Difference between revisions of "2016 AIME I Problems/Problem 6"

(Solution 3)
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==Problem==
 
==Problem==
In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
+
In <math>\triangle ABC</math> let <math>I</math> be the center of the inscribed circle, and let the bisector of <math>\angle ACB</math> intersect <math>AB</math> at <math>L</math>. The line through <math>C</math> and <math>L</math> intersects the circumscribed circle of <math>\triangle ABC</math> at the two points <math>C</math> and <math>D</math>. If <math>LI=2</math> and <math>LD=3</math>, then <math>IC=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
=Solution=
 
=Solution=
 
==Solution 1==
 
==Solution 1==
It is well known that <math>DA = DI = DB</math> and so we have <math>DA = DB = 5</math>. Then <math>\triangle DLB \sim \triangle ALC</math> and so <math>\frac{AL}{AC} = \frac{3}{5}</math> and from the angle bisector theorem <math>\frac{CI}{IL} = \frac{5}{3}</math> so <math>CI = \frac{10}{3}</math> and our answer is <math>\boxed{013}</math>
+
Suppose we label the angles as shown below.
 +
<asy>
 +
size(150);
 +
import olympiad;
 +
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
 +
pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));
 +
pair C=intersectionpoints(circle(A,b),circle(B,a))[0];
 +
pair I=incenter(A,B,C);
 +
pair L=extension(C,D,A,B);
 +
dot(I^^A^^B^^C^^D);
 +
draw(C--D);
 +
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
 +
draw(A--B--D--cycle);
 +
draw(circumcircle(A,B,D));
 +
draw(A--C--B);
 +
draw(A--I--B^^C--I);
 +
draw(incircle(A,B,C));
 +
label("$A$",A,SW,fontsize(8));
 +
label("$B$",B,SE,fontsize(8));
 +
label("$C$",C,N,fontsize(8));
 +
label("$D$",D,S,fontsize(8));
 +
label("$I$",I,NE,fontsize(8));
 +
label("$L$",L,SW,fontsize(8));
 +
label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8));
 +
label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8));
 +
label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8));
 +
label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8));
 +
label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8));
 +
label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8));
 +
</asy>
 +
As <math>\angle BCD</math> and <math>\angle BAD</math> intercept the same arc, we know that <math>\angle BAD=\gamma</math>. Similarly, <math>\angle ABD=\gamma</math>. Also, using <math>\triangle ICA</math>, we find <math>\angle CIA=180-\alpha-\gamma</math>. Therefore, <math>\angle AID=\alpha+\gamma</math>. Therefore, <math>\angle DAI=\angle AID=\alpha+\gamma</math>, so <math>\triangle AID</math> must be isosceles with <math>AD=ID=5</math>. Similarly, <math>BD=ID=5</math>. Then <math>\triangle DLB \sim \triangle ALC</math>, hence <math>\frac{AL}{AC} = \frac{3}{5}</math>. Also, <math>AI</math> bisects <math>\angle LAC</math>, so by the Angle Bisector Theorem <math>\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}</math>. Thus <math>CI = \frac{10}{3}</math>, and the answer is <math>\boxed{013}</math>.
  
 
==Solution 2==
 
==Solution 2==
 
This is a cheap solution.
 
 
   
 
   
