## Tuesday, August 31, 2021

### Sub Pool

(Click on the comic if you can't see the full image.)
(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

And if you didn't get the reference, really? And you're reading this?

I'm pretty sure most people got the reference to the math sub named Quinn Jonas as mentioned in TV Tropes.

As noted recently, Ken is otherwise occupied at the moment.

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. Come back often for more funny math and geeky comics. ### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

Part II: Each correct answer will receive 2 credits. Partial credit is possible.

32. A right prism has a square base with an area of 12 square meters. The volume of the prism is 84 cubic meters. Determine and state the height of the prism, in meters.

The Volume of a prism is the Area of the Base times the height. You are given the Volume and the Area of the Base. Plug these in and solve.

84 = 12 h
7 = h

33. State whether the lines represented by the equations y = 1/2x - 1 and y + 4 = -1/2(x - 2) are parallel, perpendicular, or neither. Explain your answer.

The lines are not parallel because they do not have the same slope. One line has a slope of 1/2 while the other has a slope of -1/2.

The liner are not perpendicular either. The products of their slopes is (1/2)(-1/2) = -1/4, not -1. The slopes are NOT inverse reciprocals.

The two lines are neither parallel nor perpendicular.

You need to include the word "neither" (not worth any points, but you will lose points if you don't state it) AND a reason why they are not parallel and why they are not perpendicular (each reason worth 1 credit).

If you just state "neither", you will earn zero credit.

34. A tree, T, is 6 meters from a row of corn, c, as represented in the diagram below. A farmer wants to place a scarecrow 2 meters from the row of corn and also 5 meters from the tree.

Sketch both loci.

Indicate, with an X, all possible locations for the scarecrow.

Five meters away from the tree means a circle with radius of 5. Since the row of corn is 6 meters away, the circle will come close to the line but not touch it. This is a sketch, so it doesn't have to be perfect.

Two meters away from the line of corn means two parallel lines on either side of the line. Since the circle comes within one meter of the line of corn, one of the parallel lines will intersect the circle in two places. Put an X on both of those places.

More to come. Comments and questions welcome.

More Regents problems.

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. ## Monday, August 30, 2021

### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

Part II: Each correct answer will receive 2 credits. Partial credit is possible.

29. The diameter of a sphere is 5 inches. Determine and state the surface area of the sphere, to the nearest hundredth of a square inch.

The formula for the Surface Area of a Sphere is in the back of the booklet: SA = 4πr2. The radius is HALF the diameter. To the nearest hundredth of a square inch means TWO decimal places.

SA = 4πr2 = 4 (3.141592...)(2.5)2 = 78.539...

The Surface Area is 78.54 square inches.

Note that if you didn't use enough accuracy for pi, such as only using 3.14, you will make a mistake.

30. Using a compass and straightedge, construct the perpendicular bisector of AB.
[Leave all construction marks.]

To construct a perpendicular bisector, start at point A. Open you compass more than halfway to B. Make an arc centered on A. Move the compass to B without changing the size. Make an arc that overlaps the first arc. Use the straightedge to draw the perpendicular bisector.

31. The endpoints of AB are A(3,-4) and B(7,2). Determine and state the length of AB in simplest radical form.

Use the Distance Formula, or Pythagorean Theorem. (They're related.) You can even make a sketch to find the distances between the x-coordinates and the y-coordinates.

d = SQRT ( (7 - 3)2 + (-4 - 2)2 )
d = SQRT ( 42 + (-6)2 )
d = SQRT ( 16 + 36 )
d = SQRT ( 52 )

You are asked for "Simplest Radical Form" so DO NOT give a decimal, not one rounded to two places, not one with 8 significant digits followed by three dots.

The prime factors of 52 = 2 * 2 * 13 = 22 * 13

SQRT (22 * 13) = 2 * SQRT (13)

More to come. Comments and questions welcome.

More Regents problems.

## Sunday, August 29, 2021

### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

26. A circle with the equation (x + 6)2 + (y - 7)2 = 64 does not include points in Quadrant

1) I
2) II
3) III
4) IV

Once again, the formula for the equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) is the center of the circle, and r is the radius. Note that there are MINUS SIGNS in the formula.

The center of the circle is (-6, 7) which is in Quadrant II, so eliminate that. The radius is equal to the square root of 64, which is 8. If you go 8 units to the right of (-6, 7), you will be at (2, 7), which is in Quadrant I. If you go 8 units below (-6, 7), you will be at (-6, -1), which is in Quadrant III.

You have eliminated Choices (1), (2) and (3), leaving only IV. But you can check this.

If the center stretches from quadrant II to quadrant IV, it has to go past the origin, and (0, 0) would be inside the circle at a distance less than the radius from the center. That would mean:

(0 + 6)2 + (0 - 7)2 < 64
and therefore 36 + 49 < 64

This is obviously not a true statement, so the circle does not extend in Quadrant IV.

