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> | ||
Line 15: | Line 15: | ||
</math> | </math> | ||
− | == Solution == | + | == Solution 1 == |
<asy> | <asy> | ||
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defaultpen(0.8); | defaultpen(0.8); | ||
pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); | pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); | ||
− | pair ortho=rotate(- | + | pair ortho=rotate(-50)*(D-A); |
− | pair E= | + | pair E=extension(C, ortho, A, D); |
− | draw(A-- | + | draw(A--B); |
+ | draw(B--C); | ||
+ | draw(A--C); | ||
+ | draw(C--E); | ||
+ | draw(E--D); | ||
draw(A--D); | draw(A--D); | ||
− | + | draw(B--E, dotted); | |
− | draw(B--E, | + | |
− | |||
− | |||
− | |||
− | |||
− | |||
label("$A$",A,SW); | label("$A$",A,SW); | ||
label("$B$",B,SE); | label("$B$",B,SE); | ||
Line 51: | Line 50: | ||
Combining the previous two observations we get that <math>AE=EC</math>, and as <math>\angle AEC=90^\circ</math>, this means that <math>\angle CAE = \angle ACE = 45^\circ</math>. | Combining the previous two observations we get that <math>AE=EC</math>, and as <math>\angle AEC=90^\circ</math>, this means that <math>\angle CAE = \angle ACE = 45^\circ</math>. | ||
− | Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{ | + | Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = 75^\circ \boxed{D}</math>. |
+ | |||
+ | == 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>. | |
− | Using Law of Sines on triangle <math>ADB</math>, we find that < | + | 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 < | + | 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 < | + | 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 | + | 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. | Note that we can also confirm that <math>\angle ACB \neq 105^\circ</math> by computing <math>\angle CAB</math> with Law of Sines. |
Latest revision as of 11:56, 27 December 2020
Problem
In , . Point is on so that and . Find
Solution 1
We start with the observation that , and .
We can draw the height from onto . In the triangle , we have . Hence .
By the definition of , we also have , therefore . This means that the triangle is isosceles, and as , we must have .
Then we compute , thus and the triangle is isosceles as well. Hence .
Now we can note that , hence also the triangle is isosceles and we have .
Combining the previous two observations we get that , and as , this means that .
Finally, we get .
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.
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 |
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