WLOG assume <math>\triangle ABC</math> is isosceles. Then, <math>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</math>, and let it meet <math>CB</math> at <math>E</math>. Since <math>IL=2</math>, <math>IE</math> is also <math>2</math> (they are both inradii). Set <math>BD</math> as <math>x</math>. Then, triangles <math>BLD</math> and <math>CEI</math> are similar, and <math>\tfrac{2}{3}=\tfrac{CI}{x}</math>. Thus, <math>CI=\tfrac{2x}{3}</math>. <math>\triangle CBD~\triangle CEI</math>, so <math>\tfrac{IE}{DB}=\tfrac{CI}{CD}</math>. Thus <math>\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}</math>. Solving for <math>x</math>, we have:
+
WLOG assume <math>\triangle ABC</math> is isosceles. Then, <math>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</math>, and let it meet <math>CB</math> at <math>E</math>. Since <math>IL=2</math>, <math>IE</math> is also <math>2</math> (they are both inradii). Set <math>BD</math> as <math>x</math>. Then, triangles <math>BLD</math> and <math>CEI</math> are similar, and <math>\tfrac{2}{3}=\tfrac{CI}{x}</math>. Thus, <math>CI=\tfrac{2x}{3}</math>. <math>\triangle CBD \sim \triangle CEI</math>, so <math>\tfrac{IE}{DB}=\tfrac{CI}{CD}</math>. Thus <math>\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}</math>. Solving for <math>x</math>, we have:
 
<math>x^2-2x-15=0</math>, or <math>x=5, -3</math>. <math>x</math> is positive, so <math>x=5</math>. As a result, <math>CI=\tfrac{2x}{3}=\tfrac{10}{3}</math> and the answer is <math>\boxed{013}</math>
 
<math>x^2-2x-15=0</math>, or <math>x=5, -3</math>. <math>x</math> is positive, so <math>x=5</math>. As a result, <math>CI=\tfrac{2x}{3}=\tfrac{10}{3}</math> and the answer is <math>\boxed{013}</math>
  
 
==Solution 3==
 
==Solution 3==
  
WLOG assume <math>\triangle ABC</math> is isosceles (with vertex <math>C</math>). Let <math>O</math> be the center of the circumcircle, <math>R</math> the circumradius, and <math>r</math> the inradius. A simple sketch will reveal that <math>\triangle ABC</math> must also be obtuse (as an acute triangle will result in <math>LI</math> being greater than <math>DL</math>) and that <math>O</math> and <math>I</math> are collinear. Next, if <math>OI=d</math>, <math>DO+OI=R+d</math> and <math>R+d=DL+DI=5</math>. Euler gives us that <math>d^{2}=R(R-2r)</math>, and in this case, <math>r=LI=2</math>. Thus, <math>d=\sqrt{R^{2}-4R}</math>. Solving for <math>d</math>, we have <math>R+\sqrt{R^{2}-4R}=5</math>, then <math>R^{2}-4R=25-10R+R^{2}</math>, yielding <math>R=\frac{25}{6}</math>. Next, <math>R+d=5</math> so <math>d=\frac{5}{6}</math>. Finally, <math>OC=OI+CI</math> gives us <math>R=d+CI</math>, and <math>CI=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}</math>. Our answer is then <math>\boxed{013}</math>.
+
WLOG assume <math>\triangle ABC</math> is isosceles (with vertex <math>C</math>). Let <math>O</math> be the center of the circumcircle, <math>R</math> the circumradius, and <math>r</math> the inradius. A simple sketch will reveal that <math>\triangle ABC</math> must be obtuse (as an acute triangle will result in <math>LI</math> being greater than <math>DL</math>) and that <math>O</math> and <math>I</math> are collinear. Next, if <math>OI=d</math>, <math>DO+OI=R+d</math> and <math>R+d=DL+LI=5</math>. Euler gives us that <math>d^{2}=R(R-2r)</math>, and in this case, <math>r=LI=2</math>. Thus, <math>d=\sqrt{R^{2}-4R}</math>. Solving for <math>d</math>, we have <math>R+\sqrt{R^{2}-4R}=5</math>, then <math>R^{2}-4R=25-10R+R^{2}</math>, yielding <math>R=\frac{25}{6}</math>. Next, <math>R+d=5</math> so <math>d=\frac{5}{6}</math>. Finally, <math>OC=OI+IC</math> gives us <math>R=d+IC</math>, and <math>IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}</math>. Our answer is then <math>\boxed{013}</math>.
 +
 