27. Trapezoid QRST is graphed on the set of axes below.

Under which transformation will there be no invariant points?

1) ry = 0
2) rx = 0
3) r(0,0)
4) ry = x

Invariant points are points on a line or shape which do not move when a specific transformation is applied. Under these transformations, if a point remains in the same place, it is invariant.

In Choice (1), y = 0 is the x-axis. Point S is on the x-axis and will reflect to itself.

In Choice (2), x = 0 is the y-axis. Point R is on the y-axis and will reflect to itself.

In Choice (3), (0,0) means a reflection about the origin. The only put that would remain in place is a point on the origin, but the trapezoid does not pass through the origin. This is the correction answer.

In Choice (4), y = x is a diagonal line through quadrants I and III. It goes through point Q and through an unlabeled point on line RS. Both of those points will reflect onto themselves.

23. How many common tangent lines can be drawn to the circles shown below??

1) 1
2) 2
3) 3
4) 4

A common tangent line means that the line will be tangent to both circles.

There will be one on top, one on the bottom, and two in-between making an "X". Look at the illustration below:

The radius is the distance from (1, 2) to (-1, 2), which is 2 units. So r = 2, and r2 = 4. Eliminate Choices (3) and (4).

Since there are minus signs in the formula, we want (x + 1) and (y - 2) in the equation. That means Choice (2).

More to come. Comments and questions welcome.

More Regents problems.

## Saturday, August 28, 2021

### The Biggest Bosses

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(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Life isn't fair.

Think About It: Where in the diagram would you find the subset "Me"?

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. Come back often for more funny math and geeky comics. ### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

21. The coordinates of point P are (7,1). What are the coordinates of the image of P after R90° about the origin?

1) (1, 7)
2) (-7, -1)
3) (1, -7)
4) (-1, 7)

First thing to remember is if you are given a rotation, but not a direction, then the directionis counterclockwise. That is to say, the rotation is in the same order as the Quadrants I, II, III, IV.

Forget the formulas. Picture it in your head, or make a scratch drawing of the problem.

Choice (1) is in Quadrant I like the original point. It is a reflection over the line y = x.

Choice (2) is is Quadrant III. It is a 180 degree rotation, or a reflection across the origin.

Choice (3) is in Quandrant IV. It is a 90 degree clockwise rotation.

22. Lines p and q are intersected by line r, as shown below.

If m∠1 = 7x - 36 and m∠2 = 5x + 12, for which value of x would p || q?

1) 17
2) 24
3) 83
4) 97

The lines will be parallel if and only if angles 1 and 2 are supplementary. That is, they add up to 180 degrees.

So

7x - 36 + 5x + 12 = 180
12x - 24 = 180
12x = 204
x = 17

They want to know x, not the size of either angle, so you are done.

Note that if you wrote the equation 7x - 36 = 5x + 12 (assuming the angles were congruent when the illustration suggests that they may not be), then you would have gotten x = 24, which is Choice (2), and you wouldn't have known you made a mistake.

23. What is the equation of the circle with its center at (-1,2) and that passes through the point (1,2)?

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

Answer: 1) (x + 1)2 + (y - 2)2 = 4

Once again, the formula for the equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) is the center of the circle, and r is the radius. Note that there are MINUS SIGNS in the formula.

The radius is the distance from (1, 2) to (-1, 2), which is 2 units. So r = 2, and r2 = 4. Eliminate Choices (3) and (4).

Since there are minus signs in the formula, we want (x + 1) and (y - 2) in the equation. That means Choice (2).

24. In the diagram below, diameter AB bisects chord CD at point E in circle F.

If AE = 2 and FB = 17, then the length of CE is

1) 7
2) 8
3) 15
4) 16

If AE = 2 and FB = 17, then AF = 17 and EF = 15. This is obviously NOT drawn to scale.

The length of CE = the length of ED because the diameter bisects it. Also, when a diameter bisects a chord, it intersects it perpendicularly, meaning that it creates right angles.

Since you know that the radius FB is 17, you can draw CF and DF and know that they also each have a length of 17.

Now you have two right triangles with a leg of 15 and a hypotenuse of 17. If you don't know your basic Pythagorean Triples (and you should!), you can use the Pythagorean Theorem

152 + x2 = 172
225 + x2 = 289
x2 = 64
x = 8

25.Which quadrilateral does not always have congruent diagonals?

1) isosceles trapezoid
2) rectangle
3) rhombus
4) square

You could sketch these are see that it's obvious that the rhombus does not have to have congruent diagonals. When it does have them, the rhombus is a square.

If the other cases, you can show through SAS that the diagonals create two congruent triangles. And therefore, the diagonals will be congruent because corresponding parts of congruent triangles are congruent.

More to come. Comments and questions welcome.

More Regents problems.

