The numbers *a*, *b*, and *c* are positive integers. An apple cost $*a*, a banana costs $*b*, and a cherry costs $*c*. The cost of *b* apples, *b* bananas, and *a* + *b* cherries is $77. What would be the cost for one apple, two bananas, and one cherry?

This problem was part of the early rounds in the Norwegian Mathematical Olympiad. Students had about 5 minutes on average to solve each problem.

It seems like it’s impossible to solve, and yet there is enough information to be able to solve it. Can you figure it out? Give it a try, and watch the video for a solution.

**Delightfully Fruitful Problem From The Norwegian Math Olympiad**

Or keep reading.

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**Answer To Seemingly Impossible Problem From Norwegian Math Olympiad – The Cost Of Fruit**

In these problems it is a good idea to write out the equations and then try to see if there are any nice patterns from factoring or re-writing. Let’s start out that *b* apples, *b* bananas, and *a* + *b* cherries cost 77 dollars. I will omit the dollar signs so the equation is easier to read.

(cost apple)(# apples) + (cost bananas)(# bananas) + (cost cherries)(# cherries) = total cost

*a*(*b*) + *b*(*b*) + *c*(*a* + *b*) = 77

Now we can further simplify because the first two terms have a common factor of *b*.

*b*(*a* + *b*) + *c*(*a* + *b*) = 77

Now we can factor *a* + *b* from both terms.

(*a* + *b*)(*b* + *c*) = 77

Since *a*, *b*, and *c* are positive integers, we have two positive whole numbers that multiply to be 77. So let us consider the factors of 77.

The number 77 only factors into two ways: either 1 × 77 or 7 × 11. But could *a* + *b* or *b* + *c* equal 1? No. Because each unknown term is a positive integer, the smallest value for *a* + *b* or *b* + *c* must be 2. Therefore, we know one factor is 7 and another factor has to be 11.

But how does this help us? We want to evaluate the price of one apple, two bananas, and one cherry. The cost of that is:

*a*(1) + *b*(2) + *c*(1)

*a* + 2*b* + *c*

Now we have to recognize a pattern: we can split the *b* term into *b* + *b*, and then we can group each pair of sums:

*a* + *b* + *b* + *c*

(*a* + *b*) + (*b* + *c*)

From our factoring above, we know one of *a* + *b* and *b* + *c* is 7 and the other is 11. So we can evaluate:

(*a* + *b*) + (*b* + *c*)

= 7 + 11 or 11 + 7 = 18

So we have found the answer! One apple, two bananas, and one cherry would cost $18.

**Source**

Thanks to Oskar for sending me the puzzle. This problem appeared on an early round of the Norwegian Mathematical Olympad.

]]>The text reads: The kite GDBE is placed in the square ACHF with DG = GB = EG. Calculate the size, *x*, of angle DBE. Justify your answer with clear geometric reasoning.

I posted a solution which is probably what test-makers intended. But the problem turned out to be even more interesting! My solution actually had a subtle flaw based strictly on how the problem was worded.

So while I got the correct answer, I would say my proof was flawed and not justified. Interestingly the correct answer is the same. So in this post I’ll explain my flawed proof and then give a corrected proof, along with another way to find the answer.

Can you figure it out? Give the problem a try and then watch the video for a solution.

**ORIGINAL VIDEO: “Impossible” Math Problem Leaves 15 Year Olds In Tears – New Zealand (Kite Question)**

**CORRECTED VIDEO: I Was Wrong – The Kite Problem Correct Proof**

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**Answer To Kite Question**

The key to the problem is considering the diagonals of a kite, which are perpendicular. Furthermore, the vertical diagonal also bisects the horizontal diagonal. This division creates some special triangles whose angles are known.

**Flawed proof method to find a 30-60-90 right triangle**

Here are the steps of my original proof, along with the part where the proof is flawed…

Here’s the outline of the proof in text:

1. Draw diagonals DE and GB and say they meet at point I.

2. The diagonals are perpendicular and GB also bisects DE because GD = EG.

3. Then DE is the length of the square, and DI and EI are 1/2 a length of the square. DG = GB = the length of the square.

