Geometry Problems of the Day (Geometry Regents, June 2024 Part I)

This exam was adminstered in June 2024.

More Regents problems.

June 2024 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.


1. In the diagram below, △BRI is the image of △JOE after a translation. Triangle CAT is the image of △BRI after a line reflection.

Which statement is always true?

(1) ∠R ≅ ∠T
(2) ∠J ≅ ∠A
(3) JE ≅ RI
(4) OE ≅ AT

Answer: (4) OE ≅ AT

You can follow the transformations, and if you wish to, you can mark up the book to show which points are moving to which new poisitions. Then you can see which angles are corresponding (and therefore congruent) and which sides are corresponding (and congruent).

But here's the little secret for state exams. Keep in mind that this doesn't always work in your classroom because teachers are sloppy (and I can be guilty of this myself sometimes if I'm not paying attention or if I'm just tired or overworked). The order of the letter is the triangle name is important. They will always correspond on state exams, and they should in textbook examples. (Textbook writers aren't always 100% at the top of their game.).

So B goes to J goes to C, R goes to O goes to A, and I goes to E goes to T.

You can eliminate Choices (1) and (2) immediately. Careful checking shows that Choice (3) is wrong and Choice (4) is the correct answer because JOE goes to CAT.

2. right cylinder is cut parallel to its base. The shape of this cross section is a

(1) cone
(2) circle
(3) triangle
(4) rectangle

Answer: (2) circle

If it is cut parallel to the base, then the cross-section has to be a circle, which is the same shape as the base of the cylinder. The correct answer is Choice (2).

A rectangle could occur if the cross section was perpendicular to the base.

A triangle would not happen, no matter how you slice it. And a cross section is a 2-D shape, so it could not be a cone, which is 3-D.

3. What is the minimum number of degrees that a regular hexagon must rotate about its center to carry it onto itself?

(1) 45°
(2) 72°
(3) 60°
(4) 120°

Answer: (3) 60°

Any figure will carry onto itself if you rotate it 360 degrees. Because regular polygons have symmetry, you only have to rotate them a minimum of 360 divided by the number of sides that the polygon has, which in this case is 6.

360 / 6 = 60 degrees, so the correct answer is Choise (3).

120 degrees is the size of each internal angle, but that's not the question that's being asked.

4. In the diagram below, a sphere is inscribed inside a cube. The cube has edge lengths of 18.

What is the volume of the sphere, in terms of π?

(1) 108π
(2) 432π
(3) 972π
(4) 7776π

Answer: (3) 972π

Use the formula for the volume of a sphere. The length of the box is the diameter of the ball. You need to cut that in half to get the radius.

Note that one of the wrong choices will always be the result if a student uses the diameter instead of the radius. They expect some students to do this. As a result, you can be fairly certain that the largest choise will not be the correct answer.

V = 4/3 π r³ = 4/3 π (9)³ = 972 π, which is Choice (3).

5. In the diagram below, EM intersects HA at J, EA ⟂ HA, and EM ⟂ HM.
If EA = 7, EJ = 9, AJ = 5.4, and HM = 3.29, what is the length of MJ, to the nearest hundredth?

(1) 2.47
(2) 2.63
(3) 4.11
(4) 4.39

Answer:(1) 2.47

You can prove that the two traingles are similar in shape and therefore their corresponding sides are proportional. You don't have to prove anything though because if this wasn't true, there wouldn't be any way to solve this question. So in this particular problem if you just assumed that they were similar, you would be okay. Because they are.

Briefly, each triangle has a right angle, which are congruent to each other, and the vertical angles are congruent. Therefore, by AA, the triangles must be similar. (They are not congruent.)

Write down the lengths that you know and put an x next to the side you're looking for, MJ.

Comparing sides, you will see that

5.4 / x = 7.2 / 3.29

so 7.2 x = (5.4)(3.29)

and x = (5.4)(3.29)/7.2 = 2.4675

To the nearest hundredth, MJ = 2.47, which is Choice (1).