## Friday, August 27, 2021

### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

16. For which measures of the sides of tiangle ABC is angle B the largest angle of the triangle?

1) AB = 2, BC = 6, AC = 7
2) AB = 6, BC = 12, AC = 8
3) AB = 16, BC = 9, AC = 10
4) AB = 18, BC = 14, AC = 5

Answer: 1) AB = 2, BC = 6, AC = 7

The longest side in a triangle is opposite the largest angle, so for B to be the largest angle, it has to be across from the largest side. The side opposite angle B is side AC.

In Choice (1), AC is the longest side, so B is the largest angle.

In Choice (2), BC is the longest side, so A is the largest angle.

In Choice (3), AB is the longest side, so C is the largest angle.

In Choice (4), AB is the longest side, so C is the largest angle.

17. What is the measure of the largest exterior angle that any regular polygon can have?

1) 60°
2) 90°
3) 120°
4) 360°

The sum of the exterior angles of a polygon is ALWAYS 360°. The more sides that a polygon has, the more exterior angles it has. The more exterior angles a regular polygon has, the smaller each individual exterior angle will be.

What you need is the polygon with the fewest sides, and that is a triangle. A regular triangle (an equilateral triangle) has three 60° interior angles. Therefore, the exterior angles will be 120° each.

18. As shown in the diagram below, a landscaper uses a cylindrical lawn roller on a lawn. The roller has a radius of 9 inches and a width of 42 inches

To the nearest square inch, the area the roller covers in one complete rotation is

1) 2,374
2) 2,375
3) 10,682
4) 10,688

They are asking for the lateral area of a cylinder. That is, if you imagine a soup can, they want the area of the rectangular label wrapped around it, but not that top or bottom of the can.

The area of a rectangle is length times width. The width is 42 inches. The length if the circumference of a circle with radius 9, with is 2 * pi * (9).

SA = 42 * 18 * 3.141592 = 2375.04...

This is a slighty EVIL question because if you used 3.14 for pi, the rounding error would give you 2,374 as an answer because you would not have used enough precision. Thankfully, your calculator should have a "PI" key.

19. In the diagram below, AC and BC are tangent to circle O at A and B, respectively, from external point C.

If m∠ACB = 38, what is m∠AOB?

1) 71
2) 104
3) 142
4) 161

AOBC is a quadrilateral and its interior angles sum up to 360 degrees. Angles A and B are right angles, 90 degrees each, because they are tangent to the circle.

So 38 + 90 + 90 + x = 360
218 + x = 360
x = 142

∠AOB is 142 degrees.

I don't see how they arrived at the incorrect choices. I don't see any obvious mistakes that might cause them, except that 71 is half of 142.

15. What is the perimeter of a square whose diagonal is 3 SQRT(2)?

1) 18
2) 12
3) 9
4) 6

(side)2 + (side)2 = (diagonal)2

So 2x2 = (3 * SQRT(2))2
2x2 = 9 * 2
2x2 = 18
x2 = 9
x = 3

Since one side = 3, then the perimeter = 12.

More to come. Comments and questions welcome.

More Regents problems.

## Thursday, August 26, 2021

### (x, why?) Mini: ASA

font size="2">(Click on the comic if you can't see the full image.)
(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

"Give us two angles and a side, and we can prove it's a match!"

There are competitors out there, but don't fall for those SSA claims!

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. Come back often for more funny math and geeky comics. ### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

11. In the diagram below of quadrilateral ABCD, E and F are points on AB and CD, respectively, BE ≅ DF, and AE ≅ CF.

Which conclusion can be proven?

1) ED ≅ FBF
2) AB ≅ CD
3) ∠A ≅ ∠C
4) ∠AED ≅ ∠CFB

If BE = DF and AE = CF that AE + EB = CF + FD, or AB = CD.

Choice (1) would be true in a parallelogram, but there is nothing to suggest that BEDF is a parallelogram. In fact, it doesn't even look like one.

Choice (3) could only be proven if ABCD were a parallelogram or triangles ADE and CBF could be shown to be congruent. Neither of those is true, or can be shown to be true.

Choice (4) would require the triangles to be congruent, which they cannot be shown to be.

12. In the diagram below, four pairs of triangles are shown. Congruent corresponding parts are labeled in each pair

1) A
2) B
3) C
4) D

Each set of triangles shows a postulate or theorem of congruency, except for one.

Choice (1) Figure A shows SSA, two pairs of congruents sides and an angle that is NOT between the two sides. This is NOT a test for congruent triangles. The only exception is the Hypotenuse-Leg Theorem for right triangles (see below), but this is not labeled as a right triangle. Choice (1) is the correct answer.

Choice (2) Figure B shows the ASA Postulate for congruent triangles.

Choice (3) Figure C shows the AAS Theorem for congruent triangles.

Choice (4) Figure D shows the HL Theorem for congruent right triangles.

Each of those lines will cross the circle twice. So there would be four points that fit the criteria, one in each quadrant. You can make a little sketch to see it.

