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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 18, 2011"

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==Solution==
 
==Solution==
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{{potd_solution}}
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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>.
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<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>.
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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>.
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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 $AD = x$. Since we know that the perimeter of $ABCD$ is $42$, $BC = 42 - 9 - 15 - x = 18 - x$. Construct a perpendicular to $DC$ from point $B$. Call the intersection of $DC$ and this perpendicular point $E$. Since angle $A$ is right, $ABED$ is a rectangle, and thus angle $BEC$ is right, and $AB = DE = 9$. $EC = 15 - 9 = 6$; by the Pythagorean Theorem, $BE = x = 8$, so $AD = 8$ and $BC = 18 - 8 = 10$. Again by the Pythagorean Theorem, $AC = 17$. Since $AB$ and $DC$ are bases, $AB$ is parallel to$DC$, and so angle $BAX$ = angle $DCX$; by vertical angles, angle $AXB$ = angle $DXC$. By AA, triangle $DXC$ ~ triangle $BXA$ in a ratio of $\frac{9}{15}$. Thus, $\frac{AX}{XC} = \frac{9}{15}$, so $XC$ is $\frac{15}{24}$ of $AC$. Note that the distance from $X$ to $DC$ is really the length of the perpendicular to $DC$ drawn from $X$ by definition. Drawing this perpendicular and letting the point of intersection of $DC$ and $X$ be point $F$, we have that triangle $XFC$ is similar to triangle $ADC$ by AA in a ratio of $\frac{15}{24}$. Thus $\frac{XF}{AD} = \frac{15}{24}$, so $\frac{XF}{8} = \frac{15}{24}$. Solving, we get $XF = \boxed{5}$.

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