Difference between revisions of "1995 AIME Problems/Problem 9"
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− | == Solution == | + | == Solution 1 == |
− | Let <math>x=\angle CAM</math>, so <math>3x=\angle CDM</math>. Then, <math>\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11</math>. Expanding <math>\tan 3x</math> using the angle sum identity gives <cmath>\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^ | + | Let <math>x=\angle CAM</math>, so <math>3x=\angle CDM</math>. Then, <math>\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11</math>. Expanding <math>\tan 3x</math> using the angle sum identity gives <cmath>\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.</cmath> |
Thus, <math>\frac{3-\tan^2x}{1-3\tan^2x}=11</math>. Solving, we get <math>\tan x= \frac 12</math>. Hence, <math>CM=\frac{11}2</math> and <math>AC= \frac{11\sqrt{5}}2</math> by the [[Pythagorean Theorem]]. The total perimeter is <math>2(AC + CM) = \sqrt{605}+11</math>. The answer is thus <math>a+b=\boxed{616}</math>. | Thus, <math>\frac{3-\tan^2x}{1-3\tan^2x}=11</math>. Solving, we get <math>\tan x= \frac 12</math>. Hence, <math>CM=\frac{11}2</math> and <math>AC= \frac{11\sqrt{5}}2</math> by the [[Pythagorean Theorem]]. The total perimeter is <math>2(AC + CM) = \sqrt{605}+11</math>. The answer is thus <math>a+b=\boxed{616}</math>. | ||
+ | |||
+ | == 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>. | ||
+ | |||
+ | == 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{BE^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 both 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 in the equations for <math>BA</math> and <math>BM</math> 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>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | The triangle is symmetrical so we can split it in half (<math>\triangle ABM</math> and <math>\triangle ACM</math>). | ||
+ | |||
+ | Let <math>\angle BAM = y</math> and <math>\angle BDM = 3y</math>. By the Law of Sines on triangle <math>BAD</math>, <math>\frac{10}{\sin 2y} = \frac{BD}{\sin y}</math>. Using <math>\sin 2y = 2\sin y\cos y</math> we can get <math>BD = \frac{5}{\cos y}</math>. We can use this information to relate <math>BD</math> to <math>DM</math> by using the Law of Sines on triangle <math>BMD</math>. | ||
+ | |||
+ | <cmath>\frac{\frac{5}{\cos y}}{\sin BMD} = \frac{1}{\sin 90^\circ - 3y}</cmath> | ||
+ | |||
+ | <math>\sin BMD = 1</math> (as <math>\angle BMD</math> is a right angle), so <math>\frac{1}{\sin 90^\circ - 3y} = \frac{5}{\cos y}</math>. Using the identity <math>\sin 90^\circ - x = \cos x</math>, we can turn the equation into:: | ||
+ | |||
+ | <cmath>\frac{1}{\cos 3y} = \frac{5}{\cos y}</cmath> | ||
+ | |||
+ | <cmath>5\cos 3y = \cos y</cmath> | ||
+ | |||
+ | <cmath>5(4\cos ^3 y - 3\cos y) = \cos y</cmath> | ||
+ | |||
+ | <cmath>20\cos ^3 y = 16 \cos y</cmath> | ||
+ | |||
+ | <cmath>5\cos ^3 y = 4\cos y</cmath> | ||
+ | |||
+ | <cmath>5\cos ^2 y = 4</cmath> | ||
+ | |||
+ | <cmath>\cos ^2 y = \frac{4}{5}</cmath> | ||
+ | |||
+ | Now that we've found <math>\cos y</math>, we can look at the side lengths of <math>BM</math> and <math>AB</math> (since they are symmetrical, the perimeter of <math>\triangle ABC</math> is <math>2(BM + AB)</math>. | ||
+ | |||
+ | We note that <math>BM = 11\tan y</math> and <math>AB = 11\sec y</math>. | ||
+ | |||
+ | <cmath>\sin ^2 y = 1 - \cos ^2 y</cmath> | ||
+ | |||
+ | <cmath>\sin ^2 y = \frac{1}{5}</cmath> | ||
+ | |||
+ | <cmath>\tan ^2 y = \frac{1}{4}</cmath> | ||
+ | |||
