Difference between revisions of "2001 AMC 12 Problems/Problem 24"
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== Problem == | == Problem == | ||
− | In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that <math>2\cdot BD=CD</math> and <math>\angle DAB=15^\circ</math>. Find <math>\angle ACB</math> | + | In <math>\triangle ABC</math>, <math>\angle ABC=45^\circ</math>. Point <math>D</math> is on <math>\overline{BC}</math> so that <math>2\cdot BD=CD</math> and <math>\angle DAB=15^\circ</math>. Find <math>\angle ACB.</math> |
<math> | <math> | ||
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</math> | </math> | ||
− | == Solution == | + | == Solution 2 == |
− | < | + | Draw a good diagram! Now, let's call <math>BD=t</math>, so <math>DC=2t</math>. Given the rather nice angles of <math>\angle ABD = 45^\circ</math> and <math>\angle ADC = 60^\circ</math> as you can see, let's do trig. Drop an altitude from <math>A</math> to <math>BC</math>; call this point <math>H</math>. We realize that there is no specific factor of <math>t</math> we can call this just yet, so let <math>AH=kt</math>. Notice that in <math>\triangle{ABH}</math> we get <math>BH=kt</math>. Using the 60-degree angle in <math>\triangle{ADH}</math>, we obtain <math>DH=\frac{\sqrt{3}}{3}kt</math>. The comparable ratio is that <math>BH-DH=t</math>. If we involve our <math>k</math>, we get: |
− | + | ||
− | + | <math>kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t</math>. Eliminating <math>t</math> and removing radicals from the denominator, we get <math>k=\frac{3+\sqrt{3}}{2}</math>. From there, one can easily obtain <math>HC=3t-kt=\frac{3-\sqrt{3}}{2}t</math>. Now we finally have a desired ratio. Since <math>\tan\angle ACH = 2+\sqrt{3}</math> upon calculation, we know that <math>\angle ACH</math> can be simplified. Indeed, if you know that <math>\tan(75)=2+\sqrt{3}</math> or even take a minute or two to work out the sine and cosine using <math>\sin(x)^2+\cos(x)^2=1</math>, and perhaps the half- or double-angle formulas, you get <math>\boxed{75^\circ}</math>. | |
− | + | ||
− | + | == Solution 3== | |
− | \ | + | Without loss of generality, we can assume that <math>BD = 1</math> and <math>CD = 2</math>. As above, we are able to find that <math>\angle ADC = 60^\circ</math> and <math>\angle ADB = 120^\circ</math>. |
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− | </ | ||
− | + | Using Law of Sines on triangle <math>ADB</math>, we find that <cmath>\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}.</cmath> Since we know that <cmath>\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4},</cmath> <cmath>\sin 45^\circ = \frac{\sqrt{2}}{2},</cmath> <cmath>\sin 120^\circ = \frac{\sqrt{3}}{2},</cmath> we can compute <math>AD</math> to equal <math>1+\sqrt{3}</math> and <math>AB</math> to be <math>\frac{3\sqrt{2}+\sqrt{6}}{2}</math>. | |
− | + | Next, we apply Law of Cosines to triangle <math>ADC</math> to see that <cmath>AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ).</cmath> Simplifying the right side, we get <math>AC^2 = 6</math>, so <math>AC = \sqrt{6}</math>. | |
− | + | Now, we apply Law of Sines to triangle <math>ABC</math> to see that <cmath>\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}.</cmath> After rearranging and noting that <math>\sin 45^\circ = \frac{\sqrt{2}}{2}</math>, we get <cmath>\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}.</cmath> | |
− | + | Dividing the right side by <math>\sqrt{3}</math>, we see that <cmath>\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4},</cmath> so <math>\angle ACB</math> is either <math>75^\circ</math> or <math>105^\circ</math>. Since <math>105^\circ</math> is not a choice, we know <math>\angle ACB = \boxed{75^\circ}</math>. | |
− | + | Note that we can also confirm that <math>\angle ACB \neq 105^\circ</math> by computing <math>\angle CAB</math> with Law of Sines. | |
− | + | ==Solution 4(FAST)== | |
