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Difference between revisions of "2019 AMC 10B Problems/Problem 16"

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label ("$D$", (2.5,3),NE);
 
label ("$D$", (2.5,3),NE);
 
</asy>
 
</asy>
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~ By Little Mouse
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~Little Mouse (Diagram)
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<!-- EDITED BY LVLUO -->
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~On da train to math(Edits)
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==Solution 1==
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~Lvluo (Edits)
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==Solutions==
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===Solution 1===
 
Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</math>.
 
Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</math>.
  
 
Then <math>CB = 5+3 = 8</math>, and <math>\triangle ABC</math> is a <math>1-2-\sqrt{5}</math> triangle.
 
Then <math>CB = 5+3 = 8</math>, and <math>\triangle ABC</math> is a <math>1-2-\sqrt{5}</math> triangle.
  
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In isosceles triangles <math>\triangle ACD</math> and <math>\triangle DEB</math>, drop altitudes from <math>C</math> and <math>E</math> onto <math>AB</math>; denote the feet of these altitudes by <math>P_C</math> and <math>P_E</math> respectively. Then <math>\triangle ACP_C \sim \triangle ABC</math> by AAA similarity, so we get that <math>AP_C = P_CD = \frac{4}{\sqrt{5}}</math>, and <math>AD = 2 \times \frac{4}{\sqrt{5}}</math>. Similarly we get <math>BD = 2 \times \frac{6}{\sqrt{5}}</math>, and <math>AD:DB = \boxed{\textbf{(A) } 2:3}</math>.
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In isosceles triangles <math>\triangle ACD</math> and <math>\triangle DEB</math>, drop altitudes from <math>C</math> and <math>E</math> onto <math>AB</math>; denote the feet of these altitudes by <math>P_C</math> and <math>P_E</math> respectively. Then <math>\triangle ACP_C \sim \triangle ABC</math> by AAA similarity, so we get that <math>AP_C = P_CD = \frac{4}{\sqrt{5}}</math>, and <math>AD = 2 \times \frac{4}{\sqrt{5}}</math>. Similarly, we get <math>BD = 2 \times \frac{6}{\sqrt{5}}</math>, and <math>AD:DB = \boxed{\textbf{(A) } 2:3}</math>.
  
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===Solution 2===
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Alternatively, once finding the length of <math>AD</math> one could use the Pythagorean Theorem to find <math>AB</math> and consequently <math>DB</math>, and then compute the ratio.
 
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==Solution 2==
 
 
Let <math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, <math>A</math> is above <math>C</math>, and <math>B</math> is to the right of <math>C</math>). Also let <math>\angle A = t^{\circ}</math>, so <math>\angle ACD = \left(180-2t\right)^{\circ}</math>, which implies <math>\angle DCB = \left(2t - 90\right)^{\circ}</math>. Similarly, <math>\angle B = \left(90-t\right)^{\circ}</math>, which implies <math>\angle BED = 2t^{\circ}</math>. This further implies that <math>\angle DEC = \left(180 - 2t\right)^{\circ}</math>.  
 
Let <math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, <math>A</math> is above <math>C</math>, and <math>B</math> is to the right of <math>C</math>). Also let <math>\angle A = t^{\circ}</math>, so <math>\angle ACD = \left(180-2t\right)^{\circ}</math>, which implies <math>\angle DCB = \left(2t - 90\right)^{\circ}</math>. Similarly, <math>\angle B = \left(90-t\right)^{\circ}</math>, which implies <math>\angle BED = 2t^{\circ}</math>. This further implies that <math>\angle DEC = \left(180 - 2t\right)^{\circ}</math>.  
  
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By the Law of Cosines in <math>\triangle BED</math>, if <math>BD = d</math>, we have <cmath>\begin{split}&d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &d = \frac{12x}{\sqrt{5}}\end{split}</cmath> Now <math>AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}</math>.
 
By the Law of Cosines in <math>\triangle BED</math>, if <math>BD = d</math>, we have <cmath>\begin{split}&d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &d = \frac{12x}{\sqrt{5}}\end{split}</cmath> Now <math>AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}</math>.
  
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==Solution 3==
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===Solution 3===
 
WLOG, let <math>AC=CD=4</math>, and <math>DE=EB=3</math>. <math>\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}</math>. Because of this, <math>\triangle DEC</math> is a 3-4-5 right triangle. Draw the altitude <math>DF</math> of <math>\triangle DEC</math>. <math>DF</math> is <math>\frac{12}{5}</math> by the base-height triangle area formula. <math>\triangle ABC</math> is similar to <math>\triangle DBF</math> (AA).  So <math>\frac{DF}{AC} = \frac{BD}{AB} = \frac35</math>. <math>DB</math> is <math>\frac35</math> of <math>AB</math>. Therefore, <math>AD:DB</math> is <math>\boxed{\textbf{(A) } 2:3}</math>.
 