13. In ABC shown below, L is the midpoint of BC, M is the midpoint of AB, and N is the midpoint of AC.

If MN = 8, ML = 5, and NL = 6, the perimeter of trapezoid BMNC is

1) 35
2) 31
3) 28
4) 26

MN is congruent to BL and LC. ML is congruent to AN and NC. NL is congruent to BM and MA. Label all of these with the correct lengths.

Now add up the sides that make up the perimeter of BMNC: 6 + 8 + 5 + 8 + 8 = 35.

14. In the diagram below, and ABC are shown with m∠A = 60 and m∠ABT = 125.

What is m∠ACR?

1) 125
2) 115
3) 65
4) 55

A triangle has 180 degrees. Supplementary angles add up to 180 degrees. That's all you need to know, but the Remote Angle Theorem will save you a step.

∠ABC is supplementary to ∠ABT, which is 125 degrees. So m∠ABC = 180 - 125 = 55.

∠ACR is an exterior angle to the triangle and equal to the sum of the two remote angles, which are 60 and 55, which is 115 degrees.

You could have done these steps in the reverse order and found ∠ACB using the Remote Angle Theorem subtracting 125 - 60 = 65, and then found the supplementary angle, 180 - 65 = 115. Works both ways.

Likewise, if you forgot about the remote angle theorem, you could have just found all of the angles along the way.

15. Which equation represents circle O shown in the graph below?

1) x2 + (y - 2)2 = 10
2) x2 + (y + 2)2 = 10
3) x2 + (y - 2)2 = 25
4) x2 + (y + 2)2 = 25

Answer: 4) x2 + (y + 2)2 = 25

They do like asking questions about the equation of a circle. A lot of them.

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

The radius is 5 (measure the vertical or horizontal distance from point O to the circle). If you square 5, you get 25, not 10. Eliminate Choices (1) and (2).

The center of the circle is (0, -2), so h = 0, and k = -2. So the formula would be

(x - 0)2 + (y - -2)2 = 25

or more simply
x2 + (y + 2)2 = 25

More to come. Comments and questions welcome.

More Regents problems.

## Wednesday, August 25, 2021

### (x, why?) Mini: Three Medians Meet at the Centroid

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(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

No jokes barred. When maybe this one.

As mentioned below, I write sometimes (but not as much as I should). And I sometimes answer writing prompts I find online. While going through a bunch of these, I was surprised at the number of prompts that basically started with A Number of People of A Certain Subgroup or A Similar Number of Unique Individuals of Differing Subgroups Walk Into an Establishment of Some Variety. The worst part was that for the most part, I left these initial sentences as titles. The better ones got real titles. (Side note: "Three Thieves Walk Into a Tavern", which was a flash fiction contest entry, not a writing prompt, was published at "The Second Story Job". It's the, believe it or not, second story, in my flash anthology, In A Flash 2020

This is the 169th mini. Had I thought about it in advance, I might've done something for the number 13, although Firday the 13th was nearly two weeks ago.

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. Come back often for more funny math and geeky comics. ### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

6. A right rectangular prism is shown in the diagram below.

Which pair of edges are not coplanar?

1) BF and CG
2) BF and DH
3) EF and CD
4) EF and BC

Two lines are coplanar if they lie on the same plane. Think of it this way: if each line was a stick, could you attach a piece of paper to both sticks without twisting it?

BF and CG are both up and down lines. They are coplanar. Eliminate Choice (1).

BF and DH are both up and down lines. They are coplanar. Eliminate Choice (2).

EF and CD are both front to back lines. They are coplanar. Eliminate Choice (3).

EF is a front to back line, but BC is a side to side line. This is the answer.

To check: from A to the midpoint, you need to go up 6 units from -4 to 2. Add another 6 units, and the y-coordinate of the endpoint must be 8.

This would be obvious if you make a sketch on graph paper and count the boxes, which is a fine alternative if you don't want to deal with formulas and subtracting signed numbers.

7. How many points in the coordinate plane are 3 units from the origin and also equidistant from both the x-axis and the y-axis?
1) 1
2) 2
3) 3
4) 4

The points that are 3 units from the origin would be on a circle, centered on the origin, with a radius of three.

The points that are equidistant from the x-axis and y-axis would lie on the angle bisectors of the axes. In other words, they would be points on the two diagonal lines y = x and y = -x.

Each of those lines will cross the circle twice. So there would be four points that fit the criteria, one in each quadrant. You can make a little sketch to see it.

8. As shown below, the medians of triangle ABC intersect at D.

If the length of BE is 12, what is the length of BD?

1) 8
2) 9
3) 3
4) 4

The medians meet at a centroid. The point of concurrence (an intersection of more than two lines) splits each median so that one segment is twice as long as the other.

Looking at BE, BD is 2/3 of the length of BE, and DE is 1/3 of the length of BE. And BD is twice the length of DE. The segment attached to the vertex is always the 2/3, while the segment intersecting the midpoint of the other side is the 1/3.

