Difference between revisions of "2008 Mock ARML 1 Problems/Problem 3"
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We use coordinates. Consider the diagram in Solution 1. Let <math>D=(0,0)</math>. By 30-60-90 | We use coordinates. Consider the diagram in Solution 1. Let <math>D=(0,0)</math>. By 30-60-90 | ||
triangle ratios, we can get that <math>C=(\frac12, \frac{\sqrt3}{2})</math>, and <math>H'=(\frac12, 0)</math>. In addition, <math>B=(\frac32, \frac{\sqrt3}{2})</math> and <math>E=(\frac12, -\frac{\sqrt3}{2})</math>. The line <math>\overline{CE}</math> is represented by <math>x=\frac12</math>. The line <math>\overline{BD}</math> is represented by <math>y=\frac{\sqrt3}{3}x</math>. The intersection (<math>H</math>) is then <math>(\frac12, \frac{\sqrt3}{6})</math>. We additionaly can get that <math>G=(\frac32,0)</math>. Then, by | triangle ratios, we can get that <math>C=(\frac12, \frac{\sqrt3}{2})</math>, and <math>H'=(\frac12, 0)</math>. In addition, <math>B=(\frac32, \frac{\sqrt3}{2})</math> and <math>E=(\frac12, -\frac{\sqrt3}{2})</math>. The line <math>\overline{CE}</math> is represented by <math>x=\frac12</math>. The line <math>\overline{BD}</math> is represented by <math>y=\frac{\sqrt3}{3}x</math>. The intersection (<math>H</math>) is then <math>(\frac12, \frac{\sqrt3}{6})</math>. We additionaly can get that <math>G=(\frac32,0)</math>. Then, by | ||
− | the distance formula, the length of <math>GH</math> is <math>\sqrt{1^2+\left(\frac{\sqrt3}{6}\right)^2}=\boxed{\frac{\sqrt{39}}{6 | + | the distance formula, the length of <math>GH</math> is <math>\sqrt{1^2+\left(\frac{\sqrt3}{6}\right)^2}=\boxed{\frac{\sqrt{39}}{6}}</math>. |
== See also == | == See also == |
Revision as of 22:27, 5 December 2020
Contents
Problem
In regular hexagon with side length
,
intersects
at
, and
intersects
at
. Compute the length of
.
Solution 1
![[asy] pointpen = black; pathpen = black + linewidth(0.62); pair v(int n){ return dir(n * 60); } D(MP("A",v(0))--MP("B",v(1),N)--MP("C",v(2),N)--MP("D",v(3),SW)--MP("E",v(4))--MP("F",v(5))--cycle); D(v(0)--v(3));D(v(1)--v(5));D(v(1)--v(3));D(v(2)--v(4)); D(D(MP("G",IP(v(0)--v(3),v(1)--v(5)),NE))--D(MP("H",IP(v(1)--v(3),v(2)--v(4)),NW)),dashed); D(MP("H'",IP(v(2)--v(4),v(0)--v(3)),SW)); [/asy]](http://latex.artofproblemsolving.com/a/2/a/a2adda0608ac45c1f95b0abb8a8656cf1274ebc3.png)
Let be the foot of the perpendicular from
to
. Since
is an inscribed angle with measure
, it follows that
is a
, and
and
. Also,
. Note that
by ratio
. Thus
.
By the Pythagorean Theorem, . Thus
.
Solution 2
We use coordinates. Consider the diagram in Solution 1. Let . By 30-60-90
triangle ratios, we can get that
, and
. In addition,
and
. The line
is represented by
. The line
is represented by
. The intersection (
) is then
. We additionaly can get that
. Then, by
the distance formula, the length ofis
.
See also
2008 Mock ARML 1 (Problems, Source) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 |