 +
==Solution 4==
 +
 
 +
Since <math>\angle{LAD} = \angle{BDC}</math> and <math>\angle{DLA}=\angle{DBC}</math>, <math>\triangle{DLA}\sim\triangle{DBC}</math>. Also, <math>\angle{DAC}=\angle{BLC}</math> and <math>\angle{ACD}=\angle{LCB}</math> so <math>\triangle{DAC}\sim\triangle{BLC}</math>. Now we can call <math>AC</math>, <math>b</math> and <math>BC</math>, <math>a</math>. By angle bisector theorem, <math>\frac{AD}{DB}=\frac{AC}{BC}</math>. So let <math>AD=bk</math> and <math>DB=ak</math> for some value of <math>k</math>. Now call <math>IC=x</math>. By the similar triangles we found earlier, <math>\frac{3}{ak}=\frac{bk}{x+2}</math> and <math>\frac{b}{x+5}=\frac{x+2}{a}</math>. We can simplify this to <math>abk^2=3x+6</math> and <math>ab=(x+5)(x+2)</math>. So we can plug the <math>ab</math> into the first equation and get <math>(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3</math>. We can now draw a line through <math>A</math> and <math>I</math> that intersects <math>BC</math> at <math>E</math>. By mass points, we can assign a mass of <math>a</math> to <math>A</math>, <math>b</math> to <math>B</math>, and <math>a+b</math> to <math>D</math>. We can also assign a mass of <math>(a+b)k</math> to <math>C</math> by angle bisector theorem. So the ratio of <math>\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}</math>. So since <math>k=\frac{2}{x}</math>, we can plug this back into the original equation to get <math>\left(\frac{2}{x}\right)^2(x+5)=3</math>. This means that <math>\frac{3x^2}{4}-x-5=0</math> which has roots -2 and <math>\frac{10}{3}</math> which means our <math>CI=\frac{10}{3}</math> and our answer is <math>\boxed{013}</math>.
 +
 
 +
==Solution 5==
 +
Since <math>\angle BCD</math> and <math>\angle BAD</math> both intercept arc <math>BD</math>, it follows that <math>\angle BAD=\gamma</math>. Note that <math>\angle AID=\alpha+\gamma</math> by the external angle theorem. It follows that <math>\angle DAI=\angle AID=\alpha+\gamma</math>, so we must have that <math>\triangle AID</math> is isosceles, yielding <math>AD=ID=5</math>. Note that <math>\triangle DLA \sim \triangle DAC</math>, so <math>\frac{DA}{DL} = \frac{DC}{DA}</math>. This yields <math>DC = \frac{25}{3}</math>. It follows that <math>CI = DC - DI = \frac{10}{3}</math>, giving a final answer of <math>\boxed{013}</math>.
 +
 
 +
==Solution 6==
 +
Let <math>I_C</math> be the excenter opposite to <math>C</math> in <math>ABC</math>. By the incenter-excenter lemma <math>DI=DC \therefore</math> <math>LI_C=8,LI=2,II_C=10</math>. Its well known that <math>(I_C,I,L,C)=-1 \implies \dfrac{LI_C}{LI}.\dfrac{CI}{CI_C}=-1 \implies \dfrac{CI}{CI+10}=\dfrac{1}{4} \implies \boxed{CI=\dfrac{10}{3}}</math>.<math>\blacksquare</math>
 +
~Pluto1708
 +
 
 +
Alternate solution: "We can use the angle bisector theorem on <math>\triangle CBL</math> and bisector <math>BI</math> to get that <math>\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}</math>. Since <math>\triangle CBL \sim \triangle ADL</math>, we get <math>\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}</math>. Thus, <math>CI=\tfrac{10}{3}</math> and <math>p+q=\boxed{13}</math>."
 +
(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
 +
 