Since BE is 12, 2/3 of 12 is 8.

Choice (4) 4 is the length of DE.

To get either 9 or 3, you would've had to have used 1/4 and 3/4, I guess.

9. The solution of the system of equations y = x2 - 2 and y = x is
1) (1,1) and (-2,-2)
2) (2,2) and (-1,-1)
3) (1,1) and (2,2)
4) (-2,-2) and (-1,-1)

The line y = x means that the x-coordinate and the y-coordinate have to be the same. Every choice works, so we can ignore this condition for now. Check the other condition:

y = (1)2 - 2 = 1 - 2 = -1, not 1, so eliminate Choices (1) and (3).

y = (2)2 - 2 = 4 - 2 = 2. (2, 2) is a solution, but we already eliminated Choice (3), so Choice (2) is the answer.

Check: y = (-1)2 - 2 = 1 - 2 = -1. (-1, -1) is a solution.

You could also have graphed these two equations in your graphing calculator and looked at the Table of Values for (-1, -1) and (2, 2). Obviously, (1, 1) and (-2, -2) would not be there.

Finally, you could have solved it algebraically, using substitution.

y = x2 - 2 and y = x
x = x2 - 2
0 = x2 - x - 2
0 = (x - 2)(x + 1)
x - 2 = 0 or x + 1 = 0
x = 2 or x = -1
y = x | y = x
y = 2 | y = -1

10. Line ℓ passes through the point (5,3) and is parallel to line k whose equation is 5x + y = 6. An equation of line ℓ is

1) y = 1/5 x + 2
2) y = -5x + 28
3) y = 1/5x - 2
4) y = -5x - 28

Answer: 2) y = -5x + 28

Parallel means that it has the same slope, so first find the slope of the given line. Once you have it, you can eliminate 2 of the choices (which, in this case, will be perpendicular to the given line).

5x + y = 6
y = -5x + 6

Eliminate Choices (1) and (3). Next, substitute (5, 3) and see if you get a true statement.

3 ?= -5(5) + 28
3 ?= -25 + 28
3 = 3

You could have also figured that starting at (5, 3) with a slope of -5, to get back to the y-axis (and the y-intercept), every time you subtract 1 from the x-coordinate, you have to add 5 to the y-coordinate. You have to add 5 five times, which is 3 + 25, or 28.

More to come. Comments and questions welcome.

More Regents problems.

## Tuesday, August 24, 2021

### (x, why?) Mini: Cool Stars

(Click on the comic if you can't see the full image.)
(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Ironically, even though he's cool, he's still hot stuff!

A red dwarf is stable enough that, unlike a brown dwarf, it won't burn out for a long, long time. Age of the universe long time.

Once again, a "mini" that took way to long to create. I did a lot of distortion and then in the end, I inverted the colors of the blue star to get the other two because I kept getting patches of green in both of them.

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. Come back often for more funny math and geeky comics. ### Geometry Problems of the Day (Geometry Regents, January 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, January 2014

1. The midpoint of AB is M(4,2). If the coordinates of A are (6,-4), what are the coordinates of B?

1) (1, -3)
2) (2, 8)
3) (5, -1)
4) (14, 0)

Remember that you are working backward here. You are NOT finding the midpoint. You are finding the other endpoint.

To get from A to the midpoint, you have to subtract 2 from the x-coordinate. Therefore, to get to the other endpoint, you have to subtract another 2. Since 4 - 2 = 2, then the x-coordinate of the endpoint must be 2. There is only one possible choice.

To check: from A to the midpoint, you need to go up 6 units from -4 to 2. Add another 6 units, and the y-coordinate of the endpoint must be 8.

This would be obvious if you make a sketch on graph paper and count the boxes, which is a fine alternative if you don't want to deal with formulas and subtracting signed numbers.

2. Which diagram shows the construction of a 45° angle?

To construct a 45° angle, you have to start with a 90° angle (a perpendicular line) and then bisect it.

Choice (1) appears to be an equilateral triangle that had an angle bisected. That angle would be 30 degrees. Look at this diagram if for no other reason that it was the answer to a Part III construction problem on the June 2014 regents. (Had I realized this, I could've saved from trouble!)

3. What are the coordinates of the center and the length of the radius computations. of the circle whose equation is (x + 1)2 + (y - 5)2 = 16?

1) (1, -5) and 16
2) (-1, 5) and 16
3) (1, -5) and 4
4) (-1, 5) and 4

Answer: 4) (-1, 5) and 4

The equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) is the center of the circle.

If r2 = 16, then r = 4. Eliminate Choices (1) and (2).

Since the signs are "flipped" because of the minus signs in the formula, the center is (-1, 5).

4. If distinct planes R and S are both perpendicular to line ℓ, which statement must always be true?
1) Plane R is parallel to plane S
2) Plane R is perpendicular to plane S
3) Planes R and S and line ℓ are all parallel.
4) The intersection of planes R and S is perpendicular to line ℓ.

