ≅ LK). This means that 2(LFK) + DFK = 180 degrees because that's the measure of a striaght line. And 2(LFK) + KLF = 180 because that's the measure of a traingle. Therefore, DFK must be congurnet to KLF. A little convoluted for a multiple-choice queston, but the correct answer is Choice (4).

19. The line whose equation is 6x + 3y = 3 is dilated by a scale factor of 2 centered at the point (0,0). An equation of its image is

(1) y = -2x + 1
(2) y = -2x + 2
(3) y = -4x + 1
(4) y = -4x + 2

Answer: (2) y = -2x + 2

When you dilate a line, the image will either be the same line or a line parallel to the original line. The slope will not change. The y-intercept will only change if the center of dilation is a point that is not on the pre-image.

First, find the slope of the original line. Since it is in Standard Form, Ax + By = C, the slope is given by the formula m = -A/B = -6/3 = -2.

If you didn't remember how to find the slope that way, then you have to rewrite it in slope-intercept form.

6x + 3y = 3
3y = -6x + 3
y = -2x + 1

Elimnate Choice (3) and (4).

So (0,1) is a point on the pre-image. If that point is dilated by a factor of 2, centered on the origin, the image will be at (0,2), which will be the y-interecept of the image. Then the correct answer is Choice (2).

20. Which figure will not carry onto itself after a 120-degree rotation about its center?

(1) equilateral triangle
(2) regular hexagon
(3) regular octagon
(4) regular nonagon

Consider that 120 degrees is 1/3 of a full rotation.

If an equilateral triangle rotates 1/3 of the way, it will carry onto itself because it has 3 sides.

If a regular hexagon rotates 1/3 of the way, two sides will have rotated around and it will still look the same.

If a regular nonagon (which has nine sides) rotates 1/3 of the way, three sides will have rotated and it will still look the same.

An octagon, however, has eight sides, so if it rotates 1/3 of the way, it will not look the same because 1/3 * 8 is not a whole number.

The correct choice is Choice (3).

21.Triangle ADF is drawn and BC || DF.

Which statement must be true?

(1) AB/BC = BD/DF
(2) BC = 1/2 DF
(4) ∠ACB ≅ ∠AFD

Because BC || DF, the small triangle is similar to the larger triangle and the corresponding sides are proportional.

Choice (1) and (3) show incorrect proportions because in each case there is are non-corresponding sides. In Choice (1), if BD said AD, or in Choice (3), if CF said AF, the proportions would have been correct. Eliminate Choice (1) and (3).

Choice (2) looks like a good answer EXCEPT it is only true if B and C are the midpoints of AD and AF, respectively, which would make BC a midsegment. We are not given this piece of information and we cannot assume it. Eliminate Choice (2).

Since the two triangles have the same shape, their corresponding angles are congruent. The correct answer is Choice (4).

22. In △ABC, M is the midpoint of AB and N is the midpoint of AC. If MN = x + 13 and BC = 5x - 1, what is the length of MN?
Which statement is always true?

(1) 3.5
(2) 9
(3) 16.5
(4) 22

The thing to remember here is that BC is NOT congruent to MN. MN is a midsegment, so it is HALF the length of BC. That is, BC is TWICE the length of MN.

Write that as an equation. Then substitute and solve.

BC = 2(MN)
5x - 1 = 2(x + 13)
5x - 1 = 2x + 26
3x = 27
x = 9

If x = 9, which is NOT the final answer, then MN = x + 13 = 9 + 13 = 22. That is Choice (4), which is the correct answer.

If you forgot to multiply by 2, then you would have gotten 3.5 for x and 16.5 for MN. Incorrect.

23. In the diagram below of isosceles trapezoid STAR, diagonals AS and RT intersect at O and ST || RA, with nonparallel sides SR and TA.
Which pair of triangles are not always similar?

(1) △STO and △ARO
(2) △SOR and △TOA
(3) △SRA and △ATS
(4) △SRT and △TAS

The legs of the isosceles trapezoid are congruent and the diagonals are congruent. The bases are not congruent. The bases are parallel, which creates alternate interior angles that are congruent. And the diagonals form vertical angles that are congruent.

In Choice (1), the top and bottom triangles are always similar because of the alternate interior angles and the vertical angles.

In Choice (2), the left and right can be show to be not only similar but congruent using AAS.

