## Under a Microscope

How much will a 12° angle measure when looked at under a microscope that magnifies 8 times?

**SOLUTION**

The angle will measure the same, 12°.

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How much will a 12° angle measure when looked at under a microscope that magnifies 8 times?

The angle will measure the same, 12°.

I have a 9 letter word, 123456789.

If I lose it, I die.

If I have 234, I can 1234.

If I have 56, I am very sick.

235 is the same as 789.

What is the word?

The answer is HEARTBEAT.

The hands of my alarm clock are indistinguishable. How many times throughout the day their positioning is such that one cannot figure out which is the hour hand and which is the minute hand?

*Remark: AM-PM is not important.*

Imagine that you have a third hand which moves 12 times as fast as the minute hand. Then, at any time, if the hour hand moves to the location of the minute hand, the minute hand will move to the location of the imaginary hand. Therefore, our task is to find the number of times during the day when the hour hand and the imaginary hand are on top of each other, and the minute hand is not.

Since the imaginary hand moves 144 times faster than the hour hand, the two hands are on top of each other exactly 143 times between 12AM and 12PM. Out of these 143 times, 11 times all three arrows are on top of each other. Therefore, we have 2 × (143 – 11) = 264 times when we cannot figure out the exact time during the entire 24-hour cycle.

Consider all 1024 vectors in a 10-dimensional space with elements ±1. Show that if you change some of the elements of some of the vectors to 0, you can still choose a few vectors, such that their sum is equal to the 0-vector.

Denote the 1024 vectors with u_{i} and their transformations with f(u_{i}). Create a graph with 1024 nodes, labeled with u_{i}. Then, for every node u_{i}, create a directed edge from u_{i} to u_{i}-2f(u_{i}). This is a valid construction, since the vector u_{i}-2f(u_{i}) has elements -1, 0, and 1 only. In the resulting graph, there is a cycle:

v_{1} ⇾ v_{2} ⇾ … ⇾ v_{k} ⇾ v_{1}.

Now, if we pick the (transformed) vectors from this cycle, their sum is the 0-vector:

f(v_{1}) + f(v_{2}) + … + f(v_{k}) = (v_{2} – v_{1})/2 + (v_{3} – v_{2})/2 + … + (v_{1} – v_{k})/2 = 0.

What do the letter **T** and an **island** have in common?

Both of them are in the middle of **waTer**.

At Creepy Beasts Inc., three of the most dreaded animals, a tiger, a wolf, and a bear, sat in their boardroom in silence while they awaited their boss. Then, Mr. Tiger broke the silence.

“Isn’t it odd that our three surnames are the same as our three species, yet none of our surnames matches our own species?”

The wolf replied, “Yeah, but does anyone care?”

They sat in silence again…

Can you figure out the surname of each animal?

Since the wolf replied to Mr. Tiger, his surname can be neither Tiger nor Wolf. Therefore, the wolf’s surname is Mr. Bear. Subsequently, Mr. Tiger must be a bear, and finally, Mr. Wolf must be a tiger.

The sides of a rectangle have lengths which are odd numbers. The rectangle is split into smaller rectangles with sides which have integer lengths. Show that there is a small rectangle, such that all distances between its sides and the sides of the large rectangle have the same parity, i.e. they are all even or they are all odd.

*Source: Shortlist IMO 2017*

Split the large rectangle into small 1×1 squares and color it in black and white, chessboard-style, such that the four corner squares are black. Since the large rectangle has more black squares than white squares, one of the smaller rectangles also must have more black squares than white squares. Therefore, the four corners of that smaller rectangle are all black. Then, it is easy to see that all distances between its sides and the sides of the large rectangle have the same parity.

Which die completes the sequence?

If you look at dots in the top row, you will see that they are put together into groups of 1, 2, 3, 4, and so on.

You have ten lanterns, five of which are working, and five of which are broken. You are allowed to choose any two lanterns and make a test that tells you whether there is a broken lantern among them or not. How many tests do you need until you find a lantern you know for sure is working?

*Remark: If the test detects that there are broken lanterns, it does not tell you which ones and how many (one or two) they are.*

You need 6 tests:

(1, 2) → (3, 4) → (5, 6) → (7, 8) → (7, 9) → (8, 9)

If at least one of these tests is positive, then you have found two working lanterns.

It all of these tests are negative, then lantern #10 must be working. Indeed, since at least one lantern in each of the pairs (1, 2), (3, 4), (5, 6) is not working. Therefore, there are at least 2 working lanterns among #7, #8, #9, #10. If #10 is not working, then at least one of the pairs (7, 8), (7, 9), or (8, 9) must yield a positive test, which is a contradiction.

The sentence below is grammatically correct. Can you explain it?

**Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo.**

The sentence says that buffalo (animals) from Buffalo (city, US), which are buffaloed (intimidated) by Buffalo (city, US) buffalo (animals), themselves buffalo (intimidate) buffalo (animals) from Buffalo (city, US).

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