Answer: 1) Plane R is parallel to plane S

Consider a beam running from the floor to the ceiling. The beam is parallel to both. The floor and the ceiling are parallel.

Line ℓ can't be parallel to planes R and S. It is stated to be perpendicular. It cannot be both. Parallel means it would not intersect. Perpendicular intersects at a right angle. Eliminate Choice (3).

5. If triangle ABC and its image, triangle A'B'C', are graphed on a set of axes, triangle ≅ triangle A'B'C' under each transformation except

1) D2
2) R90
3) ry=x
4) T(-2,3)

A Dilation is not a rigid motion. It does not preserve size. The image will be similar, but not congruent.

D2 means the image will be twice the size (dilation with a scale factor of 2). It's location will depend upon the center of dilation.

R90 is a Rotation of 90 degrees counterclockwise (unless otherwise stated). Turning the preimage doesn't affect its size or shape, only its orientation.

ry = x is a reflection of the preimage over the line y = x. A reflection doesn't affect its size or shape, only its orientation.

T(-2,3) is a translation of the preimage 2 units to the left and 3 units up. A translation doesn't affect its size, shape, or orientation.

More to come. Comments and questions welcome.

More Regents problems.

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. ## Monday, August 23, 2021

### Geometry Problems of the Day (Geometry Regents, June 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, June 2014

Part IV: A correct answer will receive 6 credits. Partial credit is possible.

38.The vertices of quadrilateral JKLM have coordinates J(-3,1), K(1,–5), L(7,-2), and M(3,4).

Prove that JKLM is a parallelogram.

Prove that JKLM is not a rhombus.

[The use of the set of axes below is optional.]

You have a few options here. The graph can make it easier for you, but it isn't necessary to use.

To prove that JKLM is a parallelogram, you can show that the slopes of the opposite sides are equal, which makes them parallelogram. Two pairs of parallel sides makes a quadrilateral a parallelogram.

Slope of JK: (-5 - 1) / (1 - -3) = -6/4 = -3/2

Slope of KL: (-2 - -5) / (7 - 1) = 3/6 = 1/2

Slope of LM: (4 - -2) / (3 - 7) = 6/-4 = -3/2

Slope of MJ: (1 - 4) / (-3 - 3) = -3/-6 = 1/2

Opposite sides are parallel (same slope), therefore JKLM is a paralellogram. (You must state this! Don't just give a bunch of slopes and let the person scoring the question to mark you incomplete.)

You can also show that a quadrilateral is a parallelogram if the diagonals bisect each other. This is even to show. Find the midpoint of each diagonal. If they are the same point, then the diagonals bisect each other.

Midpoint of JL ( (-3 + 7)/2 , (1 + -2) / 2 ) = (2, -1/2)

Midpoint of KM ( (1 + 3)/2 , (-5 + 4) / 2 ) = (2, -1/2)

Since the midpoint of diagonals JL and KM is the same point (2, -1/2), the diagonals bisect each other. Therefore, JKLM is a parallelogram.

Finally, you could also find the length of the opposite sides. If they are equal, it is a parallelogram. That would require using the Distance Formula four times. The previous two methods are quicker and easier.

To show that JKLM is NOT a rhombus, you need to find a property that makes a parallelogram a rhombus and show that it isn't true. There are TWO that are easy to show. First, you could find the lengths of two consecutive sides. If they are not the same length, then it is not a rhombus. OR you could find the slopes of the diagonals because the diagonals of a rhombus are perpendicular to each other.

Method 1: (Note that you found the distances when did the slope. You can write it out again.)

Length of JK: SQRT(62 + 42) = SQRT(36 + 16) = SQRT(52)

Length of KL: SQRT(32 + 62) = SQRT(9 + 36) = SQRT(45)

Since JK is not congruent to KL, JKLM is not a rhombus.

Method 2:

Slope of JL: (-2 - 1) / (7 - -3) = -3/10

Slope of KM: (4 - -5) / (3 - 1) = 9/2

Multiply (-3/10) * (9/2) = -27/20. The product of the slopes is not -1, so the diagonals are not perpendicular. Therefore, JKLM is not a rhombus.

If you want to use the provided graph, you still need to prove the statements above. Just plotting the points and drawing the figure will not get you any credit. However, it might make it easier for you to find the slopes of the lines because you can count boxes. This also helps for showing Method 2, above.

End of Exam.

More to come. Comments and questions welcome.

More Regents problems.

## Sunday, August 22, 2021

### Geometry Problems of the Day (Geometry Regents, June 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, June 2014

Part III: Each correct answer will receive 4 credits. Partial credit is possible.

35. The graph below shows triangle A'B'C', the image of ABC after it was reflected over the y-axis.

Graph and label ABC, the pre-image of A'B'C'.

Graph and label A"B"C", the image of A'B'C' after it is reflected through the origin.

