Difference between revisions of "2001 AMC 12 Problems/Problem 24"

Latest revision as of 21:48, 7 March 2024

Problem

In $\triangle ABC$ , $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$

$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$

Solution 2

Draw a good diagram! Now, let's call $BD=t$ , so $DC=2t$ . Given the rather nice angles of $\angle ABD = 45^\circ$ and $\angle ADC = 60^\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$ ; call this point $H$ . We realize that there is no specific factor of $t$ we can call this just yet, so let $AH=kt$ . Notice that in $\triangle{ABH}$ we get $BH=kt$ . Using the 60-degree angle in $\triangle{ADH}$ , we obtain $DH=\frac{\sqrt{3}}{3}kt$ . The comparable ratio is that $BH-DH=t$ . If we involve our $k$ , we get:

$kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$ . Eliminating $t$ and removing radicals from the denominator, we get $k=\frac{3+\sqrt{3}}{2}$ . From there, one can easily obtain $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$ . Now we finally have a desired ratio. Since $\tan\angle ACH = 2+\sqrt{3}$ upon calculation, we know that $\angle ACH$ can be simplified. Indeed, if you know that $\tan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and cosine using $\sin(x)^2+\cos(x)^2=1$ , and perhaps the half- or double-angle formulas, you get $\boxed{75^\circ}$ .

Solution 3

Without loss of generality, we can assume that $BD = 1$ and $CD = 2$ . As above, we are able to find that $\angle ADC = 60^\circ$ and $\angle ADB = 120^\circ$ .

Using Law of Sines on triangle $ADB$ , we find that $\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}.$ Since we know that $\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4},$ $\sin 45^\circ = \frac{\sqrt{2}}{2},$ $\sin 120^\circ = \frac{\sqrt{3}}{2},$ we can compute $AD$ to equal $1+\sqrt{3}$ and $AB$ to be $\frac{3\sqrt{2}+\sqrt{6}}{2}$ .

Next, we apply Law of Cosines to triangle $ADC$ to see that $AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ).$ Simplifying the right side, we get $AC^2 = 6$ , so $AC = \sqrt{6}$ .

Now, we apply Law of Sines to triangle $ABC$ to see that $\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}.$ After rearranging and noting that $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , we get $\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}.$

Dividing the right side by $\sqrt{3}$ , we see that $\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4},$ so $\angle ACB$ is either $75^\circ$ or $105^\circ$ . Since $105^\circ$ is not a choice, we know $\angle ACB = \boxed{75^\circ}$ .

Note that we can also confirm that $\angle ACB \neq 105^\circ$ by computing $\angle CAB$ with Law of Sines.

Solution 4(FAST)

Note that $\angle{ADB} = 120$ and $\angle{ADC} = 60$ . Seeing these angles makes us think of 30-60-90 triangles. Let $E$ be the foot of the altitude from $A$ to $BC$ . This means $\angle{DAE} = 30$ and $\angle{BAE} = 45$ . Let $BD = x$ and $DE = y$ . This means $AE = y\sqrt{3}$ and since $AE = BE$ we know that $x + y = y\sqrt{3}$ . This means $y = \frac{x(\sqrt{3} + 1)}{2}$ . This gives $CE = \frac{4x - x(\sqrt{3} + 1)}{2}$ . Note that $\tan{\angle{ACE}} = \frac{AE}{CE} = 2 + \sqrt{3}$ . Looking that the answer options we see that $\tan{75^\circ} = 2 + \sqrt{3}$ . This means the answer is $D$ . ~coolmath_2018

Solution 5 (Law of Sines)

$\angle ADB = 120^\circ$ , $\angle ADC = 60^\circ$ , $\angle DAB = 15^\circ$ , let $\angle ACB = \theta$ , $\angle DAC = 120^\circ - \theta$

By the Law of Sines, we have $\frac{CD}{\sin(120^\circ - \theta)} = \frac{AD}{\sin \theta}$

$\space$ $\frac{BD}{\sin15^\circ} = \frac{AD}{\sin45^\circ}$

$\frac{BD}{CD} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}$

$\frac{1}{2} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}$

By the Triple-angle Identities, $\sin 45^\circ = 3\sin15^\circ -4\sin^3 15^\circ$

$\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3\ -4\sin^2 15^\circ}$

$\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$

$\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3 - 4 \cdot \frac{2 - \sqrt{3}}{4}} = \frac{\sin \theta}{1+\sqrt{3}}$

$\frac{\sin \theta}{\sin(120^\circ - \theta)} = \frac{1+\sqrt{3}}{2}$

$\sin(120^\circ - \theta) = \sin120^\circ \cos\theta - \sin\theta \cos120^\circ = \frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta$ ，so

$\frac{\sin \theta}{\frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta} = \frac{1+\sqrt{3}}{2}$

$\frac{\sqrt{3} \cdot \cos\theta + \sin\theta}{2 \sin \theta} = \frac{2}{1+\sqrt{3}}$

$\frac{\sqrt{3}}{2} \cdot \cot \theta = \frac{2}{1+\sqrt{3}} - \frac{1}{2} = \frac{3 - \sqrt{3}}{2 + 2 \sqrt{3}}$