WLOG, let <math>AC=CD=4</math>, and <math>DE=EB=3</math>. <math>\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}</math>. Because of this, <math>\triangle DEC</math> is a 3-4-5 right triangle. Draw the altitude <math>DF</math> of <math>\triangle DEC</math>. <math>DF</math> is <math>\frac{12}{5}</math> by the base-height triangle area formula. <math>\triangle ABC</math> is similar to <math>\triangle DBF</math> (AA).  So <math>\frac{DF}{AC} = \frac{BD}{AB} = \frac35</math>. <math>DB</math> is <math>\frac35</math> of <math>AB</math>. Therefore, <math>AD:DB</math> is <math>\boxed{\textbf{(A) } 2:3}</math>.
  
 
~Thegreatboy90
 
~Thegreatboy90
  
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===Solution 4 (a bit long)===
 +
WLOG, <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Notice that in <math>\triangle ACB</math>, we have <math>m\angle BAC + m\angle ABC = 90^{\circ}</math>. Since <math>AC = CD</math> and <math>DE = EB</math>, we find that <math>m\angle DAC = m\angle ADC</math> and <math>m\angle DBE = m\angle BDE</math>, so <math>m\angle ADC + m\angle BDE = 90^{\circ}</math> and <math>\angle EDC</math> is right. Therefore, <math>CE = 5</math> by 3-4-5 triangle, <math>CB = 8</math> and <math>AB = 4\sqrt{5}</math>. Define point F such that <math>CF</math> is an altitude; we know the area of the whole triangle is <math>16</math> and we know the hypotenuse is <math>4\sqrt{5}</math>, so <math>CF = \frac{16}{4\sqrt{5}}\cdot2=\frac{8}{\sqrt{5}}</math>. By the geometric mean theorem, <math>x\left(4\sqrt{5}-x\right)=4\sqrt{5}x-x^{2}=\left(\frac{8}{\sqrt{5}}\right)^{2}=\frac{64}{5}</math>. Solving the quadratic we get <math>x=\frac{10\sqrt{5}\pm6\sqrt{5}}{5}</math>, so <math>x=\frac{4\sqrt{5}}{5} or \frac{16\sqrt{5}}{5}</math>. For now, assume <math>x=\frac{4\sqrt{5}}{5}</math>. Then <math>FB=4\sqrt{5}-\frac{4\sqrt{5}}{5}=\frac{16\sqrt{5}}{5}</math>. <math>CF</math> splits <math>AD</math> into two parts (quick congruence by Leg-Angle) so <math>FD = AF = \frac{4\sqrt{5}}{5}</math> and <math>DB = FB - FD = \frac{16\sqrt{5}}{5}-\frac{4\sqrt{5}}{5}=\frac{12\sqrt{5}}{5}</math>. <math>AD = 2\cdot\frac{4\sqrt{5}}{5}=\frac{8\sqrt{5}}{5}</math>. Now we know <math>AD</math> and <math>DB</math>, we can find <math>\frac{AD}{DB}=\frac{\frac{8\sqrt{5}}{5}}{\frac{12\sqrt{5}}{5}}=\frac{8\sqrt{5}}{5}\cdot\frac{5}{12\sqrt{5}}=\frac{8}{12}=\frac{2}{3}</math> or <math>\boxed{\textbf{(A) } 2:3}</math>.
  
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==Solution 4 (a bit long)==
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===Solution 5 (Short with Trig)===
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WLOG, <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Notice that in <math>\triangle ACB</math>, we have <math>m\angle BAC + m\angle ABC = 90^{\circ}</math>. Since <math>AC = CD</math> and <math>DE = EB</math>, we find that <math>m\angle DAC = m\angle ADC</math> and <math>m\angle DBE = m\angle BDE</math>, so <math>m\angle ADC + m\angle BDE = 90^{\circ}</math> and <math>\angle EDC</math> is right. Therefore, <math>CE = 5</math> by 3-4-5 triangle, <math>CB = 8</math> and <math>AB = 4\sqrt{5}</math>. Define point F such that <math>CF</math> is an altitude; we know the area of the whole triangle is <math>16</math> and we know the hypotenuse is <math>4\sqrt{5}</math>, so <math>CF = \frac{16}{4\sqrt{5}}\cdot2=\frac{8}{\sqrt{5}}</math>. By the geometric mean theorem, <math>x\left(4\sqrt{5}-x\right)=4\sqrt{5}x-x^{2}=\left(\frac{8}{\sqrt{5}}\right)^{2}=\frac{64}{5}</math>. Solving the quadratic we get <math>x=\frac{10\sqrt{5}\pm6\sqrt{5}}{5}</math>, so <math>x=\frac{4\sqrt{5}}{5} or \frac{16\sqrt{5}}{5}</math>. For now, assume <math>x=\frac{4\sqrt{5}}{5}</math>. Then <math>FB=4\sqrt{5}-\frac{4\sqrt{5}}{5}=\frac{16\sqrt{5}}{5}</math>. <math>CF</math> splits <math>AD</math> into two parts (quick congruence by Leg-Angle) so <math>FD = AF = \frac{4\sqrt{5}}{5}</math> and <math>DB = FB - FD = \frac{16\sqrt{5}}{5}-\frac{4\sqrt{5}}{5}=\frac{12\sqrt{5}}{5}</math>. <math>AD = 2\cdot\frac{4\sqrt{5}}{5}=\frac{8\sqrt{5}}{5}</math>. Now we know <math>AD</math> and <math>DB</math>, we can find <math>\frac{AD}{DB}=\frac{\frac{8\sqrt{5}}{5}}{\frac{12\sqrt{5}}{5}}=\frac{8\sqrt{5}}{5}\cdot\frac{5}{12\sqrt{5}}=\frac{8}{12}=\frac{2}{3}</math> or <math>\boxed{\textbf{(A) } 2:3}</math>.
 