State a single transformation that will map ABC onto A"B"C".

First, you must reflect A'B'C' over the y-axis to get the pre-image (which was reflected over the y-axis to get A'B'C'). It will be in Quadrant II.

Next, reflect A'B'C' about the origin to get A"B"C", which will be in Quadrant III. Reflecting over the origin will be the same as rotating the image 180 degrees.

Do this before answering the final question.

A single transformation of rx-axis, a reflection over the x-axis, would map triangle ABC onto triangle A"B"C".

36.On the set of axes below, sketch the locus of points 2 units from the x-axis and sketch the locus of points 6 units from the point (0,4). Label with an X all points that satisfy both conditions.

The locus of points 2 units away from the x-axis would be two lines parallel to the x-axis, one above it, and one below it. (Make sure you do the x-axis and not the y-axis.)

The locus of points 6 units from a point will form a circle with a raidus of 6. You can sketch it as best as you can. Start with (0, 4) and mark points (6, 4), (-6, 4), (0, 10) and (0, -2). Make the best circle that you can. It doesn't have to be perfect, but don't let it look like a diamond or an octagon or something crazy.

When you are done with the circle, you will see where the circle intersects to the two parallel lines. Mark those 3 points off with Xs.

37. 7 Using a compass and straightedge, construct an equilateral triangle with AB as a side.

Using this triangle, construct a 30° angle with its vertex at A.
[Leave all construction marks.]

Constructing an equilateral triangle from a line segment isn't difficult. Using the compass, measure the length from point A to point B. Now swing the compass up and make an arc above the center of the line segment.

Turn the compass around without changing its size, and place the end on B. Make an arc above the line. It will intersect the first arc you drew. Label the intersection point C.

Use the straightedge to draw AC and BC. The first part is done.

To construct a 30 degree angle isn't difficult because you already have a 60 degree angle in the equilateral triangle. You just have to bisect it.

To bisect the angle, put the compass at A and make a small arc that intersects AC and AB. I'll call these D and E just for clarity, but you don't need to label these. From D make an arc in the center of the triangle. From E make the same size arc so that it intersects with the one you just made. You could call this F. Use the straightedge to draw AF. You made a 30 degree angle at vertex A.

You final image might look something like this, only a little better I hope. I can't use a compass and straightedge in Windows Paint.

UPDATE: Here is a better constructed example which appeared as an incorrect answer to a multiple-choice problem on the January 2014 Geometry Regents!

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

## Saturday, August 21, 2021

### Geometry Problems of the Day (Geometry Regents, June 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, June 2014

Part II: Each correct answer will receive 2 credits. Partial credit is possible.

32. Two prisms with equal altitudes have equal volumes. The base of one prism is a square with a side length of 5 inches. The base of the second prism is a rectangle with a side length of 10 inches.

Determine and state, in inches, the measure of the width of the rectangle.

The area of the square must equal the area of the rectangle.

5 * 5 = 10 * w
25 = 10w
w = 2.5

Yes, it's that simple. One credit for the work and one for the solution. You probably could have done the work in your head, but put it on paper anyway.

33.As shown in the diagram below, BO and tangents BA and BC are drawn from external point B to circle O. Radii OA and OC are drawn.

If OA = 7 and DB = 18, determine and state the length of AB.

First thing to remember is that every radius is the same size, so if you know one of them, you know all of them. Write them down on the paper so you have it in front of you.

Second thing is that tangents to circles make right angles with radii, so you have right triangles in this figure, and you can use Pythagorean Theorem.

OA = 7, OD = 7, DB = 18, so AB = 25.

72 + (AB)2 = 252
49 + x2 = 625
AB2 = 576
AB = 24

7, 24, 25 is a Pythagorean triple you should be familiar with if only to speed up problems like this one.

34. Triangle RST is similar to XYZ with RS = 3 inches and XY = 2 inches. If the area of triangle RST is 27 square inches, determine and state the area of trianlge XYZ, in square inches.

If RST ~ XYZ then side RS corresponds to side side XY. The scale factor between the two triangles is 3/2.

This means that that base of XYZ is 2/3 of the base of RST and the altitude of XYZ will be 2/3 of the altitude of RST. So the Area of XYZ = (2/3)(2/3)Area of RST.

Area XYZ = (4/9)(27) = 12 square inches.

End of Part II.

More to come. Comments and questions welcome.

More Regents problems.

## Friday, August 20, 2021

### Lass

(Click on the comic if you can't see the full image.)
(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Is this Lass? I mean, Who is this Lass? We'll get to that.

If anyone thought we'd see a ride to the hospital or anything in the delivery room, well, I appreciate your esitimation of my artistic abilities!

I thought this would be a good time to see more of Michele's family.

I could've saved this for Comic #1800 -- but who knows when that will be? Nothing is ever certain.

