Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 15"
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[[Triangle]] <math>ABC</math> has sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> of [[length]] 43, 13, and 48, respectively. Let <math>\omega</math> be the [[circle]] [[circumscribe]]d around <math>\triangle ABC</math> and let <math>D</math> be the [[intersection]] of <math>\omega</math> and the [[perpendicular bisector]] of <math>\overline{AC}</math> that is not on the same side of <math>\overline{AC}</math> as <math>B</math>. The length of <math>\overline{AD}</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are [[positive integer]]s and <math>n</math> is not [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>. | [[Triangle]] <math>ABC</math> has sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> of [[length]] 43, 13, and 48, respectively. Let <math>\omega</math> be the [[circle]] [[circumscribe]]d around <math>\triangle ABC</math> and let <math>D</math> be the [[intersection]] of <math>\omega</math> and the [[perpendicular bisector]] of <math>\overline{AC}</math> that is not on the same side of <math>\overline{AC}</math> as <math>B</math>. The length of <math>\overline{AD}</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are [[positive integer]]s and <math>n</math> is not [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>. | ||
==Solution== | ==Solution== | ||
− | The perpendicular bisector of any [[chord]] of any circle passes through the [[center]] of that circle. Let <math>M</math> be the midpoint of <math>\overline{AC}</math>, and <math>R</math> be the radius of <math>\omega</math>. By the [[Power of a Point Theorem]], <math>MD \cdot (2R - MD) = AM \cdot MC = 24^2</math> or <math>0 = MD^2 -2R\cdot MD 24^2</math>. By the [[Pythagorean Theorem]], <math>AD^2 = MD^2 + AM^2 = MD^2 + 24^2</math>. | + | The perpendicular bisector of any [[chord]] of any circle passes through the [[center]] of that circle. Let <math>M</math> be the midpoint of <math>\overline{AC}</math>, and <math>R</math> be the length of the [[radius]] of <math>\omega</math>. By the [[Power of a Point Theorem]], <math>MD \cdot (2R - MD) = AM \cdot MC = 24^2</math> or <math>0 = MD^2 -2R\cdot MD 24^2</math>. By the [[Pythagorean Theorem]], <math>AD^2 = MD^2 + AM^2 = MD^2 + 24^2</math>. |
Let's compute the [[circumradius]] <math>R</math>: By the [[Law of Cosines]], <math>\cos B = \frac{AB^2 + BC^2 - CA^2}{2\cdot AB\cdot BC} = \frac{43^2 + 13^2 - 48^2}{2\cdot43\cdot13} = -\frac{11}{43}</math>. By the [[Law of Sines]], <math>2R = \frac{AC}{\sin B} = \frac{48}{\sqrt{1 - \left(-\frac{11}{43}\right)^2}} = \frac{86}{\sqrt 3}</math> so <math>R = \frac{43}{\sqrt 3}</math>. | Let's compute the [[circumradius]] <math>R</math>: By the [[Law of Cosines]], <math>\cos B = \frac{AB^2 + BC^2 - CA^2}{2\cdot AB\cdot BC} = \frac{43^2 + 13^2 - 48^2}{2\cdot43\cdot13} = -\frac{11}{43}</math>. By the [[Law of Sines]], <math>2R = \frac{AC}{\sin B} = \frac{48}{\sqrt{1 - \left(-\frac{11}{43}\right)^2}} = \frac{86}{\sqrt 3}</math> so <math>R = \frac{43}{\sqrt 3}</math>. | ||
− | Now we can use this to compute <math>MD</math> and thus <math>AD</math>. By the [[quadratic formula]], <math>MD = \frac{2R + \sqrt{4R^2 - 4*24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}</math>. (We only take the positive sign because [[angle]] <math>B</math> is [[obtuse angle | obtuse]] so <math>\overline{MD}</math> is the longer of the two [[line segment | segments]] into which the chord <math>\overline{AC}</math> divides the diameter.) Then <math>AD^2 = MD^2 + 24^2 = 1548</math> so <math>AD = 6\sqrt{43}</math>, and <math>12 < 6 + \sqrt{43} < 13</math> so the answer is <math>012</math>. | + | Now we can use this to compute <math>MD</math> and thus <math>AD</math>. By the [[quadratic formula]], <math>MD = \frac{2R + \sqrt{4R^2 - 4*24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}</math>. (We only take the positive sign because [[angle]] <math>B</math> is [[obtuse angle | obtuse]] so <math>\overline{MD}</math> is the longer of the two [[line segment | segments]] into which the chord <math>\overline{AC}</math> divides the [[diameter]].) Then <math>AD^2 = MD^2 + 24^2 = 1548</math> so <math>AD = 6\sqrt{43}</math>, and <math>12 < 6 + \sqrt{43} < 13</math> so the answer is <math>012</math>. |
Revision as of 16:33, 12 February 2007
Problem
Triangle has sides
,
, and
of length 43, 13, and 48, respectively. Let
be the circle circumscribed around
and let
be the intersection of
and the perpendicular bisector of
that is not on the same side of
as
. The length of
can be expressed as
, where
and
are positive integers and
is not divisible by the square of any prime. Find the greatest integer less than or equal to
.
Solution
The perpendicular bisector of any chord of any circle passes through the center of that circle. Let be the midpoint of
, and
be the length of the radius of
. By the Power of a Point Theorem,
or
. By the Pythagorean Theorem,
.
Let's compute the circumradius : By the Law of Cosines,
. By the Law of Sines,
so
.
Now we can use this to compute and thus
. By the quadratic formula,
. (We only take the positive sign because angle
is obtuse so
is the longer of the two segments into which the chord
divides the diameter.) Then
so
, and
so the answer is
.