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Suppose $\cos \theta = k(\sqrt{3} - 1)$ , and $\sin \theta = k(\sqrt{3} + 1)$

$\sin^2 \theta + \cos^2 \theta = 1$ , $k^2(\sqrt{3} + 1)^2 + k^2(\sqrt{3} - 1)^2 = 8k^2 = 1$ , $k = \frac{1}{2 \sqrt{2}}$

$\sin \theta = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ , $\cos \theta = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

$\sin 2 \theta = 2 \cdot \frac{\sqrt{3} + 1}{2 \sqrt{2}} \cdot \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{1}{2}$

Two possible values of $2 \theta$ are $150^\circ$ and $30^\circ$ . However we can rule out $30^\circ$ because $\cos 15^\circ$ is positive, while $\cos \theta$ is negative.

Therefore $2 \theta = 150^\circ$ , $\angle ACB = \boxed{\textbf{(D) } 75^\circ }$

~isabelchen

Solution 6

For starters, we have $\angle ABD=120^\circ.$ Dropping perpendiculars $\overline{DX}$ and $\overline{CY}$ from $D$ and $C$ to $\overline{AB}$ gives $\angle ADX=120^\circ-45^\circ=75^\circ,$ since $\angle BDX=45^\circ.$

Without loss of generality, let $BD=1$ and $CD=2.$ This tells us that $BX=DX=\sqrt{2}/2.$ Using trigonometric identities, we find that

$\tan \angle ADX=\tan 75^\circ=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2}.$ Thus, $AX/DX=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2},$ which gives $AX=\dfrac{2\sqrt 2+\sqrt 6}{2}.$ Thus, $AB=AX+BX=\dfrac{3\sqrt 2+\sqrt 6}{2}.$

Now, note that $BCY$ is a $45-45-90$ triangle, so $CY=BY=\dfrac{3\sqrt 2}{2}.$ Thus, we have

$[ABC]=\dfrac{1}{2}\cdot AB\cdot CY=\dfrac{1}{2}\cdot \dfrac{3\sqrt 2+\sqrt 6}{2}\cdot \dfrac{3\sqrt 2}{2}=\dfrac{9+3\sqrt 3}{4}.$ Additionally, note that $AY=AB-BY=\sqrt{6}/2.$ Applying the Pythagorean Theorem to triangle $AYC$ then tells us that $AC=\sqrt{6}.$ By the trigonometric formula for area,

$[ABC]=\dfrac{1}{2}\cdot BC\cdot AC\cdot \sin \angle ACB=\dfrac{3\sqrt 6}{2}\sin \angle ACB.$ Setting this equal to our other area and solving gives $\sin \angle ACB=\dfrac{\sqrt 6+\sqrt 2}{4},$ so $\angle ACB=\boxed{75^\circ}.$ ~vaporwave

@@ Line 1: / Line 1: @@
 == 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>
@@ Line 15: / Line 15: @@
 </math>
-== Solution ==
+== Solution 2 ==
-<asy>
+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:
-unitsize(2cm);
-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 ortho=rotate(-90)*(D-A);
-pair E=intersectionpoint(A--D, C--(C+10*ortho));
-draw(A--B--C--cycle);
-draw(A--D);
-draw(C--E, dashed);
-draw(B--E, dashed);
-draw( rightanglemark(C,E,D) );
-draw( scale(2)*anglemark(B,A,D) );
-draw( anglemark(D,B,A) );
-label("$15^\circ$",(0.6,0),5*ENE);
-label("$45^\circ$",B,5*WNW,UnFill);
-label("$A$",A,SW);
-label("$B$",B,SE);
-label("$C$",C,N);
-label("$D$",D,NE);
-label("$E$",E,S);
-</asy>
-We start with the observation that <math>\angle ADB = 180^\circ - 15^\circ - 45^\circ = 120^\circ</math>, and <math>\angle ADC = 15^\circ + 45^\circ = 60^\circ</math>.
+<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>.
-We can draw the height <math>CE</math> from <math>C</math> onto <math>AD</math>. In the triangle <math>CED</math>, we have <math>\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12</math>. Hence <math>ED = CD/2</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>.
-By the definition of <math>D</math>, we also have <math>BD=CD/2</math>, therefore <math>BD=DE</math>. This means that the triangle <math>BDE</math> is isosceles, and as <math>\angle BDE=120^\circ</math>, we must have <math>\angle BED = \angle EBD = 30^\circ</math>.
+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>.
-Then we compute <math>\angle ABE = 45^\circ - 30^\circ = 15^\circ</math>, thus <math>\angle ABE = \angle BAE</math> and the triangle <math>ABE</math> is isosceles as well. Hence <math>AE=BE</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 can note that <math>\angle DCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ</math>, hence also the triangle <math>EBC</math> is isosceles and we have <math>BE=CE</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>
-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>.
+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>.
-Finally, we get <math>\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \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, 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 ==
 {{AMC12 box|year=2001|num-b=23|num-a=25}}
+{{MAA Notice}}

2001 AMC 12 (Problems • Answer Key • Resources)
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