 +
==Solution 7==
 +
We can just say that quadrilateral <math>ADBC</math> is a right kite with right angles at <math>A</math> and <math>B</math>. Let us construct another similar right kite with the points of tangency on <math>AC</math> and <math>BC</math> called <math>E</math> and <math>F</math> respectively, point <math>I</math>, and point <math>C</math>. Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call <math>CI</math> <math>x</math> for simplicity's sake. Based on the fact that <math>\triangle BCD</math> is similar to <math>\triangle FCI</math> we can use triangle proportionality to say that <math>BD</math> is <math>2\frac{x+5}{x}</math>. Using geometric mean theorem we can show that <math>BL</math> must be <math>\sqrt{3x+6}</math>. With Pythagorean Theorem we can say that <math>3x+6+9=4{(\frac{x+5}{x})}^2</math>. Multiplying both sides by <math>x^2</math> and moving everything to LHS will give you <math>3{x}^3+11{x}^2-40x-100=0</math> Since <math>x</math> must be in the form <math>\frac{p}{q}</math> we can assume that <math>x</math> is most likely a positive fraction in the form <math>\frac{p}{3}</math> where <math>p</math> is a factor of <math>100</math>. Testing the factors in synthetic division would lead <math>x = \frac{10}{3}</math>, giving us our desired answer <math>\boxed{013}</math>. ~Lopkiloinm
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2016|n=I|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:55, 25 August 2020

Problem

In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Solution 1

Suppose we label the angles as shown below. [asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$. Similarly, $\angle ABD=\gamma$. Also, using $\triangle ICA$, we find $\angle CIA=180-\alpha-\gamma$. Therefore, $\angle AID=\alpha+\gamma$. Therefore, $\angle DAI=\angle AID=\alpha+\gamma$, so $\triangle AID$ must be isosceles with $AD=ID=5$. Similarly, $BD=ID=5$. Then $\triangle DLB \sim \triangle ALC$, hence $\frac{AL}{AC} = \frac{3}{5}$. Also, $AI$ bisects $\angle LAC$, so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$. Thus $CI = \frac{10}{3}$, and the answer is $\boxed{013}$.

Solution 2

WLOG assume $\triangle ABC$ is isosceles. Then, $L$ is the midpoint of $AB$, and $\angle CLB=\angle CLA=90^\circ$. Draw the perpendicular from $I$ to $CB$, and let it meet $CB$ at $E$. Since $IL=2$, $IE$ is also $2$ (they are both inradii). Set $BD$ as $x$. Then, triangles $BLD$ and $CEI$ are similar, and $\tfrac{2}{3}=\tfrac{CI}{x}$. Thus, $CI=\tfrac{2x}{3}$. $\triangle CBD \sim \triangle CEI$, so $\tfrac{IE}{DB}=\tfrac{CI}{CD}$. Thus $\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}$. Solving for $x$, we have: $x^2-2x-15=0$, or $x=5, -3$. $x$ is positive, so $x=5$. As a result, $CI=\tfrac{2x}{3}=\tfrac{10}{3}$ and the answer is $\boxed{013}$

Solution 3

WLOG assume $\triangle ABC$ is isosceles (with vertex $C$). Let $O$ be the center of the circumcircle, $R$ the circumradius, and $r$ the inradius. A simple sketch will reveal that $\triangle ABC$ must be obtuse (as an acute triangle will result in $LI$ being greater than $DL$) and that $O$ and $I$ are collinear. Next, if $OI=d$, $DO+OI=R+d$ and $R+d=DL+LI=5$. Euler gives us that $d^{2}=R(R-2r)$, and in this case, $r=LI=2$. Thus, $d=\sqrt{R^{2}-4R}$. Solving for $d$, we have $R+\sqrt{R^{2}-4R}=5$, then $R^{2}-4R=25-10R+R^{2}$, yielding $R=\frac{25}{6}$. Next, $R+d=5$ so $d=\frac{5}{6}$. Finally, $OC=OI+IC$ gives us $R=d+IC$, and $IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}$. Our answer is then $\boxed{013}$.