4. The triangles DIG and EIG are 30-60-90 right triangles (the hypotenuses is 2 times a leg), with DGI and EGI equal to 30 degrees.

5. Triangles GDB and GEB are isosceles right triangles. Thus angles GDB, GBD, GEB, GBE are equal, and they are equal to *x*/2 since any two angles would combine to be *x*.

6. The angles in a triangle sum of 180 degrees, so we then have:

angle GDB + angle GBD + 30 = 180

2(angle GDB) + 30 = 180

2(angle GDB) = 150 = *x*

Thus *x* is 150 degrees.

The flaw is in step 3; in particular, the incorrect statement is:

“DG = GB = the length of the square”

While this conclusion is apparent from the diagram, it is not justified unless we know DE is perpendicular to the sides AF and CH or GB is perpendicular to the sides AC and FH.

Christopher Night, Steve’s Mathy Stuff, and Joseph Lombardo pointed out the flaw in the comments to the video I posted. Christopher Night even demonstrated how GB and DE could be longer than the sides of the square, if the kite is rotated and off-axis.

Diagram by Christopher Night with numbers on Imgur: https://imgur.com/qJxdwXW.

Somewhat incredibly the answer is still the same even if the kite is rotated. So how can we fix the proof?

**Correct proof to find an equilateral triangle**

The important fact is the two diagonals of the kite have equal length. So let’s prove GB = DE, even if the kite is rotated inside the square.

1. Draw DJ perpendicular to sides AF and CH; draw BK perpendicular to sides AC and AH.

2a. Triangles EJD and GKB are congruent by angle-side-angle, for the following justifications:

2b. Angles EJD and GKB are both right angles.

2c. Sides DJ and BK are equal as both equal the side length of the square.

2d. Angles EDJ and GBK are equal. This is a little bit harder to see. Since GB and DE are perpendicular, they intersect perpendicular lines as the same angle: GB intersects BK at the same angle that DE intersects DJ (since DJ is perpendicular to BK).

3. Since the triangles EJD and GKB are congruent, we can conclude GB = DE.

4. Now we can conclude DGE is an equilateral triangle since DG = GB = EG (given) and we proved GB = DE.

5. DGE is a 60 degree angle.

6. DGB and EGB are congruent isosceles triangles (by side-side-side; the last two sides DB = BE because GDBE is a kite and we are given GD = EG.)

7. Thus DGB and EGB are each 60/2 = 30 degree angles.

8. We can then calculate the remaining angles, which are all equal, and two of which will be equal to *x*. Since a triangle sums to 180 degrees:

GDB + GBD + 30 = 180

2(GBD) + 30 = 180

2(GBD) = 150 = *x*

**Another proof using arcs of a circle!**

Ali A. pointed out this method in a comment to the video. Let’s start from the point we know DGE is an equilateral triangle and DGE is 60 degrees.

Since GD = GB = EG, we can draw a circle centered at point G that passes through points D, B, and E.

Now we can solve for DBE as an inscribed angle of the circle, as follows.

Here is the text explanation.

Mark point P on the major arc of DE. Since an entire circle measures 360 degrees, we have the measures of the arcs and angles as follows in degrees:

arc(DBE) + arc(DPE) = 360

arc(DBE) = 60 (because DGE is a central angle 60)

arc(DBE) + 60 = 360

arc(DPE) = 300

angle(DBE) = arc(DPE)/2 (because DBE is an inscribed angle for arc(DPE))

angle(DBE) = 300/2 = 150 = *x*

So we solved for *x* primarily using circles and arcs!

I don’t think the test-writers had this proof in mind. But it’s a really neat way to see the solution too!

