Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 18, 2011"
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==Solution== | ==Solution== | ||
− | {{ | + | Let <math>AD = x</math>. Since we know that the perimeter of <math>ABCD</math> is <math>42</math>, <math>BC = 42 - 9 - 15 - x = 18 - x</math>. Construct a perpendicular to <math>DC</math> from point <math>B</math>. Call the intersection of <math>DC</math> and this perpendicular point <math>E</math>. Since angle <math>A</math> is right, <math>ABED</math> is a rectangle, and thus angle <math>BEC</math> is right, and <math>AB = DE = 9</math>. |
+ | <math>EC = 15 - 9 = 6</math>; by the Pythagorean Theorem, <math>BE = x = 8</math>, so <math>AD = 8</math> and <math>BC = 18 - 8 = 10</math>. | ||
+ | Again by the Pythagorean Theorem, <math>AC = 17</math>. Since <math>AB</math> and <math>DC</math> are bases, <math>AB</math> is parallel to<math>DC</math>, and so angle <math>BAX</math> = angle <math>DCX</math>; by vertical angles, angle <math>AXB</math> = angle <math>DXC</math>. By AA, triangle <math>DXC</math> ~ triangle <math>BXA</math> in a ratio of <math>\frac{9}{15}</math>. | ||
+ | Thus, <math>\frac{AX}{XC} = \frac{9}{15}</math>, so <math>XC</math> is <math>\frac{15}{24}</math> of <math>AC</math>. Note that the distance from <math>X</math> to <math>DC</math> is really the length of the perpendicular to <math>DC</math> drawn from <math>X</math> by definition. Drawing this perpendicular and letting the point of intersection of <math>DC</math> and <math>X</math> be point <math>F</math>, we have that triangle <math>XFC</math> is similar to triangle <math>ADC</math> by AA in a ratio of <math>\frac{15}{24}</math>. Thus <math>\frac{XF}{AD} = \frac{15}{24}</math>, so <math>\frac{XF}{8} = \frac{15}{24}</math>. Solving, we get <math>XF = \boxed{5}</math>. |
Latest revision as of 21:50, 17 June 2011
Problem
AoPSWiki:Problem of the Day/June 18, 2011
Solution
Let . Since we know that the perimeter of
is
,
. Construct a perpendicular to
from point
. Call the intersection of
and this perpendicular point
. Since angle
is right,
is a rectangle, and thus angle
is right, and
.
; by the Pythagorean Theorem,
, so
and
.
Again by the Pythagorean Theorem,
. Since
and
are bases,
is parallel to
, and so angle
= angle
; by vertical angles, angle
= angle
. By AA, triangle
~ triangle
in a ratio of
.
Thus,
, so
is
of
. Note that the distance from
to
is really the length of the perpendicular to
drawn from
by definition. Drawing this perpendicular and letting the point of intersection of
and
be point
, we have that triangle
is similar to triangle
by AA in a ratio of
. Thus
, so
. Solving, we get
.