In Choice (3), the left and the bottom together is not similar to the top and right together. They cannot be. Because SR ≅ TA, then ST must be congruent to AR for the corresponding sides to be proportional. However, we know that ST is NOT congruent to AR, because then we would have a parallelogram instead of a trapezoid. Choice (3) is the correct answer.

In Choice (4), these two triangles can be shown to be congruent (and therefore similar) through SAS.

24. The endpoints of AB are A(0,4) and B(–4,6). Which equation of a line represents the perpendicular bisector of AB?

(1) y = -1/2 x + 4
(2) y = -2x + 1
(3) y = 2x + 8
(4) y = 2x + 9

Answer: (4) y = 2x + 9

The perpendicular bisector has a slope that is the inverse reciprocal to the line connecting (0,4) and (-4,6). It also must contain the midpoint of AB, which is (-2,5), which is half the distance from A to B.

(6 - 4) / (-4 - 0) = 2/-4 = -1/2
Slope of AB = -1/2
Perpendicular slope = 2

The slope of the bisector is 2, so y = mx + b, and 5 = (2)(-2) + b.

5 = -4 + b, so b = 9.

End of Part I. How did you do?

More Regents problems.

## Friday, August 04, 2023

### Geometry Problems of the Day (Geometry Regents, June 2023)

This exam was adminstered in June 2023.

More Regents problems.

### Part I

9. An equation of circle M is x2 + y2 + 6x - 2y + 1 = 0. What are the coordinates of the center and the length of the radius of circle M?

(1) center (3, -1) and radius 9
(2) center (3, -1) and radius 3
(3) center (-3, 1) and radius 9
(4) center (-3, 1) and radius 3

You have to complete the squares for the x2 term and the y2 term. Thankfully, the choices give you hints.

Rewrite the equation grouping the x terms and the y terms and then move the constant to the right side of the equation.

x2 + 6x + y2 - 2y + 1 - 1 = 0 - 1

x2 + 6x + y2 - 2y = -1

To complete the squares, take have of 6, which is 3, and then square it, which is 9. Take half of -2, which is -1, and square it, which is 1. Add 9 + 1 to both sides of the equation.

x2 + 6x + 9 + y2 - 2y + 1 = -1 + 10

(x + 3)2 + (y - 1)2 = 9

The equation of a circle is given by the formula (x - h)2 + (y - k)2 = r2, where (h,k) is the center of the circle and r is the radius.

So the center of the circle is (-3,1), because the signs flip to make 0, and the radius is 3, because r2 is 9. Choice (4) is the correct answer.

10. Parallelogram BETH, with diagonals BT and HE, is drawn below.

Which additional statement is sufficient to prove that BETH is a rectangle?

(1) BT ⊥ HE
(2) BE || HT
(3) BT ≅ HE
(4) BE ≅ ET

If a parallelogram has diagonals that are congruent then it is a rectangle. Choice (3) is the correct answer.

Perpendicular diagonals make a parallelogram a rhombus, not a rectangle. Eliminate Choice (1).

Parallel sides are part of the definition of a parallelogram. Eliminate Choice (2).

Consecutive sides being congruent makes a parallelogram into a rhombus, not a rectangle. Eliminate Choice (4).

11. A gardener wants to buy enough mulch to cover a rectangular garden that is 3 feet by 10 feet. One bag contains 2 cubic feet of mulch and costs \$3.66. How much will the minimum number of bags cost to cover the garden with mulch 3 inches deep?

(1) \$3.66
(2) \$10.98
(3) \$14.64
(4) \$29.28

Volume of a cone is V = L * W * H, but make sure that you are using the correct units, which must be the same.

V = 3 * 10 * .25 = 7.5. Since one bag covers 2 cubic feet, he needs 4 bags (because three isn't enough and they won't sell half a bag.)

The cost is then \$3.66 * 4 = \$14.64, which is Choice (3).

12. In the diagram below, &9651;DOG ~ &9651;CAT, where ∠G and ∠T are right angles.

Which expression is always equivalent to sin D?

(1) cos A
(2) sin A
(3) tan A
(4) cos C

Sin D = sin C, but that isn't one of the choices.

However, the sine of one complementary angle is equal to the cosine of the other complementary angle in a right triangle.

That means that sin D = cos O = cos A. Cos A is Choice (1), which is the correct answer.

