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Difference between revisions of "2008 Mock ARML 1 Problems/Problem 8"

(solution by e^(pi*i)=-1)
 
 
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Compute <math>ab + ac + ad</math>.
 
Compute <math>ab + ac + ad</math>.
  
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== Solution ==
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== Solution 1==
 
We consider a geometric interpretation, specifically with an equilateral triangle. Let the distances from the vertices to the incenter be <math>x</math>, <math>y</math>, and <math>z</math>, and the tangents to the incircle be <math>b</math>, <math>c</math>, and <math>d</math>.  Then use Law of Cosines to express the sides in terms of <math>x</math>, <math>y</math>, and <math>z</math>, and Pythagorean Theorem to express <math>x</math>, <math>y</math>, and <math>z</math> in terms of <math>b</math>, <math>c</math>, <math>d</math>, and the inradius <math>a</math>.  This yields the first three equations.  The fourth is the result of the sine area formula for the three small triangles, and gives the area as <math>\frac {\sqrt {3}}{2}</math>. The desired expression is <math>rs</math>, which is also the area, so the answer is <math>\boxed{\frac {\sqrt {3}}{2}}</math>.
 
We consider a geometric interpretation, specifically with an equilateral triangle. Let the distances from the vertices to the incenter be <math>x</math>, <math>y</math>, and <math>z</math>, and the tangents to the incircle be <math>b</math>, <math>c</math>, and <math>d</math>.  Then use Law of Cosines to express the sides in terms of <math>x</math>, <math>y</math>, and <math>z</math>, and Pythagorean Theorem to express <math>x</math>, <math>y</math>, and <math>z</math> in terms of <math>b</math>, <math>c</math>, <math>d</math>, and the inradius <math>a</math>.  This yields the first three equations.  The fourth is the result of the sine area formula for the three small triangles, and gives the area as <math>\frac {\sqrt {3}}{2}</math>. The desired expression is <math>rs</math>, which is also the area, so the answer is <math>\boxed{\frac {\sqrt {3}}{2}}</math>.
  
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Note that since the equations are symmetric in <math>b,c,d</math>, we may consider <math>b=c=d</math>; the system reduces quickly, and we find that the desired sum is <math>\frac{\sqrt{3}}{2}</math>.
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== Solution 2==
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Since the equations are symmetric in <math>b,c,d</math>, we may consider <math>b=c=d</math>; the system reduces and we find that the desired sum is <math>\boxed{\frac {\sqrt {3}}{2}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 19:17, 4 December 2016

Problem

For positive real numbers $a,b,c,d$,

\begin{align*}2a^2 + \sqrt {(a^2 + b^2)(a^2 + c^2)} &= 2bc\\ 2a^2 + \sqrt {(a^2 + c^2)(a^2 + d^2)} &= 2cd\\ 2a^2 + \sqrt {(a^2 + d^2)(a^2 + b^2)} &= 2db\end{align*} \[\sqrt {(a^2 + b^2)(a^2 + c^2)} + \sqrt {(a^2 + c^2)(a^2 + d^2)} + \sqrt {(a^2 + d^2)(a^2 + b^2)} = 2\]

Compute $ab + ac + ad$.

Solution 1

We consider a geometric interpretation, specifically with an equilateral triangle. Let the distances from the vertices to the incenter be $x$, $y$, and $z$, and the tangents to the incircle be $b$, $c$, and $d$. Then use Law of Cosines to express the sides in terms of $x$, $y$, and $z$, and Pythagorean Theorem to express $x$, $y$, and $z$ in terms of $b$, $c$, $d$, and the inradius $a$. This yields the first three equations. The fourth is the result of the sine area formula for the three small triangles, and gives the area as $\frac {\sqrt {3}}{2}$. The desired expression is $rs$, which is also the area, so the answer is $\boxed{\frac {\sqrt {3}}{2}}$.

Solution 2

Since the equations are symmetric in $b,c,d$, we may consider $b=c=d$; the system reduces and we find that the desired sum is $\boxed{\frac {\sqrt {3}}{2}}$.

See also

2008 Mock ARML 1 (Problems, Source)
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