+ | Note that <math>\angle{ADB} = 120</math> and <math>\angle{ADC} = 60</math>. Seeing these angles makes us think of 30-60-90 triangles. Let <math>E</math> be the foot of the altitude from <math>A</math> to <math>BC</math>. This means <math>\angle{DAE} = 30</math> and <math>\angle{BAE} = 45</math>. Let <math>BD = x</math> and <math>DE = y</math>. This means <math>AE = y\sqrt{3}</math> and since <math>AE = BE</math> we know that <math>x + y = y\sqrt{3}</math>. This means <math>y = \frac{x(\sqrt{3} + 1)}{2}</math>. This gives <math>CE = \frac{4x - x(\sqrt{3} + 1)}{2}</math>. Note that <math>\tan{\angle{ACE}} = \frac{AE}{CE} = 2 + \sqrt{3}</math>. Looking that the answer options we see that <math>\tan{75^\circ} = 2 + \sqrt{3}</math>. This means the answer is <math>D</math>. | ||
+ | ~coolmath_2018 | ||
− | + | == Solution 5 (Law of Sines) == | |
− | == | + | <math>\angle ADB = 120^\circ</math>, <math>\angle ADC = 60^\circ</math>, <math>\angle DAB = 15^\circ</math>, let <math>\angle ACB = \theta</math>, <math>\angle DAC = 120^\circ - \theta</math> |
− | + | By the [[Law of Sines]], we have <math>\frac{CD}{\sin(120^\circ - \theta)} = \frac{AD}{\sin \theta}</math> | |
+ | |||
+ | <math>\space</math> <math>\frac{BD}{\sin15^\circ} = \frac{AD}{\sin45^\circ}</math> | ||
+ | |||
+ | <math>\frac{BD}{CD} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}</math> | ||
+ | |||
+ | <math>\frac{1}{2} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}</math> | ||
+ | |||
+ | By the [https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities#Triple-angle_identities Triple-angle Identities], <math>\sin 45^\circ = 3\sin15^\circ -4\sin^3 15^\circ</math> | ||
+ | |||
+ | <math>\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3\ -4\sin^2 15^\circ}</math> | ||
+ | |||
+ | <math>\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}</math> | ||
+ | |||
+ | <math>\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3 - 4 \cdot \frac{2 - \sqrt{3}}{4}} = \frac{\sin \theta}{1+\sqrt{3}}</math> | ||
+ | |||
+ | <math>\frac{\sin \theta}{\sin(120^\circ - \theta)} = \frac{1+\sqrt{3}}{2}</math> | ||
+ | |||
+ | <math>\sin(120^\circ - \theta) = \sin120^\circ \cos\theta - \sin\theta \cos120^\circ = \frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta</math>,so | ||
+ | |||
+ | <math>\frac{\sin \theta}{\frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta} = \frac{1+\sqrt{3}}{2}</math> | ||
+ | |||
+ | <math>\frac{\sqrt{3} \cdot \cos\theta + \sin\theta}{2 \sin \theta} = \frac{2}{1+\sqrt{3}}</math> | ||
+ | |||
+ | <math>\frac{\sqrt{3}}{2} \cdot \cot \theta = \frac{2}{1+\sqrt{3}} - \frac{1}{2} = \frac{3 - \sqrt{3}}{2 + 2 \sqrt{3}}</math> | ||
+ | |||
+ | <math>\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}</math> | ||
+ | |||
+ | Suppose <math>\cos \theta = k(\sqrt{3} - 1)</math>, and <math>\sin \theta = k(\sqrt{3} + 1)</math> | ||
+ | |||
+ | <math>\sin^2 \theta + \cos^2 \theta = 1</math>, <math>k^2(\sqrt{3} + 1)^2 + k^2(\sqrt{3} - 1)^2 = 8k^2 = 1</math>, <math>k = \frac{1}{2 \sqrt{2}}</math> | ||
+ | |||
+ | <math>\sin \theta = \frac{\sqrt{3} + 1}{2 \sqrt{2}}</math>, <math>\cos \theta = \frac{\sqrt{3} - 1}{2 \sqrt{2}}</math> | ||
+ | |||
+ | <math>\sin 2 \theta = 2 \cdot \frac{\sqrt{3} + 1}{2 \sqrt{2}} \cdot \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{1}{2}</math> | ||
+ | |||
+ | Two possible values of <math>2 \theta</math> are <math>150^\circ</math> and <math>30^\circ</math>. However we can rule out <math>30^\circ</math> because <math>\cos 15^\circ</math> is positive, while <math>\cos \theta</math> is negative. | ||
+ | |||
+ | Therefore <math>2 \theta = 150^\circ</math>, <math>\angle ACB = \boxed{\textbf{(D) } 75^\circ }</math> | ||
− | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | |
− | == | + | ==Solution 6== |
