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iver bank<br />
Ak*,<br />
A*-,<br />
Ak<br />
Figure 3<br />
case where AD < AC. The same argument<br />
will hold If AD > AC. Of.<br />
course/ when s : 1 or l, A, _, or A, _<br />
, should be replaced byA,.)Triangles<br />
ABC and ABD are congruent to the<br />
faces A, _ rArP and Ar_rAtP of the<br />
pyramid, respectively, so BC : PA,<br />
and BD = PAt.By the Triangle Inequality,<br />
BD - BC < DC : IAD ACI-that is, PA, - PA,. I PAj_ \ -<br />
PAs_Ll 3 PAt - PAr, because PA., is<br />
the longest arrd PA, the shortest of<br />
the edges PAo-and this is a contradiction<br />
that proves that all these<br />
edges are the same length. Now drop<br />
the height PH of the pyramid (fig.<br />
3). By the Pythagorean Theorem,<br />
HAt, : PAo, - PH2, so HAr = HA2<br />
HAn. Therefore, the vertices<br />
Ay, ..., Arall lie on a circle with<br />
center H and so divide it into equal<br />
arcs (subtended by equal chords<br />
A*A* * ,). This means that<br />
AtA2...An is a regular n-gon, and<br />
the base H of the pyramid's height<br />
is its center, so we're done.<br />
Ml20<br />
The answer is a : B. If f(xl : A* +<br />
Bx+ C, thenl'(1) =2A+ B. Now<br />
so we have<br />
flot: c,<br />
f(rl2l: Al4 + Bfz + c,<br />
f(Il=A+B+C,<br />
V'llll: l2A + Bl<br />
= tsflll - +f(tl2l + flolt<br />
< 3ll(1)l + atf$lzlt + l/(0)l<br />
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