The Bank Robber

Harry Ape robbed a bank yesterday and has been hiding in the bitterly cold forest. Max Mouse found Harry's clothes near a hole in the pond ice. Why does Slylock Fox believe that Harry is nearby?

Slylock Fox Puzzle

Slylock thinks Harry fell thought the ice, and after clibing out of the water, took off his wet clothes due to the frigid temperatures. The seuth suspects the ape is nearby because the wet clothes are still unfrozen.

Chess at the Olympics

For the 2020 Olympic Games, the World Chess Federation asked the Olympic committee to include chess as official sport. The committee agreed under one condition:

The Chess Federation must design a chess-colored black and white logo, consisting of three intersecting Olympic rings with areas 1 each, such that the black area - covered by odd number of rings, is less than 1. It is not necessary that each of the rings intersects all of the others.

Below you can see an example of one chess-colored black and white logo.

Is chess going to be part of the 2020 Olympic Games?


It is easy to check that if two of the circles do not intersect each other, the total black area is at least 1. Therefore now we will consider the case when each of the circles intersects both of the others.

Denote the intersection points of the circles with A, B, C, D, E, F, as shown on the diagram above. It is straightforward to check that the following 4 conditions are equivalent:

1.S(AEC)+S(BFA)+S(CDB)+S(DEF)≥1

2.S(AEC)≥S(DFB)

3.S(BFA)≥S(DCE)

4.S(CDB)≥S(AFE)

First, notice that DC+EA+BF=BD+CE+AF, regarded as arcs (we stick to this notation throughout the solution). Simple combinatoric argument shows that there are two consecutive arcs among these, touching at D, E, or F, such that each of them is at least as large as its opposite. Let without of generality DC≥AF and BD≥EA.

Now draw the line AD. Let it intersects the arcs EF and BC at points K and L.

Notice that:

•LC≥CD

•LB≥BD

•AE≥EK

•AF≥KF

•∠LCD=∠AFK

•∠DBL=∠KEA

All of these relations follow from simple symmetries and angles in the circles. For example, ∠LCD=∠DFB=∠KFA, LC=∠CKD≥CD, and AF=∠FDA≥KF.Combining the relations, we get:

1.LC≥AF, DC≥KF, ∠LCD=∠KFA

2.LB≥AE, DB≥KE, ∠LBD=∠KEA

Therefore the light green piece AFK can fully fit inside the dark green piece CDL, and the light blue piece KEA can fully fit inside the dark blue piece DBL. This implies that S(CDB)≥S(AFE), which is exactly point 4. from the conditions above.

Every acute triangle is isosceles

Consider an arbitrary acute triangle ABC. Let E be the intersection of the bisector at vertex C and the bisection of the side AB. Let F and G be the projections of E on AC and BC respectively.

Since E belongs to the bisection of AB, we must have AE = BE. Also, since E belongs to the bisector of C, we must have EF = EG. However, this would imply that triangles AEF and BGF are identical, and then AF = BF. We also have that CF = CG, which implies that AC = BC. The arbitrary chosen triangle ABC is isosceles!

Can you find where the logic fails?

isoceles triangle puzzle

The bisector of C and the bisection of AB always intersect outside the triangle, on the circumcircle. One of the points F/G always lies on the segment AC/BC, and the other one does not.

Abbreviations

Can you figure out what the following abbreviations stand for?
 

24 H in a D = 24 Hours in a Day

26 L of the A = ???

7 D of the W = ???

7 W of the W = ???

12 S of the Z = ???

66 B of the B = ???

52 C in a P (W J) = ???

13 S in the U S F = ???

18 H on a G C = ???

39 B of the O T = ???

5 T on a F = ???

90 D in a R A = ???

3 B M (S H T R) = ???

32 is the T in D F at which W F = ???

15 P in a R T = ???

3 W on a T = ???

100 C in a R = ???

11 P in a F (S) T = ???

12 M in a Y = ???

13 is U F S = ???

8 T on a O = ???

29 D in F in a L Y = ???

27 B in the N T = ???

365 D in a Y = ???

13 L in a B D = ???

52 W in a Y = ???

9 L of a C = ???

60 M in a H = ???

23 P of C in the H B = ???

64 S on a C B = ???

9 P in S A = ???

6 B to an O in C = ???

1000 Y in a M = ???

15 M on a D M C = ???


24 H in a D = 24 Hours in a Day

26 L of the A = 26 Letters of the Alphabet

7 D of the W = 7 Days of the Week

7 W of the W = 7 Wonders of the World

12 S of the Z = 12 Signs of the Zodiac

66 B of the B = 66 Books of the Bible

52 C in a P (W J) = 52 Cards in a Pack (Without Jokers)

13 S in the U S F = 13 Stripes in the United States Flag

18 H on a G C = 18 Holes on a Golf Course

39 B of the O T = 39 Books of the Old Testament

5 T on a F = 5 Toes on a Foot

90 D in a R A = 90 Degrees in a Right Angle

3 B M (S H T R) = 3 Blind Mice (See How They Run)

32 is the T in D F at which W F = 32 Degrees is the Temperature in Fahrenheit at which Water Freezes

15 P in a R T = 15 Players in a Rugby Team

3 W on a T = 3 Wheels on a Tricycle

100 C in a R = 100 Cents in a Rand

11 P in a F (S) T = 11 Players in a Football (Soccer) Team

12 M in a Y = 12 Months in a Year

13 is U F S = 13 is Unlucky For Some

8 T on an O = 8 Tentacles on an Octopus

29 D in F in a L Y = 29 Days in February in a Leap Year

27 B in the N T = 27 Books in the New Testament

365 D in a Y = 365 Days in a Year

13 L in a B D = 13 Loaves in a Baker's Dozen

52 W in a Y = 52 Weeks in a Year

9 L of a C = 9 Lives of a Cat

60 M in an H = 60 Minutes in an Hour

23 P of C in the H B = 23 Pairs of Chromosomes in the Human Body

64 S on a C B = 64 Squares on a Chess Board

9 P in S A = 9 Provinces in South Africa

6 B to an O in C = 6 Balls to an Over in Cricket

1000 Y in a M = 1000 Years in a Millennium

15 M on a D M C = 15 Men on a Dead Man's Chest

Donald "Rusty" Rust

Donald "Rusty" Rust is an American artist, who has created numerous paintings in variety of styles. Even though he is famous mostly for his pin-up artwork, his optical illusions are some of our all-time favorites. In this special interview, he shares with us his thoughts on painting, op art, and everything which drives him forward in his long and successful career. If you like Rusty's artwork, you can order some of it through his official store on Etsy.