6. Which equation represents the line that passes through the point (2,-7) and is perpendicular to the line whose equation is y = 3/4x + 4?

(1) y + 7 = 3/4(x - 2)
(2) y - 7 = 3/4(x + 2)
(3) y + 7 = -4/3(x - 2)
(4) y - 7 = -4/3(x + 2)

Answer: (3) y + 7 = -4/3(x - 2)

Parallel lines have the same slope. Perpendicular lines have slopes that are NOT the same but are inverse reciprocals -- the product of the slopes is -1.

Since they want perpendicular, you can cross out Choices (1) and (2).

The question uses slope-intercept form, which gives you the slopes and the y-intercept. The choices are given using point-slope form, which gives you the slope and a point on the line.

The thing to remember is that the formula has minus signs in it (as do many, many formulas in Algebra and Geometry): y - k = m(x - h).

You may have learned that formula as y - y₀ = m(x - x₀), which is the same thing. However, if you get used to seeing h and k, you'll notice that (h,k) is all over the place.

Since the point is (2,-7) and the signs are "flipped" in the equation, then the correct answer is Choice (3).

7. In △RHM below, m∠R = 110° and m∠M = 40°.

If △RHM os reflected over side HM to form quadrilateral RHR'M, which statement is always true?

(1) Quadrilateral RHR'M is a parallelogram.
(2) m∠MHR' = 40°
(3) m∠HMR' = 40°
(4) MR ≅ HR'

Answer: (3) m∠HMR' = 40°

You can make a sketch in the book underneath the original image and mark it up.

You can see that the correct answer is Choice (3) because HMR and HMR' are corresponding angles of congruent triangles.

The figure is not a parallelogram because the opposite sides are not parallel. It's a kite. Eliminate Choice (1).

The measure of angle MHR' is the same as angle MHR, both of which can be calculated to be 30 degrees. (180 - 110 - 40 = 30). Choice (2) is incorrect.

Since the figure is a kite and not a parallelogram, the opposite sides are not congruent, so MR is not congruent to HR'. This eliminates Choice (4).

8. The funnel shown below can be used to decorate cookies with melted chocolate. The funnel can be modeled by a cone whose radius is 6 cm and height is 13 cm.
The baker uses 2 cubic centimeters of chocolate to decorate each cookie. When the funnel is completely filled, what is the maximum number of cookies that can be decorated with the melted chocolate?

(1) 78
(2) 245
(3) 490
(4) 735

Answer: (2) 245

Find the volume of the funnel using the formula for the volume of a cone: V = 1/3 π r² h.

V = (1/3) π (6)² (13) = 490.08845396...

If the baker uses 2 cm³ per cookie, then divide the result by 2: 490 / 2 = 245.

The correct answer is Choice (2).

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now order my newest book Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!
Also, check out Devilish & Divine, an anthology filled with stories of angels and devils by 13 different authors, and In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

June 2024 Geometry Regents Part IV

This exam was adminstered in June 2024 .

June 2024 Geometry, Part IV

Each correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.

35. Triangle JOE has vertices whose coordinates are J(4,6), O(-2,4), and E(6,0).
Prove that △JOE is isosceles.
[The use of the set of axes on the next page is optional.]

Point Y(2,2) is on OE.
Prove that JY is the perpendicular bisector of OE.

Answer:

As far as Part IV questions go, this one isn't terrible. In fact, it's pretty simple compared to some of the ones they ask involving parallelograms.

To show that JOE is an isosceles triangle, you have to find the lengths of all three sides, state which two of them are congruent. Then state JOE is an isosceles triangle because it has to congruent sides. (YES! You need to write this concluding statement!)

You can use the Distance Formula, or, if you graph JOE, you can just count boxes and use Pythagorean Theorem.

