... bare the fangs like a dog
laglerolarpok (Inuit) the gnashing of teeth
kashr (Persian) displaying the teeth in laughter
zhaghzhagh (Persian) the chattering of the teeth fromthe cold
or from ... rights reserved
THE LIBRARY OF CONGRESS HAS CATALOGED THE HARDCOVER EDITION AS FOLLOWS:
Jacot de Boinod, Adam.
The meaning of tingo and other extraordinary words from
around theworld / Adam ... hands pressed together in salutation
Legging it
U
ndue attention is put on their shapeliness but the bottom line is
it’s good to have two of them and they should, ideally, be the same
length:...
... APQ.
15. There are cards labeled from 1 to 2000. The cards are shuffled and placed in a pile. The
top card is placed on the table, then the next card at the bottom of the pile. Then the next ... ☺ The best problems fromaroundtheworld Cao Minh Quan
g
47
15. Two circles touch the x-axis and the line y = mx (m > 0). They meet at (9,6) and another
point and the product of their ... placed on the table to the right of the first card, and the next card is placed at the bottom of
the pile. This process is continued until all the cards are on the table. The final order (from left...
... and centers aroundthe world. I am deeply indebted to the many and diverse contributions
that each and every one of them made.
The project would not have been possible without the tremendous ... like
the U.S.; (2) The OECD data represent what companies tell their regulators, and what these regulators in
turn tell the OECD; the concern is that companies sometimes misreport to their ...
Looking purely at the 3G levels of subscription as reported by the OECD, the United States would not
rank in the top 20, and this is so also the case in that report for otherwise high performing...
... Stephenson, was in the habit of seeing, from his oce
window, English ships daily passing to and fro on thegreat
canal, by which the old roundabout route from England to
India by the Cape of Good ... how, from
the description you have, you will be able to recognise your
man, even if he is on board the Mongolia.’
‘A man rather feels the presence of these fellows, consul,
than recognises them. ... Englishmen on their travels, and the
hospitable eorts of the purser, the time passed quickly on
the Mongolia. e best of fare was spread upon the cabin
tables at breakfast, lunch, dinner, and the eight...
... Report
Conficker Worm Infects Millions AroundtheWorld
April 14, 2009
Introduction
The major news of the first quarter
was the rapid propagation of the
Conficker worm. Research indicates ... like the
one pictured here. In the case where an account was compromised and used to
perpetuate the scheme, the real Twitter “proactively reset the passwords of the
accounts” and offered the ... In the example here, the green “Order Now”
button and the “Find Exclusive Deals Online!” tab are both images hosted fromthe
Pizza Hut site.
In this case, the spam provider also masked the...
... with UNAIDS, PEPFAR, the
countries suering the greatest burden
and other partners, virtually eliminate
the transmission of HIV from mother-to-
child by 2015 worldwide. The Business
Leadership ...
diarrheal deaths in the two regions of the
world where they are most concentrated
— the northern states of India and Nigeria.
In Phase 2 this approach will be extended
to the other high child ... management in both
the public and private sectors.
NEAR-ZERO TRANSMISSION OF HIV
FROM MOTHER-TO-CHILD: Virtually
eliminate the transmission of HIV from
mother-to-child by the end of 2015
In...
... whose father has died but
whose mother is alive; 2) a maternal orphan is a child whose mother has died but whose father is
alive; and 3) a double orphan is a child whose mother and father have ... During the first year of the program, 2,743 parents and sponsors signed up. As of
December 2009, there were 6,929 Fundisa account owners. At the end of October 2009, the
total volume of the program ... one
parent. Eighty-three percent of the girls did not own any assets—neither on their own nor
together with someone else. The three most common reasons they reported for saving were
for personal...
... P in the interior of the triangle P QR, there exists
a tetrahedron ABCD such that P is the point of the face ABC at
the greatest distance (measured along the surface of the tetrahedron)
from ... |a
i
n
|. Note that either all of the terms are
positive, or they alternate in sign; in the latter case, the terms of
either sign form a geometric progression by themselves.
There cannot be three ... 1)
and the edges are parallel to the coordinate axes. If r is the radius
of the sphere, then (r, r, r) is its center, and (r, 1, 1) is the point of
tangency of one of the edges at B. Therefore...
... the parallelogram. It intersects the line AC in two
points; let the point farther from C be P
2
. then the locus is the
intersection of the segments AP
1
and CP
2
.
We begin by proving that the ... prefer B to C, but only the first prefers A
to C.
2. Let a, b, c be the sides, m
a
, m
b
, m
c
the lengths of the altitudes, and
d
a
, d
b
, d
c
the distances fromthe vertices to the orthocenter in ... > 6
the center O of the n-gon is at a distance greater than 1 from each
vertex, so if the P
i
are sufficiently close to the A
i
then we will also
have OP
i
> 1 for each i. Thus we can add the...
... Ramsey’s Theorem, either three of them
(call them X, Y, Z) are all adjacent or all not adjacent. But then in
the group {V, X, Y, Z}, none of the players draws exactly once with
the other players, a ... contradiction.
On the other hand, 2
n
2
−1 days suffice. Suppose that n = 2t + 1
or 2t + 2; number the teams from 1 to n and the Sundays from 1 to
2t + 1. On the i-th Sunday, let team i either sit out ... whenever there is a 0 in the binary
representation of n − k, there is a 1 in the corresponding digit of k.
Then the corresponding
(n−k)
i
k
i
equals 0, and by Lucas’s Theorem
n−k
k
is even.
Therefore,...
... given
operations preserve the product of the values of the balls in the box.
This product is initially i
2000
= 1. If three balls were left in the box,
none of them green, then the product of their values ... position and then arranging the other n − m − 1
big numbe rs into the rest of chosen places. We then arrange all the
small numbers in the remaining m − 1 places. Hence, there are
x
m
=
n − 1
n ... another eight are of the same type. Consider
the set A of these 24 answer sheets. Given any two answer sheets in
A, two of them must be of the same type, that is, their solutions for
the last 4 problems...