Difference between revisions of "2021 AMC 12B Problems/Problem 17"
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Let <math>X</math> and <math>Y</math> be the feet of the perpendiculars from <math>P</math> to <math>AB</math> and <math>CD</math>, respectively. Observe that <math>PX = \tfrac 8r</math> and <math>PY = \tfrac 4s</math>. Now using the formula for the area of a trapezoid yields | Let <math>X</math> and <math>Y</math> be the feet of the perpendiculars from <math>P</math> to <math>AB</math> and <math>CD</math>, respectively. Observe that <math>PX = \tfrac 8r</math> and <math>PY = \tfrac 4s</math>. Now using the formula for the area of a trapezoid yields | ||
− | + | <cmath>14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr.</cmath> | |
− | 14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr. | + | Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>2\rho + \rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{2+\sqrt 2\textbf{ (B)}}</math>. - djmathman |
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Revision as of 15:56, 11 February 2021
Problem 17
Let be an isoceles trapezoid having parallel bases
and
with
Line segments from a point inside
to the vertices divide the trapezoid into four triangles whose areas are
and
starting with the triangle with base
and moving clockwise as shown in the diagram below. What is the ratio
Solution
Without loss let have vertices
,
,
, and
, with
and
. Also denote by
the point in the interior of
.
Let and
be the feet of the perpendiculars from
to
and
, respectively. Observe that
and
. Now using the formula for the area of a trapezoid yields
Thus, the ratio
satisfies
; solving yields
. - djmathman