  
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==Solution 5==
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Let <math>\angle B=\theta_1</math>, then <math>\angle A=90-\theta_1</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta_1</math>. Similarly, <math>\angle BDE=\theta_1</math>. Then, <math>\angle EDC=180-\theta_1-(90-\theta_1)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\theta_2</math>. We know that <math>\cos \theta_2=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math> so the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</math>.
  
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Let <math>\angle B=\theta</math>, then <math>\angle A=90-\theta</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta</math>. Similarly, <math>\angle BDE=\theta</math>. Then, <math>\angle EDC=180-\theta-(90-\theta)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\alpha</math>. We know that <math>\cos \alpha=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]
  
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==Video Solution 1==
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==Video Solution by TheBeautyofMath==
 
https://youtu.be/_0YaCyxiMBo
 
https://youtu.be/_0YaCyxiMBo
  

Latest revision as of 15:46, 2 November 2024

Problem

In $\triangle ABC$ with a right angle at $C$, point $D$ lies in the interior of $\overline{AB}$ and point $E$ lies in the interior of $\overline{BC}$ so that $AC=CD,$ $DE=EB,$ and the ratio $AC:DE=4:3$. What is the ratio $AD:DB?$

$\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2$

Diagram

[asy] draw((0,0)--(0,8),black); dot((0,0)); dot((0,8)); draw((0,0)--(4,0),black); dot((4,0)); draw((0,8)--(4,0),black); draw((0,0)--(2.5,3),black); dot((2.5,3)); draw((0,5)--(2.5,3),black); dot((0,5)); label ("$B$", (0,8),N); label ("$C$", (0,0),SW); label ("$A$", (4,0),SE); label ("$E$", (0,5),W); label ("$D$", (2.5,3),NE); [/asy] ~Little Mouse (Diagram) ~On da train to math(Edits) ~Lvluo (Edits)

Solutions

Solution 1

Without loss of generality, let $AC = CD = 4$ and $DE = EB = 3$. Let $\angle A = \alpha$ and $\angle B = \beta = 90^{\circ} - \alpha$. As $\triangle ACD$ and $\triangle DEB$ are isosceles, $\angle ADC = \alpha$ and $\angle BDE = \beta$. Then $\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}$, so $\triangle CDE$ is a $3-4-5$ triangle with $CE = 5$.

Then $CB = 5+3 = 8$, and $\triangle ABC$ is a $1-2-\sqrt{5}$ triangle.

In isosceles triangles $\triangle ACD$ and $\triangle DEB$, drop altitudes from $C$ and $E$ onto $AB$; denote the feet of these altitudes by $P_C$ and $P_E$ respectively. Then $\triangle ACP_C \sim \triangle ABC$ by AAA similarity, so we get that $AP_C = P_CD = \frac{4}{\sqrt{5}}$, and $AD = 2 \times \frac{4}{\sqrt{5}}$. Similarly, we get $BD = 2 \times \frac{6}{\sqrt{5}}$, and $AD:DB = \boxed{\textbf{(A) } 2:3}$.

Solution 2

Let $AC=CD=4x$, and $DE=EB=3x$. (For this solution, $A$ is above $C$, and $B$ is to the right of $C$). Also let $\angle A = t^{\circ}$, so $\angle ACD = \left(180-2t\right)^{\circ}$, which implies $\angle DCB = \left(2t - 90\right)^{\circ}$. Similarly, $\angle B = \left(90-t\right)^{\circ}$, which implies $\angle BED = 2t^{\circ}$. This further implies that $\angle DEC = \left(180 - 2t\right)^{\circ}$.