13. On the set of axes below, nDEF is the image of nABC after a dilation of scale factor 1/3.

The center of dilation is at

(1) (0,0)
(2) (2,-3)
(3) (0,-2)
(4) (-4,0)

Take a straightedge and draw line AD, extending it past D. Do the same for BE and CF.

The point of concurrence is (2,-3), which is the center of dilation. Choice (2) is the correct answer.

You can see that from (2,-3) to point F is a translation of -2,+1, and from (2,-3) to C is -6,+3, which is 3 times farther. Or F is only 1/3 of the distance to C.

You can check the distance to D and A and you will find that D is 1/3 of the way to A. Likewise, E is 1/3 of the way to B.

14. In the diagram below of isosceles triangle AHE with the vertex angle at H, CB ⊥ AE and FD ⊥ AE.
Which statement is always true?

(1) AH/AC = EH/EF
(2) AC/EF = AB/ED
(3) AB/ED = CB/FE

Because AHE is isosceles, angles A is congruent to angle E. Angle ABC and EDF are right angles, so they are congruent to each other. Therefore, by AA, triangles ABC and EDF are similar (not congruent), and that makes the corresponding sides proportional.

Choice (1) uses point H, comparing part of the side to the length of the entire side of triangle AHE. This would imply that AC ≅ EF, which isn't necessarily true. (It doesn't look it.) Eliminate Choice (1).

Choice (2) compares the ratio of the two hypotenuses to the ratio of the two bases. Choice (2) is the correct answer.

Choice (3) is incorrect because the first ratio compares the bases of the two triangles but the second ratio compares the vertical leg of one triangle to the hypotenuse of the other.

Choice (4) uses line segment BD, which is not part of either of the two similar triangles. Eliminate Choice (4).

15. Rectangle ABCD has two vertices at coordinates A(-1,-3) and B(6,5). The slope of BC is

(1) -7/8
(2) 7/8
(3) -8/7
(4) 8/7

Sides AB and BC must be perpendicular to each other, so the slopes of the two sides must be inverse reciprocals. That is, flip the fraction over, and switch the sign.

The slope of AB is (-3 - 5) / (-1 - 6) = -8/-7 = 8/7. That means that the negative reciprocal is -7/8, which is Choice (1).

16. In right triangle ABC, m∠A = 90o, m∠B = 18o, and AC = 8. To the nearest tenth, the length of BC is

(1) 2.5
(2) 8.4
(3) 24.6
(4) 25.9

AC is Opposite of angle B, and BC is the hypotenuse of the triangle. Opposite and hypotenuse means that we need to use the sine ratio.

sin 18 = 8 / x
so x = 8 / (sin 18) = 25.88...

This rounds to 25.9, so Choice (4) is the correct answer.

More to come. Comments and questions welcome.

More Regents problems.

## Thursday, August 03, 2023

### Geometry Problems of the Day (Geometry Regents, June 2023)

This exam was adminstered in June 2023.

More Regents problems.

### Part I

1. A square pyramid is intersected by a plane passing through the vertex and perpendicular to the base.
Which two-dimensional shape describes this cross section?

(1) square
(2) triangle
(3) pentagon
(4) rectangle

If this were a cake and you were to slice it straight down the middle so that you cut through the highest point (the vertex) and sliced straight down, when you separated the two pieces, you would have a triangle slice, which is Choice (2).

If you missed the vertex, any other vertical cut would result in a trapezoid (which is not one of the choices).

If the plane was parallel to the base, the cross section would have been a square. If the was not parallel but passed through all four sides, it would result is a rectangle. Note that since a square is a rectangle, you know that square MUST BE incorrect because rectangle would also be correct.

There is no way for a place to intersect the pyramid and result in a pentagon.

2. Trapezoid ABCD is drawn such that AB||DC. Trapezoid A'B'C'D' is the image of trapezoid ABCD after a rotation of 110o counterclockwise about point P.

Which statement is always true?

(1) ∠A ≅ ∠D'
(2) AC ≅ B'D'
(3) A'B' || D'C'
(4) B'A' ≅ C'D'

The iamge after the rotation will have the same shape. That means that parallel lines are still parallel. The image is still a trapezoid. Therefore, A'B' || D'C'. Choice (3) is the correct choice.

∠A doesn't correspond to ∠D', so there is no reason for the two angles to be congruent. Eliminate Choice (1).

Since it is not stated that this is an isosceles trapezoid, the diagonals (not drawn) are not necessarily congruent. Eliminate Choice (2).

