Difference between revisions of "1995 AIME Problems/Problem 9"
(Added third solution. Methods found in official AIME 1995 solutions packet.) |
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== Solution 2 == | == Solution 2 == | ||
In a similar fashion, we encode the angles as complex numbers, so if <math>BM=x</math>, then <math>\angle BAD=\text{Arg}(11+xi)</math> and <math>\angle BDM=\text{Arg}(1+xi)</math>. So we need only find <math>x</math> such that <math>\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)</math>. This will happen when <math>\frac{363x-x^3}{1331-33x^2}=x</math>, which simplifies to <math>121x-4x^3=0</math>. Therefore, <math>x=\frac{11}{2}</math>. By the Pythagorean Theorem, <math>AB=\frac{11\sqrt{5}}{2}</math>, so the perimeter is <math>11+11\sqrt{5}=11+\sqrt{605}</math>, giving us our answer, <math>\boxed{616}</math>. | In a similar fashion, we encode the angles as complex numbers, so if <math>BM=x</math>, then <math>\angle BAD=\text{Arg}(11+xi)</math> and <math>\angle BDM=\text{Arg}(1+xi)</math>. So we need only find <math>x</math> such that <math>\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)</math>. This will happen when <math>\frac{363x-x^3}{1331-33x^2}=x</math>, which simplifies to <math>121x-4x^3=0</math>. Therefore, <math>x=\frac{11}{2}</math>. By the Pythagorean Theorem, <math>AB=\frac{11\sqrt{5}}{2}</math>, so the perimeter is <math>11+11\sqrt{5}=11+\sqrt{605}</math>, giving us our answer, <math>\boxed{616}</math>. | ||
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+ | == Solution 3 == | ||
+ | Let <math>\angle BAD=\alpha</math>, so <math>\angle BDM=3\alpha</math>, <math>\angle BDA=180-3\alpha</math>, and thus <math>\angle ABD=2\alpha.</math> We can then draw the angle bisector of <math>\angle ABD</math>, and let it intersect <math>\overline{AM}</math> at <math>E.</math> Since <math>\angle BAE=\angle ABE</math>, <math>AE=BE.</math> Let <math>AE=x</math>. Then we see by the Pythagorean Theorem, <math>BM=\sqrt{BM^2-ME^2}=\sqrt{x^2-(11-x)^2}=\sqrt{22x-121}</math>, <math>BD=\sqrt{BM^2+1}=\sqrt{22x-120}</math>, <math>BA=\sqrt{BM^2+121}=\sqrt{22x}</math>, and <math>DE=10-x.</math> By the angle bisector theorem, <math>BA/BD=EA/ED.</math> Substituting in what we know for the lengths of those segments, we see that <cmath>\frac{\sqrt{22x}}{\sqrt{22x-120}}=\frac{x}{10-x}.</cmath> multiplying by moth denominators and squaring both sides yields <cmath>22x(10-x)^2=x^2(22x-120)</cmath> which simplifies to <math>x=\frac{55}{8}.</math> Substituting this in for x yields <math>BA=\frac{\sqrt{605}}{2}</math> and <math>BM=\frac{11}{2}.</math> Thus the perimeter is <math>11+\sqrt{605}</math>, and the answer is <math>\boxed{616}</math>. | ||
== See also == | == See also == |
Revision as of 00:45, 28 November 2015
Problem
Triangle is isosceles, with
and altitude
Suppose that there is a point
on
with
and
Then the perimeter of
may be written in the form
where
and
are integers. Find
Solution 1
Let , so
. Then,
. Expanding
using the angle sum identity gives
Thus,
. Solving, we get
. Hence,
and
by the Pythagorean Theorem. The total perimeter is
. The answer is thus
.
Solution 2
In a similar fashion, we encode the angles as complex numbers, so if , then
and
. So we need only find
such that
. This will happen when
, which simplifies to
. Therefore,
. By the Pythagorean Theorem,
, so the perimeter is
, giving us our answer,
.
Solution 3
Let , so
,
, and thus
We can then draw the angle bisector of
, and let it intersect
at
Since
,
Let
. Then we see by the Pythagorean Theorem,
,
,
, and
By the angle bisector theorem,
Substituting in what we know for the lengths of those segments, we see that
multiplying by moth denominators and squaring both sides yields
which simplifies to
Substituting this in for x yields
and
Thus the perimeter is
, and the answer is
.
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.