Difference between revisions of "2008 Mock ARML 1 Problems/Problem 8"
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Compute <math>ab + ac + ad</math>. | Compute <math>ab + ac + ad</math>. | ||
− | == Solution == | + | == 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>. | ||
− | + | == Solution 2== | |
+ | 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 20:17, 4 December 2016
Contents
Problem
For positive real numbers ,
Compute .
Solution 1
We consider a geometric interpretation, specifically with an equilateral triangle. Let the distances from the vertices to the incenter be ,
, and
, and the tangents to the incircle be
,
, and
. Then use Law of Cosines to express the sides in terms of
,
, and
, and Pythagorean Theorem to express
,
, and
in terms of
,
,
, and the inradius
. 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
. The desired expression is
, which is also the area, so the answer is
.
Solution 2
Since the equations are symmetric in , we may consider
; the system reduces and we find that the desired sum is
.
See also
2008 Mock ARML 1 (Problems, Source) | ||
Preceded by Problem 7 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 |