+ | <cmath>\tan y = \frac{1}{2}</cmath> | ||
+ | |||
+ | (Note it is positive since <math>BM > 0</math>). | ||
+ | |||
+ | <cmath>\sec ^2 y = \frac{5}{4}</cmath> | ||
+ | |||
+ | <cmath>\sec y = \frac{\sqrt{5}}{2}</cmath> | ||
+ | |||
+ | <cmath>BM + AB = 11\frac{\sqrt{5}+1}{2}</cmath> | ||
+ | |||
+ | <cmath>2(BM + AB) = 11(\sqrt{5} + 1)</cmath> | ||
+ | |||
+ | <cmath>2(BM + AB) = 11\sqrt{5} + 11</cmath> | ||
+ | |||
+ | <cmath>2(BM + AB) = \sqrt{605} + 11</cmath> | ||
+ | |||
+ | The answer is <math>\boxed{616}</math>. | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | Suppose <math>\angle BAM=\angle CAM =x</math>, since <math>\angle BDC=3\angle BAC</math>, we have <math>\angle BDM=\angle MDC = 3x</math>. Therefore, <math>\angle DBC=\angle DCB = 90^\circ -3x</math> and <math>\angle ABD=\angle DCA=2x</math>. As a result, <math>\triangle KAC</math> is isosceles, <math>KC=KA</math>. | ||
+ | |||
+ | Let <math>H</math> be a point on the extension of <math>CD</math> through <math>D</math> such that <math>\overline{HB}\perp\overline{BC}</math> and denote the intersection of <math>\overline{HC}</math> and <math>\overline{AB}</math> as <math>K</math>. Then, <math>BH=2DM=2, \overline{HB}\parallel\overline{DM}</math>, and <math>HD=DC</math> by the [[Midpoint]] Theorem. So, <math>\angle HBA=x</math> and <math>\angle CDM=\angle CHB=\angle HDA= 3x</math>. | ||
+ | |||
+ | Consequently, <math>\triangle HBK\sim \triangle DAK</math>, <cmath>\frac{BK}{KA}=\frac{HK}{KD}=\frac{1}{5}</cmath> Assume <math>BK=a</math> and <math>HK=b</math>, then <math>KA=5a</math> and <math>KD = 5b</math>. Since <math>KC=KA, KC=5a</math>, and since <math>HD=DC</math>, <math>KC=11b</math>. Therefore, <math>a=\frac{11}{5}b</math>. | ||
+ | |||
+ | In <math>\triangle BDM</math>, by the [[Pythagorean Theorem]], <math>BM=\sqrt{36b^2-1}</math>. Similarly in <math>\triangle BAM</math>, <math>BM=\sqrt{36a^2-121}</math>. So <cmath>\sqrt{36a^2-121}=\sqrt{36b^2-1}</cmath> Since <math>a=\frac{11}{5}b</math>, we have <math>b=\frac{5\sqrt{5}}{12}</math> and <math>a=\frac{11\sqrt{5}}{12}</math>. Consequently, <math>BM=\frac{11}{2}</math> and <math>AB=\frac{11\sqrt{5}}{2}</math>. Thus, the perimeter of <math>\triangle ABC</math> is <math>11+\sqrt{605}</math>, and the answer is <math>\boxed{616}</math>. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:47, 3 February 2023
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 both denominators and squaring both sides yields
which simplifies to
Substituting this in for x in the equations for
and
yields
and
Thus the perimeter is
, and the answer is
.
Solution 4
The triangle is symmetrical so we can split it in half ( and
).
Let and
. By the Law of Sines on triangle
,
. Using
we can get
. We can use this information to relate
to
by using the Law of Sines on triangle
.
(as
is a right angle), so
. Using the identity
, we can turn the equation into::
Now that we've found , we can look at the side lengths of
and
(since they are symmetrical, the perimeter of
is
.
We note that and
.
(Note it is positive since ).
The answer is .
Solution 5
Suppose , since
, we have
. Therefore,
and
. As a result,
is isosceles,
.
Let be a point on the extension of
through
such that
and denote the intersection of
and
as
. Then,
, and
by the Midpoint Theorem. So,
and
.
Consequently, ,
Assume
and
, then
and
. Since
, and since
,
. Therefore,
.
In , by the Pythagorean Theorem,
. Similarly in
,
. So
Since
, we have
and
. Consequently,
and
. Thus, the perimeter of
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.