TWENTY-NINE contains 29 straight lines — if you don’t count the hyphen.
At the 1939 World’s Fair, San Francisco Seals catcher Joe Sprinz tried to catch a baseball dropped from the Goodyear blimp 1,200 feet overhead.
Sprinz knew baseball but he hadn’t studied physics — he lost five teeth and spent three months in the hospital with a fractured jaw.
3685 = (36 + 8) × 5
What do these sentences have in common?
They’re all precisely the same length.
You’re overseeing a murder trial. The defendant will be hanged if his crime is judged to be both willful and premeditated. You poll the jurors:
A majority think it was willful, and a majority think it was premeditated, so you order the death penalty. As he’s dragged off to the gallows, the defendant screams that this is unfair and swears that his ghost will return for revenge.
You think nothing more of this until the evening, when a strange thought occurs to you. If you’d simply asked the jurors, “Should this man receive the death penalty?”, most would have voted no — only one of the three jurors believed that the crime was both willful and premeditated. Was your own reasoning unsound?
And who’s that behind you — ?
2502 = 2 + 502
Zhuangzi and Huizi were strolling along the dam of the Hao Waterfall when Zhuangzi said, ‘See how the minnows come out and dart around where they please! That’s what fish really enjoy!’
Huizi said, ‘You’re not a fish — how do you know what fish enjoy?’
Zhuangzi said, ‘You’re not I, so how do you know I don’t know what fish enjoy?’
Huizi said, ‘I’m not you, so I certainly don’t know what you know. On the other hand, you’re certainly not a fish — so that still proves you don’t know what fish enjoy!’
Zhuangzi said, ‘Let’s go back to your original question, please. You asked me how I know what fish enjoy — so you already knew I knew it when you asked the question. I know it by standing here beside the Hao.’
— Zhuangzi, China, fourth century B.C.
Lewis Carroll offered this proof that all triangles are isosceles:
Let ABC be any triangle. Bisect BC at D, and from D draw DE at right angles to BC. Bisect the angle BAC.
(1) If the bisector does not meet DE, they are parallel. Therefore the bisector is at right angles to BC. Therefore AB = AC, i.e., ABC is isosceles.
(2) If the bisector meets DE, let them meet at F. Join FB, FC, and from F draw FG, FH, at right angles to AC, AB.
Then the triangles AFG, AFH are equal, because they have the side AF in common, and the angles FAG, AGF equal to the angles FAH, AHF. Therefore AH = AG, and FH = FG.
Again, the triangles BDF, CDF are equal, because BD = DC, DF is common, and the angles at D are equal. Therefore FB = FC.
Again, the triangles FHB, FGC are right-angled. Therefore the square on FB = the [sum of the] squares on FH, HB; and the square on FC = the [sum of the] squares on FG, GC. But FB = FC, and FH = FG. Therefore the square on HB = the square on GC. Therefore HB = GC. Also, AH has been proved equal to AG. Therefore AB = AC; i.e., ABC is isosceles.
Therefore the triangle ABC is always isosceles. Q.E.D.