**Thanks to all patrons! Special thanks to:**

Shrihari Puranik

Kyle

If you like my videos, you can support me at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

**Sources**

My original flawed proof

Blog post: https://wp.me/p6aMk-5nC

Video: https://www.youtube.com/watch?v=z3zjyCZFzDo

NZQA Level 1 Exam 2017, Question Three (b)

http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2017/91031-exm-2017.pdf

Exam on Scribd

https://www.scribd.com/document/365042448/91031-exm-2017

New Zealand Herald. “How good is your maths? The questions that drove 15-year-olds to tears”

http://www.nzherald.co.nz/nz/news/article.cfm?c_id=1&objectid=11946237

I was requested to cover another problem about a kite. I have re-created the question:

The text reads: The kite GDBE is placed in the square ACHF with DG = GB = EG. Calculate the size, *x*, of angle DBE. Justify your answer with clear geometric reasoning.

Can you figure it out? Give the problem a try and then watch the video for a solution.

**ORIGINAL VIDEO: “Impossible” Math Problem Leaves 15 Year Olds In Tears – New Zealand (Kite Question)**

**CORRECTED VIDEO: I Was Wrong – The Kite Problem Correct Proof**

*December 6, 2017 added link to corrected video proof.

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**Answer To Kite Question**

The key to the problem is considering the diagonals of a kite, which are perpendicular. Furthermore, the vertical diagonal also bisects the horizontal diagonal. This division creates some special triangles whose angles are known.

*As pointed out in the comments, there is a missing assumption to my original proofs! The kite could be positioned so it is skewed in the square, see: https://imgur.com/qJxdwXW.

To correct the proofs, we can add a first *assumption* that the diagram is accurate–that the kite is placed so its diagonals are parallel to the sides of the square. Then the proofs below work.

**BUT the assumption is not justified as the “diagram is not to scale.” So please read a new post with the correct proof:**

**I have kept the old flawed proofs up for documentation**

**Flawed proof 1: Find a 30-60-90 right triangle**

Here are the steps.

Here’s the outline of the proof in text:

1. Draw diagonals DE and GB and say they meet at point I.

2. The diagonals are perpendicular and GB also bisects DE because GD = EG.

3. Then DE is the length of the square, and DI and EI are 1/2 a length of the square. DG = EG = GB = the length of the square. **This step GB = DE = length of the square depends on the diagram being to scale which is technically incorrect! I present the correct proof here.**

4. The triangle DIG is a right triangle whose hypotenuse DG is twice the length of DI.

5. This means DIG is a 30-60-90 right triangle, and the angle opposite DI is 30 degrees. So DGI is 30 degrees.

6. Similarly EIG is a 30-60-90 right triangle and angle EGI is 30 degrees.

7. Triangles GDB and GEB are isosceles right triangles. Thus angles GDB, GBD, GEB, GBE are equal, and they are equal to *x*/2 since any two angles would combine to be *x*.

8. The angles in a triangle sum of 180 degrees, so we then have:

angle GDB + angle GBD + 30 = 180

2(angle GDB) + 30 = 180

2(angle GDB) = 150 = *x*

Thus *x* is 150 degrees.

**Flawed proof 2: find an equilateral triangle**

Here is a very similar proof that some people might have found.

1. Draw the diagonals DE and GB which are perpendicular, and GB also bisects DE.

2. DI is a length of the square, as is GB. Since GB = DG = EG, DEG is an equilateral triangle and angle DGE is 60 degrees. **This step GB = DE = length of the square depends on the diagram being to scale which is technically incorrect! I present the correct proof here.**

3. GB also bisects the angle. This is because DIG and EIG are congruent triangles (by side-side-side since DI = IE, DG = EG, and IG = IG).

4. Angles DGI and EGI are 60/2 = 30 degrees.

5. Now we proceed as in proof 1 to the congruent isosceles triangles DGB and EGB.

Did you figure out the problem, and was it too difficult for 15 year olds?