Solution 4

Since $\angle{LAD} = \angle{BDC}$ and $\angle{DLA}=\angle{DBC}$, $\triangle{DLA}\sim\triangle{DBC}$. Also, $\angle{DAC}=\angle{BLC}$ and $\angle{ACD}=\angle{LCB}$ so $\triangle{DAC}\sim\triangle{BLC}$. Now we can call $AC$, $b$ and $BC$, $a$. By angle bisector theorem, $\frac{AD}{DB}=\frac{AC}{BC}$. So let $AD=bk$ and $DB=ak$ for some value of $k$. Now call $IC=x$. By the similar triangles we found earlier, $\frac{3}{ak}=\frac{bk}{x+2}$ and $\frac{b}{x+5}=\frac{x+2}{a}$. We can simplify this to $abk^2=3x+6$ and $ab=(x+5)(x+2)$. So we can plug the $ab$ into the first equation and get $(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3$. We can now draw a line through $A$ and $I$ that intersects $BC$ at $E$. By mass points, we can assign a mass of $a$ to $A$, $b$ to $B$, and $a+b$ to $D$. We can also assign a mass of $(a+b)k$ to $C$ by angle bisector theorem. So the ratio of $\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}$. So since $k=\frac{2}{x}$, we can plug this back into the original equation to get $\left(\frac{2}{x}\right)^2(x+5)=3$. This means that $\frac{3x^2}{4}-x-5=0$ which has roots -2 and $\frac{10}{3}$ which means our $CI=\frac{10}{3}$ and our answer is $\boxed{013}$.

Solution 5

Since $\angle BCD$ and $\angle BAD$ both intercept arc $BD$, it follows that $\angle BAD=\gamma$. Note that $\angle AID=\alpha+\gamma$ by the external angle theorem. It follows that $\angle DAI=\angle AID=\alpha+\gamma$, so we must have that $\triangle AID$ is isosceles, yielding $AD=ID=5$. Note that $\triangle DLA \sim \triangle DAC$, so $\frac{DA}{DL} = \frac{DC}{DA}$. This yields $DC = \frac{25}{3}$. It follows that $CI = DC - DI = \frac{10}{3}$, giving a final answer of $\boxed{013}$.

Solution 6

Let $I_C$ be the excenter opposite to $C$ in $ABC$. By the incenter-excenter lemma $DI=DC \therefore$ $LI_C=8,LI=2,II_C=10$. Its well known that $(I_C,I,L,C)=-1 \implies \dfrac{LI_C}{LI}.\dfrac{CI}{CI_C}=-1 \implies \dfrac{CI}{CI+10}=\dfrac{1}{4} \implies \boxed{CI=\dfrac{10}{3}}$.$\blacksquare$ ~Pluto1708

Alternate solution: "We can use the angle bisector theorem on $\triangle CBL$ and bisector $BI$ to get that $\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}$. Since $\triangle CBL \sim \triangle ADL$, we get $\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}$. Thus, $CI=\tfrac{10}{3}$ and $p+q=\boxed{13}$." (https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)

Solution 7

We can just say that quadrilateral $ADBC$ is a right kite with right angles at $A$ and $B$. Let us construct another similar right kite with the points of tangency on $AC$ and $BC$ called $E$ and $F$ respectively, point $I$, and point $C$. Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call $CI$ $x$ for simplicity's sake. Based on the fact that $\triangle BCD$ is similar to $\triangle FCI$ we can use triangle proportionality to say that $BD$ is $2\frac{x+5}{x}$. Using geometric mean theorem we can show that $BL$ must be $\sqrt{3x+6}$. With Pythagorean Theorem we can say that $3x+6+9=4{(\frac{x+5}{x})}^2$. Multiplying both sides by $x^2$ and moving everything to LHS will give you $3{x}^3+11{x}^2-40x-100=0$ Since $x$ must be in the form $\frac{p}{q}$ we can assume that $x$ is most likely a positive fraction in the form $\frac{p}{3}$ where $p$ is a factor of $100$. Testing the factors in synthetic division would lead $x = \frac{10}{3}$, giving us our desired answer $\boxed{013}$. ~Lopkiloinm

See also

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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