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. Come back often for more funny math and geeky comics. ### Geometry Problems of the Day (Geometry Regents, June 2014)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

### Geometry Regents, June 2014

Part II: Each correct answer will receive 2 credits. Partial credit is possible.

29. The coordinates of the endpoints of BC are B(5,1) and C(-3,-2). Under the transformation R90, the image of BC is B'C'. State the coordinates of points B' and C'.

The first thing to remember here is if the do not state the direction of the rotation, the default is COUNTER-Clockwise. Don't think of the clock. Think of the Quadrants. Which way do you go to get from I to II? You go counterclockwise.

Second, this is an unusual question because there really isn't work to be shown. You could show formulas, but it isn't necessary. It could be assumed you found it on the graph paper in the back of the book. Basically, each correct endpoint is worth one credit.

The formula for a rotation of R90 ccw is as follows: P(x, y) --> P'(-y, x).

That is, whatever the y coordinate is, change its sign and make it the new x coordinate. Whatever the x coordinate is, make it the y coordinate without changing it. It is better to visualize this and make a sketch than to try and blindly memorize a formula.

B(5, 1) --> B'(-1, 5). Goes from Quadrant I to II.

C(-3, -2) --> C'(2, -3). Goes from Quadrant III to IV.

27. As shown in the diagram below, AS is a diagonal of trapezoid STAR, RA || ST, m∠ATS = 48, m∠RSA = 47, and m∠ARS = 68.

Determine and state the longest side of triangle SAT

You don't need to find any lenght, you just need to determine which side, SA, AT, or TS, is the longest side of the triangle.

The longest side of any triangle is the side opposite the greatest angle. In a right triangle, the hypotenuse is always the longest side, and it is across from the right angle, which is the largest angle in a right triangle.

You are given that angle T is 48 degrees. Therefore, SA is NOT the longest side because 48 can be the largest angle. (You know this is because 48 * 3 < 180, right?)

Since RA || ST then RS is a transversal across parallel lines. That means ∠ARS and ∠RST are supplementary angles. Therefore, m∠ARS + m∠RSA + m∠AST = 180

68 + 47 + m∠AST = 180

115 + m∠AST = 180
m∠AST = 65

65 + 48 + m∠SAT = 180
113 + m∠SAT = 180
m∠SAT = 67

The measure of angle SAT is 67, which makes it the biggest angle. Therefore, the side opposite SAT, which is ST, is the longest side.

Be careful with your computations. The numbers are so close together that getting one incorrect number will carry through and affect your final answer. However, if you are consistent, you will only lose a single point for the one error.

Note that guessing ST without showing any work is worth ZERO points even though it is correct. The Regents do not award credit for a guess that is basically a flip of a coin or eeny-meeny-miney.

31. In right triangle ABC shown below, altitude BD is drawn to hypotenuse AC.

If AD = 8 and DC = 10, determine and state the length of AB.

The altitude divides the right triangle into two smaller right triangles. What you need to realize is that all three of these right triangles are Similar and therefore the corresponding sides are proportional. Write a proportion and solve it.

AB is the hypotenuse of ABD. AC is the hypotenuse of ABC. AC has a length equal to AD + DC.

In ABD, the leg adjacent angle A is AD. In ABC, the leg adjacent angle A is AB.

So AD / AB = AB / AC.
8 / AB = AB / 18
AB2 = 144
AB = 12

Alternatively, you could have done the following to find the length of altitude BD first. Since ADB is proportional to BDC, we can set up the following proportion:

AD / BD = BD / DC (This is also known as the Right Triangle Altitude Theorem)
8 / BD = DB / 10
BD2 = 80
BD = SQRT(80)

You can now use Pythagorean Theorem to find AB:
82 + SQRT(80)2 = AB2
64 + 80 = AB2
144 = AB2
12 = AB

This second methold is longer but it still a valid and complete answer.

More to come. Comments and questions welcome.

More Regents problems.

## Thursday, August 19, 2021

### School Life #23: Beach Time

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(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Fan service for the Svenessa fans.

There's at least one, right? (If I include myself?)

Going through the archives, the two of them haven't appeared together (other than an allusion to falling down a rabbit hole) over a year. (They technically weren't together during the Video Call Hour: Going Live or Remote.)

Now if I could actually draw decent cartwheels and backflips, that would be great. At least now there's a reason Vanessa, who before she started doing handstands in swimwear was kind of quiet and shy, was drawn to Sven, other than his quietness and shyness. So maybe in the future, they'll each make the other more outgoing if only to make themselves more outgoing. Also, why not mention Vanessa's interests in tumbling and gymnastics, right?

On the other hand, that all sounds like fanfic, even if it turns out to be fanfic that I write myself at Archive of Our Own. There's nothing (x, why?)-related posted there -- yet. And there are no cartwheel videos posted by Sven, either. That's up to Vanessa, it's her phone.

### I also write Fiction!

You can now preorder 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.
Preorder the softcover or ebook at Amazon. Come back often for more funny math and geeky comics. 