JO = √(6² + 2²) = √(40)

OE = √(8² + 4²) = √(80)

EJ = √(2² + 6²) = √(40)

Since JO ≅ EJ, triangle JOE is an isosceles triangle.

To prove that JY is the perpendicular bisector of OE, you have to prove that JY perpendicular to OE and that JY is the bisector of OE. Yes, you need to do both things.

If the lines are perpendicular, then the slopes will be inverse reciprocals (that is, they have a product of -1). If JY is a bisector, then it goes through the midpoint of OE.

The midpoint of OE is ( (-2+6)/2, (4+0)/2 ) or (2,2), which is point Y. So Y is the midpoint of OE and JY is a bisector of OE.

The slope of OE is (0-4)/(6-(-2)) = -4/8 = -1/2. The slope of JY is (6-2)/(4-2) = 4/2 = 2/1. The slopes are inverse reciprocals, and (-1/2)(2/1) = -1, so OE is perpendicular to JY.

Therefore JY is the perpendicular bisector of OE.

End of Part Exam

How did you do?

Questions, comments and corrections welcome.

I also write Fiction! You can now order my newest book Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!
Also, check out Devilish & Divine, an anthology filled with stories of angels and devils by 13 different authors, and In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

June 2024 Geometry Regents Part III

This exam was adminstered in June 2024 .

June 2024 Geometry, Part III

Each correct answer is worth up to 4 credits. Partial credit can be given. Work must be shown or explained.

32. A building is composed of a rectangular pyramid on top of a rectangular prism, as shown in the diagram below. The rectangular prism has a length of 38 feet, a width of 15 feet, and a height of 22 feet. The rectangular pyramid sits directly on top of the rectangular prism, and its height is 12 feet.

An air purification filter was installed that will clean all the air in the building at a rate of 2400 cubic feet per minute. Determine and state how long it will take, to the nearest tenth of a minute, for the filter to clean the air contained in the building.

Answer:

First you need to find the volume of the building. The building is a pyramid sitting on top of a prism. You have to find the volumes for each of these separately and add them together.

The volume of a rectangular prism is V = (L)(W)(H), and the volume of a rectangular pyramid is V = 1/3(L)(W)(H). Note that the length and width of the two objects are the same but the height are different.

V(prism) = (38)(15)(22) = 12,540 and v(pyramid) = 1/3(38)(15)(12) = 2280. Total volume = 14820.

Next, divide this about by 2400 ft³/min. 14820/2400 = 6.175

To the nearest tenth of a minute, it will take 6.2 minutes.

33. Given: △ABC, △DEF, AB ⟂ BC, DE ⟂ EF, AE ≅ DB, and AC || FD.

Prove: △ABC ≅ △DEF

Answer:

They are expecting a two-column, or a paragraph, proof. Even as a paragraph, all the statements have to be made and the dots connected (so to speak). ou can't just BS a one- or two-sentence rationale which doesn't prove a thing.

Yes, I see a lot of those for every exam.

As always, start with the Given information. Your last statement should be the thing that you are trying to prove, along with a final reason.

Notice that there are two parallel lines, so you will likely have alternate interior angles. There are perpendicular lines, which create right angles, and all right angles are congruent. That means you only need to show that one other pair of sides are congruent to use either ASA or AAS. Since you're given AE = BD, you can pretty much figure out that you'll need to show that AB = DE.

1. △ABC, △DEF, AB ⟂ BC, DE ⟂ EF, AE ≅ DB, and AC || FD.	Given.
2. ∠ABC and ∠DEF are right angles.	Def of perpendicular lines
3. ∠ABC ≅ ∠DEF	All right angles are congruent.
4. ∠A ≅ ∠D	Alternate interior angles
5. BE ≅ BE	Reflexive property
6. AB ≅ DE	Addition postulate of congruence
7. △ABC ≅ △DEF	ASA

34. In the diagram below, a boat at point A is traveling toward the most powerful waterfall in North America, the Horseshoe Falls. The Horseshoe Falls has a vertical drop of 188 feet. The angle of elevation from point A to the top of the waterfall is 15°.