− | + | For starters, we have <math>\angle ABD=120^\circ.</math> Dropping perpendiculars <math>\overline{DX}</math> and <math>\overline{CY}</math> from <math>D</math> and <math>C</math> to <math>\overline{AB}</math> gives <math>\angle ADX=120^\circ-45^\circ=75^\circ,</math> since <math>\angle BDX=45^\circ.</math> | |
− | + | Without loss of generality, let <math>BD=1</math> and <math>CD=2.</math> This tells us that <math>BX=DX=\sqrt{2}/2.</math> Using trigonometric identities, we find that | |
− | + | <cmath>\tan \angle ADX=\tan 75^\circ=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2}.</cmath> | |
+ | Thus, <math>AX/DX=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2},</math> which gives <math>AX=\dfrac{2\sqrt 2+\sqrt 6}{2}.</math> Thus, <math>AB=AX+BX=\dfrac{3\sqrt 2+\sqrt 6}{2}.</math> | ||
− | Now, | + | Now, note that <math>BCY</math> is a <math>45-45-90</math> triangle, so <math>CY=BY=\dfrac{3\sqrt 2}{2}.</math> Thus, we have |
− | + | <cmath>[ABC]=\dfrac{1}{2}\cdot AB\cdot CY=\dfrac{1}{2}\cdot \dfrac{3\sqrt 2+\sqrt 6}{2}\cdot \dfrac{3\sqrt 2}{2}=\dfrac{9+3\sqrt 3}{4}.</cmath> | |
+ | Additionally, note that <math>AY=AB-BY=\sqrt{6}/2.</math> Applying the Pythagorean Theorem to triangle <math>AYC</math> then tells us that <math>AC=\sqrt{6}.</math> By the trigonometric formula for area, | ||
− | + | <cmath>[ABC]=\dfrac{1}{2}\cdot BC\cdot AC\cdot \sin \angle ACB=\dfrac{3\sqrt 6}{2}\sin \angle ACB.</cmath> | |
+ | Setting this equal to our other area and solving gives <math>\sin \angle ACB=\dfrac{\sqrt 6+\sqrt 2}{4},</math> so <math>\angle ACB=\boxed{75^\circ}.</math> ~vaporwave | ||
== See Also == | == See Also == |
Latest revision as of 22:48, 7 March 2024
Contents
Problem
In ,
. Point
is on
so that
and
. Find
Solution 2
Draw a good diagram! Now, let's call , so
. Given the rather nice angles of
and
as you can see, let's do trig. Drop an altitude from
to
; call this point
. We realize that there is no specific factor of
we can call this just yet, so let
. Notice that in
we get
. Using the 60-degree angle in
, we obtain
. The comparable ratio is that
. If we involve our
, we get:
. Eliminating
and removing radicals from the denominator, we get
. From there, one can easily obtain
. Now we finally have a desired ratio. Since
upon calculation, we know that
can be simplified. Indeed, if you know that
or even take a minute or two to work out the sine and cosine using
, and perhaps the half- or double-angle formulas, you get
.
Solution 3
Without loss of generality, we can assume that and
. As above, we are able to find that
and
.
Using Law of Sines on triangle , we find that
Since we know that
we can compute
to equal
and
to be
.
Next, we apply Law of Cosines to triangle to see that
Simplifying the right side, we get
, so
.
Now, we apply Law of Sines to triangle to see that
After rearranging and noting that
, we get
Dividing the right side by , we see that
so
is either
or
. Since
is not a choice, we know
.
Note that we can also confirm that by computing
with Law of Sines.
Solution 4(FAST)
Note that and
. Seeing these angles makes us think of 30-60-90 triangles. Let
be the foot of the altitude from
to
. This means
and
. Let
and
. This means
and since
we know that
. This means
. This gives
. Note that
. Looking that the answer options we see that
. This means the answer is
.
~coolmath_2018
Solution 5 (Law of Sines)
,
,
, let
,
By the Law of Sines, we have
By the Triple-angle Identities,
,so
Suppose , and
,
,
,
Two possible values of are
and
. However we can rule out
because
is positive, while
is negative.
Therefore ,
Solution 6
For starters, we have Dropping perpendiculars
and
from
and
to
gives
since
Without loss of generality, let and
This tells us that
Using trigonometric identities, we find that
Thus,
which gives
Thus,
Now, note that is a
triangle, so
Thus, we have
Additionally, note that
Applying the Pythagorean Theorem to triangle
then tells us that
By the trigonometric formula for area,
Setting this equal to our other area and solving gives
so
~vaporwave
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.