**Thanks to all patrons! Special thanks to:**

Shrihari Puranik

Kyle

If you like my videos, you can support me at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

**Sources**

NZQA Level 1 Exam 2017, Question Three (b)

http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2017/91031-exm-2017.pdf

Exam on Scribd

https://www.scribd.com/document/365042448/91031-exm-2017

New Zealand Herald. “How good is your maths? The questions that drove 15-year-olds to tears”

http://www.nzherald.co.nz/nz/news/article.cfm?c_id=1&objectid=11946237

This is the latest tricky math puzzle to go viral. Can you figure it out?

Watch the video for a solution.

**Math Problem From China Stumps Internet – Shoes, Cat, Whistle Puzzle Explained**

Or keep reading for a text explanation.

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**Answer To Shoes, Cat, Whistle Puzzle Explained**

The tricky part is two of the pictures in the last line have slight differences from the previous equations.

The third equation has a term with a pair of whistles. The last line involves a single whistle.

Furthermore, the cats in the second and third lines are wearing a whistle, but the cat in the last line is not wearing a whistle. Presumably the value of the whistle should be accounted for to get the correct answer.

The pictures can be translated into the following equations:

shoes + shoes + shoes = 30

shoes + (cat + whistle) + (cat + whistle) = 20

(cat + whistle) + 2(whistles) + 2(whistles) = 13

shoes + (cat) x (whistle) = ?

From the first equation we can solve for the shoes value:

shoes + shoes + shoes = 30

3(shoes) = 30

shoes = 10

We can then solve the second equation for the (cat + whistle) value:

shoes + (cat + whistle) + (cat + whistle) = 20

10 + 2(cat + whistle) = 20

2(cat + whistle) = 10

cat + whistle = 5

Then we solve the third equation for the whistle:

(cat + whistle) + 2(whistles) + 2(whistles) = 13

5 + 4(whistles) = 13

4(whistles) = 8

whistle = 2

We also need to solve for the value of the cat:

cat + whistle = 5

cat + 2 = 5

cat = 3

Now we can evaluate the final expression, remembering the order of operations that multiplication should be evaluated before addition:

shoes + (cat) x (whistle) = ?

10 + 3 x 2

= 10 + 3 x 2

= 10 + 6

= 16

The value of 16 is what many people believe to be the correct answer.

**Sources**

https://twitter.com/ChinaDailyUSA/status/934753309117751297

https://theconservativetreehouse.com/2017/11/26/chinese-math-quiz/

]]>**Mental Floss**: Can You Guess the Secret Word in This Brain Teaser?**Readers Digest**: If You Can Solve the Secret Word In This Logic Puzzle, You Might Be a Genius**MSN**If You Can Solve the Secret Word In This Logic Puzzle, You Might Be a Genius

I wanted to share my longer explanation to the Secret Word Puzzle. Here’s a video where I explain everything in detail visually.

**Can You Solve The Secret Word Logic Puzzle?**

And here’s a link to my blog post which is an alternate explanation in text:

**Can You Solve The Secret Word Logic Puzzle?**

**The Puzzle**

A teacher writes six words on the board:

cat

dog

has

max

dim

tag

The teacher hands a piece of paper to Albert, another to Bernard, and another to Cheryl. The teacher explains each paper contains a different letter from one of the words written on the board.

The teacher asks if Albert knows the secret word, and [*after a moment] he replies aloud, “Yes.” The teacher then asks Bernard, and after a moment of thinking, he also says, “Yes.” And finally Cheryl is asked and she takes a moment and then confidently replies, “Yes,” she also knows the word.

Albert, Bernard and Cheryl always ace their logic exams. What was the secret word?

*The original puzzle caused some confusion about the “moment of thinking.” I meant to say each person took a moment to think so I added that to Albert’s reply.

Watch the video for a solution.

**Can You Solve The Secret Word Logic Puzzle?**

Or keep read the solution here in my blog post:

If you like this puzzle, I post about similar puzzles every week. Do check out my YouTube channel for many more puzzles! If you “subscribe” to my YouTube channel you will also get notified of new videos (it’s free to subscribe to my channel).

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