After the boat travels toward the falls, the angle of elevation at point B to the top of the waterfall is 23°. Determine and state, to the nearest foot, the distance the boat traveled from point A to point B.

Answer:

Call the base of the waterfall C. To find the length of AB, you have to find the lengths of AC and BC and then subtract AC - BC. You can find both of those lengths using tangent. (Make sure your calculator is in DEGREE mode, or you will get crazy answers.)

Tan 18 = 188 / x

x = 188 / tan 18 = 701.625...

Tan 23 = 188 / y

y = 188 / tan 23 = 442.900...

701.625 - 442.900 = 258.725, or about 259 feet to the nearest foot.

End of Part III

How did you do?

Questions, comments and corrections welcome.

I also write Fiction! You can now order my newest book Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!
Also, check out Devilish & Divine, an anthology filled with stories of angels and devils by 13 different authors, and In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

June 2024 Geometry Regents Part II

This exam was adminstered in June 2024 .

June 2024 Geometry, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.

25. In △ABC below, m∠C = 90^o, AC = 11, and AB = 18.

Determine and state the measure of angle A, to the nearest degree.

Answer:

It's a right triangle. For angle A, you have the adjacent side and the hypotenuse. This means that you can find the size of the angle using the cosine ratio.

Cos A = 11/18

A = cos^-1 (11/18) = 52.3

Angle A is 52^o to the nearest degree. Make sure the calculator is in DEGREE Mode.

26. Use a compass and straightedge to construct an equilateral triangle inscribed in circle A below. [Leave all construction marks.]

Answer:

There are multiple methods that can be used. If you recall that a hexagon inscribed in a circle can be split up into 6 congruent equilateral triangles.

[IMAGE OMITTED]

Here's a simple solution, and it really is simple.

Use your straightedge to draw a diameter (or a radius). Use the compass to measure the distance from the point on the circle to the center of the circle. Make arcs that cross the circle above and below. Move the compass to the new points and make another arc and then another. When you are finished, you will have six points on the circle.

Use the straightedge to draw three lines connecting every other point. The result will be an equilateral triangle.

27. Quadirlateral DEAR and its image, quadrilateral D'E'A'R', are graphed on the set on axes below.

Described a sequence of transformations that maps quadrilateral DEAR onto quadrilateral D'E'A'R'.

Answer:

There are several possibilities but each of them will involve a reflection and a translation. The reflection must include the line of reflection and the translation must state how far is each direction.

Note that under the new curriculum, you could draw a line, say EE', and say something like "translate DEAR along directed line segement EE'". I won't do that here. I'm just saying that it is valid.

One valid answer is to translate DEAR two spaces to the left and seven spaces down, followed by a reflection over the y-axis.

Or you could reflect over the y-axis and then translate two spaces to the right and seven spaces down.

28. In circle P below, tangent AL and secant AKE are drawn.

If AK = 12 and KE = 36, determine and state the length of AL.

Answer:

This was a little bit of an "irrelevant" problem. By this, I mean, when grading these problems, the scorer is instructed to give 1 credit to a correct answer without any work at all. However, a correct answer that was found through an obviously incorrect, incoherent or irrelevant manner is supposed to be scored with 0 credits.

In this problem, if you subtracted 36 - 12 = 24, you got the correct answer through an incorrect method. You shouldn't have gotten any points. In reality, scorers are probably looking for excuses to give you points if you did something so desparate.

The correct method involves using the following formula: (AL)² = (AK)(AE)

x² = (12)(12 + 36)
x² = 576
x = 24

AL has a length of 24.

29. The equation of a circle is x² + y² + 8x - 6y + 7 = 0. Determine and state the coordinates of the center and the length of the radius of the circle.