In a trapezoid, the bases do not have to be congruent. In fact, in standard HS Geometry, the bases cannot be congruent, because that would make it a parallelogram instead of a trapezoid. (When you get to advanced mathematics, parallelograms will become special cases of trapezoids the way that equilateral triangles are also isosceles.)

3. What is the volume of a right circular cone that has a height of 7.2 centimeters and a radius of 2.5 centimeters, to the nearest tenth of a cubic centimeter?

(1) 37.7
(2) 47.1
(3) 113.1
(4) 141.4

The Volume of a cone is V = 1/3 π r2 h.

V = 1/3 π (2.5)2 (7.2) = 47.1239..., which is approximately 47.1. Choice (2) is the correct answer.

I think the other two choices come from using (2.5)(2) instead of (2.5)2, with and without the 1/3 in front.

4. In the diagram below of right triangle SUN, where ∠N is a right angle, SU = 13.6 and SN = 12.3.
What is m∠S, to the nearest degree?

(1) 25o
(2) 42o
(3) 48o
(4) 65o

You are given the hypotenuse and the side adjacent to the angle, so you need to use the cosine ratio.

Cos x = 12.3/13.6, so x = cos-1 (12.3/13.6) = 25.26 degrees, so Choice (1) is the correct answer.

The other choices are for sine and tangent (using the hypotenuse instead of the opposite side) and tangent flipping the order of the numbers.

5. In the diagram below of circle O, diameter AOB and chord CB are drawn, and m/B = 28°

What is mBC (arc BC)?

(1) 56°
(2) 124°
(3) 152°
(4) 166°

Angle B is an inscribed angle, so it intercepts an arc that is equal to twice the size of its angle. AOB is a diameter which marks off two semicircles, each having arcs of 180 degrees. Arc BC is therefore 180 - 2(28), which is 124 degrees, which is Choice (2).

Choice (1) is the size of arc AC, not BC. Choice (3) is 180 - 28, instead of 2(28). Choice (4) is 180 - 1/2(28).

6. In the diagram below of parallelogram ABCD, diagonal BED and EF are drawn, EF ⊥ DFC, m∠DAB = 111°, and m∠DBC = 39°.

What is m∠DEF?

(1) 11°
(2) 51°
(3) 60°
(4) 120°

Work your way around the parallelogram. Angle C = 111°. Angle EFC = 90°. So m∠BEF = 360 - (39 + 111 + 90) = 120°.

This makes m∠DEF = 180 - 120 = 60°.

7. In the diagram below of nACT, ES is drawn parallel to AT such that E is on CA and S is on CT.

Which statement is always true?

(1) CE/CA = CS/ST
(2) CE/ED = EA/AT
(3) CE/EA = CS/ST
(4) CE/ST = EA/CS

Triangle CES is similar to triangle CAT, so the corresponding sides are proportional. Look for the correct proportion from the choices.

Choice (1) is incorrect because it uses ST instead of CT.

Choice (2) is incorrect because it uses EA instead of CA.

Choice (3) compares segments split by parallel lines. Choice (3) is the correct answer.

Choice (4) is inconsisent with the triangles, comparing a part of one triangle with a part of the a different triangle.

8. On the set of axes below, congruent triangles ABC and DEF are drawn.

Which sequence of transformations maps △ABC onto △DEF?

(1) A counterclockwise rotation of 90 degrees about the origin, followed by a translation 8 units to the right.
(2)A counterclockwise rotation of 90 degrees about the origin, followed by a reflection over the y-axis.
(3) A counterclockwise rotation of 90 degrees about the origin, followed by a translation 4 units down.
(4) A clockwise rotation of 90 degrees about the origin, followed by a reflection over the x-axis.

Answer: (1) A counterclockwise rotation of 90 degrees about the origin, followed by a translation 8 units to the right.

Look at each of the choices, one at a time.

Choice (1) moves point A to (-6,-2) and B to (-2,-2). If you translate the triangle 8 units to the right, it will line up with DE. Likewise, you can follow point C and it will move onto point F. Choice (1) is the correct answer.

Choice (2) is incorrect because a reflection over the y-axis would reverse the positions of D and E. Furthermore, the image of point C would be must closer to the y-axis (at (2,-5)).

Choice (3) has a translation down, instead of to the right.

Choice (4) has a clockwise rotation, which will not line up with DEF after a reflection over the x-axis.

More to come. Comments and questions welcome.

More Regents problems.