Now we see that $\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2t^{\circ} + 90^{\circ} - 180^{\circ} + 2t^{\circ} = 90^{\circ}$. Thus $\triangle CDE$ is a right triangle, with side lengths of $3x$, $4x$, and $5x$ (by the Pythagorean Theorem, or simply the Pythagorean triple $3-4-5$). Therefore $AC=4x$ (by definition), $BC=5x+3x = 8x$, and $AB=4\sqrt{5}x$. Hence $\cos{\left(2t^{\circ}\right)} = 2 \cos^{2}{t^{\circ}} - 1$ (by the double angle formula), giving $2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 = -\frac{3}{5}$.

By the Law of Cosines in $\triangle BED$, if $BD = d$, we have \[\begin{split}&d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &d = \frac{12x}{\sqrt{5}}\end{split}\] Now $AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}$. Thus the answer is $\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}$.

Solution 3

WLOG, let $AC=CD=4$, and $DE=EB=3$. $\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}$. Because of this, $\triangle DEC$ is a 3-4-5 right triangle. Draw the altitude $DF$ of $\triangle DEC$. $DF$ is $\frac{12}{5}$ by the base-height triangle area formula. $\triangle ABC$ is similar to $\triangle DBF$ (AA). So $\frac{DF}{AC} = \frac{BD}{AB} = \frac35$. $DB$ is $\frac35$ of $AB$. Therefore, $AD:DB$ is $\boxed{\textbf{(A) } 2:3}$.

~Thegreatboy90

Solution 4 (a bit long)

WLOG, $AC = CD = 4$ and $DE = EB = 3$. Notice that in $\triangle ACB$, we have $m\angle BAC + m\angle ABC = 90^{\circ}$. Since $AC = CD$ and $DE = EB$, we find that $m\angle DAC = m\angle ADC$ and $m\angle DBE = m\angle BDE$, so $m\angle ADC + m\angle BDE = 90^{\circ}$ and $\angle EDC$ is right. Therefore, $CE = 5$ by 3-4-5 triangle, $CB = 8$ and $AB = 4\sqrt{5}$. Define point F such that $CF$ is an altitude; we know the area of the whole triangle is $16$ and we know the hypotenuse is $4\sqrt{5}$, so $CF = \frac{16}{4\sqrt{5}}\cdot2=\frac{8}{\sqrt{5}}$. By the geometric mean theorem, $x\left(4\sqrt{5}-x\right)=4\sqrt{5}x-x^{2}=\left(\frac{8}{\sqrt{5}}\right)^{2}=\frac{64}{5}$. Solving the quadratic we get $x=\frac{10\sqrt{5}\pm6\sqrt{5}}{5}$, so $x=\frac{4\sqrt{5}}{5} or \frac{16\sqrt{5}}{5}$. For now, assume $x=\frac{4\sqrt{5}}{5}$. Then $FB=4\sqrt{5}-\frac{4\sqrt{5}}{5}=\frac{16\sqrt{5}}{5}$. $CF$ splits $AD$ into two parts (quick congruence by Leg-Angle) so $FD = AF = \frac{4\sqrt{5}}{5}$ and $DB = FB - FD = \frac{16\sqrt{5}}{5}-\frac{4\sqrt{5}}{5}=\frac{12\sqrt{5}}{5}$. $AD = 2\cdot\frac{4\sqrt{5}}{5}=\frac{8\sqrt{5}}{5}$. Now we know $AD$ and $DB$, we can find $\frac{AD}{DB}=\frac{\frac{8\sqrt{5}}{5}}{\frac{12\sqrt{5}}{5}}=\frac{8\sqrt{5}}{5}\cdot\frac{5}{12\sqrt{5}}=\frac{8}{12}=\frac{2}{3}$ or $\boxed{\textbf{(A) } 2:3}$.

Solution 5 (Short with Trig)

Let $\angle B=\theta_1$, then $\angle A=90-\theta_1$. Since $AC=AD$, $\angle ADC=90-\theta_1$. Similarly, $\angle BDE=\theta_1$. Then, $\angle EDC=180-\theta_1-(90-\theta_1)=90$. Therefore $\bigtriangleup CDE$ is right. Let $AC=CD=4$ and $DE=EB=3$, then $EC=5$. Let $\angle DEC=\angle ACD=\theta_2$. We know that $\cos \theta_2=\frac{3}{5}$ so we can apply the Law of Cosines on $\bigtriangleup ACD$ to find $AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}$. Doing Pythagorean for $BA$, we get $4\sqrt{5}$. Then, $BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}$ so the requested ratio is $8:12=\boxed{\textbf{(A) } 2:3}$.

~Magnetoninja

Video Solution by TheBeautyofMath

https://youtu.be/_0YaCyxiMBo

~IceMatrix

Video Solution by OmegaLearn

https://youtu.be/4_x1sgcQCp4?t=4245

~ pi_is_3.14

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 10 Problems and Solutions

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