Answer:
You need to Complete the Square, twice. Note that you can find the center without completing the square, but you need to complete them to find the radius.

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

Reminder: (x + 5)² is x² + 10x + 25. So if I had x² + 10x, I could factor the expression if I add 25 to it first.

x² + y² + 8x - 6y + 7 = 0
x² + 8x + y² - 6y = -7
x² + 8x + 16 + y² - 6y + 9 = -7 + 16 + 9
(x + 4)² + (y - 3)² = 18

The center of the circle is (-4,3) and the radius is &SQRT;(18).

30. On the set of axes below, △ABC is drawn with vertices that have coordinates A(2,-3), B(4,5) and c(-5,1)

Determine and state the area of △ABC.

Answer:
This problem is deceptively simple. Draw a rectangle about the triangle, so that that C and A are points on the rectangle and B is a vertex. The rectangle has a length of 9 and a width of 8, and an Area of (9)(8) = 72.

This creates three triangles. Find the areas of the three triangles and subtract those from 72.

1/2(4)(7) = 14, 1/2(4)(9) = 18, 1/2(2)(8) = 8.

72 - 14 - 18 - 8 = 32.

The area of ABC is 32.

31. In the diagram below, AE = 15, EB = 27, AF = 20, and FC = 36.

Explain why EF || BC.

Answer:
If the proportion AE/AB = AF/AC is true, then the two triangles are similar by SAS since they share angle A. This means that corresponding angles AEF and AEB are congruent. That makes the EF || BC.

Check: 15/27 = 5/9, and 20/36 = 5/9

Therefore, AE/AB = AF/AC, so the triangles are similar and the lines are parallel.

You could also have used the side-splitter theorem, if you remembered that one. It states that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. Same idea, but you don't need to mention angle A or SAS.

You didn't need to write a complete proof. If you did write a complete proof, any mistake could cost 1 of the 2 points.

End of Part II

How did you do?

Questions, comments and corrections welcome.

I also write Fiction! You can now order my newest book Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!
Also, check out Devilish & Divine, an anthology filled with stories of angels and devils by 13 different authors, and In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

School Life #42: Oh Well

(Click on the comic if you can't see the full image.)

(C)Copyright 2024, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

It's amazing the little tidbits I remember from odd Jeopard categories or other game shows.

It works both ways, when some odd thing I remember from my life turns out to be an answer.

One quick Final Jeopardy stories: years ago my brother Stephen told me a philosophy joke (seriously). I've since heard it repeated numerous times using Plato and Socrates. But Stephen used Satre and Camus. I've encountered Satre's name since then but really know nothing about him. The clue on Jeopardy wanted to know who Satre was refering to when he said (some quote). I'm sitting at home thinking, "How the hell should I know? Camus!"

And the answer was Camus. And none of the contestants got it.

As for this joke, it was something that occurred to me in class, and then I realized that I do have a character named Mark. Except that Mark has been shown, I believe, to be one of the smarter students.

I also write Fiction! A Bucket Full of Moonlight is funding NOW on Kickstarter. A long of stretch goals have been unlocked, meaning that there are a lot of freebies and extras to be had! Check out my book or any (or all!) of the six other books being launched!
You can now order my latest Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!

Come back often for more funny math and geeky comics.

School Life #41: Do or Do Not

(Click on the comic if you can't see the full image.)

(C)Copyright 2024, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Sometimes I realize that I need to be cautious with what movie quotes I use. The students might interpret them in a way other than what was intended.

I wanted to get back to what the students were up to before school is out for summer. (In NYC, we're still in school.)

I have to confess to getting a bit disorganized in my 17th year of (x, why?). I wasn't sure of either the Comic number or the School Life number. And I keep referring to "School Life" as "Student Life", privately. Neither one works for the summer, but I'll use it regardless. At least, I didn't go forward with the "(x, why?) After Hours" tag that I'd thought about many years ago.

I also write Fiction! A Bucket Full of Moonlight is funding NOW on Kickstarter. A long of stretch goals have been unlocked, meaning that there are a lot of freebies and extras to be had! Check out my book or any (or all!) of the six other books being launched!
You can now order my latest Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!

Come back often for more funny math and geeky comics.

Venn There, Done That

(Click on the comic if you can't see the full image.)

(C)Copyright 2024, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

You could just stay home, shop online line, and just tell everyone that you've done that.

This comic illustrates one of my problems with "quick comics", and that's humor creep. It's like mission creep, but with humor. This comic was supposed to be just a Venn diagram. But the more I thought about it -- first mistake: thinking too much about a "quick joke" -- the less I could decide what the diagram should look like. So I devised a comic using two of them.

Second mistake was realizing that once you applied logiv, the entire joke falls apart, so Michele had to step in and tell both the guys that they're wrong.

For once, Ken is being the overly-analytical one, instead of Mike, but it balances the comic better with him standing by Michele.

I also write Fiction! A Bucket Full of Moonlight is funding NOW on Kickstarter. A long of stretch goals have been unlocked, meaning that there are a lot of freebies and extras to be had! Check out my book or any (or all!) of the six other books being launched!
You can now order my latest Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!

Come back often for more funny math and geeky comics.

(x, why?) Mini: Radian

(Click on the comic if you can't see the full image.)

(C)Copyright 2024, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

Not the same thing at all.

They should invite her out for pie.

So I'm back. For how long? I don't know. I do't have any reasons for being away. This post has been a "draft" since May 31, and the comic is based on a joke told in class nearly a month ago.

Obviously, I needed to get back to the eclipse story becasue I had a two-part follow-up, but that's the kind of thing that takes time to work on. This comic, on the other hand, can be knocked out and posted within an hour. So why'd it take over a month? Again, I don't know.

This just hasn't been a great year scholastically even as things are going pretty well on the writing front. I'd been counting down to 2,000 comics for so long, and then I just came to a halt.

There probably are actual reasons, but not that I can (or will) identify right now.

Enjoy the ride. I hope to have more updates.

I also write Fiction! You can now order my newest book Burke's Lore, Briefs: Portrait of a Lady Vampire & Other Vampiric Cravings, written by Christopher J. Burke, which contains the aforementioned story and three other stories. Order the softcover or ebook at Amazon. And don't forget that Burke's Lore, Briefs:A Heavenly Date / My Damned Best Friend is still available!
Also, check out Devilish & Divine, an anthology filled with stories of angels and devils by 13 different authors, and In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.

Come back often for more funny math and geeky comics.

The Kickstarter for "A Bucket Full of Moonlight" has begun!

I'm really late in this announcement. But you can see that I've been off my blogs for sometime now.

There are seven books in this Tenth Anniversary of eSpec Books Kickstarter event:

https://www.kickstarter.com/projects/e-specbooks/something-for-everyone

The books being launched are Aeros & Heroes by Ef Deal, A Curse of Time and Vengeance by Christine Norris, Checkmate by Ty Drago, A Bucket Full of Moonlight by Christopher J. Burke, Feat of Clay by Keith R.A. DeCandido, Aleyara’s Descent by Christopher L. Bennett, and The Pharaoh's Cat by Lisanne Norman.

A Bucket Full of Moonlight has 30+ stories, most of which started as ideas or paragraphs on reddit in r/WritingPrompts, which were later imported into my own subreddit r/xwhy. (That’s my username on reddit.)

Several stretch goals have already been unlocked, including an additional short story, previously published, but with an additional, never-before-seen ending to it.

Final note: YOU CAN BE IN MY BOOK! One of the Pledge Levels is a "Tuckerization" of my book. That means that one of the side characters will be renamed for the person who pledges this amount. There are up to THREE characters available to be renamed in the book.

Check it out!