Difference between revisions of "2019 AIME I Problems/Problem 11"

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Problem 11

In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$ .

Solution 1

Let the tangent circle be $\omega$ . Some notation first: let $BC=a$ , $AB=b$ , $s$ be the semiperimeter, $\theta=\angle ABC$ , and $r$ be the inradius. Intuition tells us that the radius of $\omega$ is $r+\frac{2rs}{s-a}$ (using the exradius formula). However, the sum of the radius of $\omega$ and $\frac{rs}{s-b}$ is equivalent to the distance between the incenter and the the $B/C$ excenter. Denote the B excenter as $I_B$ and the incenter as $I$ . Lemma: $I_BI=\frac{2b*IB}{a}$ We draw the circumcircle of $\triangle ABC$ . Let the angle bisector of $\angle ABC$ hit the circumcircle at a second point $M$ . By the incenter-excenter lemma, $AM=CM=IM$ . Let this distance be $\alpha$ . Ptolemy's theorem on $ABCM$ gives us $a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}$ Again, by the incenter-excenter lemma, $II_B=2IM$ so $II_b=\frac{2b*IB}{a}$ as desired. Using this gives us the following equation: $\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}$ Motivated by the $s-a$ and $s-b$ , we make the following substitution: $x=s-a, y=s-b$ This changes things quite a bit. Here's what we can get from it: $a=2y, b=x+y, s=x+2y$ It is known (easily proved with Heron's and a=rs) that $r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}$ Using this, we can also find $IB$ : let the midpoint of $BC$ be $N$ . Using Pythagorean's Theorem on $\triangle INB$ , $IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y}$ We now look at the RHS of the main equation: $r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}$ Cancelling some terms, we have $\frac{r(x+4y)}{x}=IB$ Squaring, $\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)$ Expanding and moving terms around gives $(x-8y)(x+2y)=0\to x=8y$ Reverse substituting, $s-a=8s-8b\to b=\frac{9}{2}a$ Clearly the smallest solution is $a=2$ and $b=9$ , so our answer is $2+9+9=\boxed{020}$ -franchester

Solution 2 (Lots of Pythagorean Theorem)

First, assume $BC=2$ and $AB=AC=x$ . The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$ , $BC$ , and $CA$ be denoted $M$ , $N$ , and $O$ , respectively.

Next, we calculate $r$ in terms of $x$ . Note the right triangle formed by $A$ , $I$ , and $M$ . The length $IM$ is equal to $r$ . Using the Pythagorean Theorem, the length $AN$ is $\sqrt{x^2-1}$ , so the length $AI$ is $\sqrt{x^2-1}-r$ . Note that $BN$ is half of $BC=2$ , and by symmetry caused by the incircle, $BN=BM$ and $BM=1$ , so $MA=x-1$ . Applying the Pythagorean Theorem to $AIM$ , we get $r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.$ Expanding yields $r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,$ which can be simplified to $2r\sqrt{x^2-1}=2x-2.$ Dividing by $2$ and then squaring results in $r^2(x^2-1)=(x-1)^2,$ and isolating $r^2$ gets us $r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},$ so $r=\sqrt{\frac{x-1}{x+1}}$ .

We then calculate the radius of the excircle tangent to $BC$ . We denote the center of the excircle $E_N$ and the radius $r_N$ .

Consider the quadrilateral formed by $M$ , $I$ , $E_N$ , and the point at which the excircle intersects the extension of $AB$ , which we denote $H$ . By symmetry caused by the excircle, $BN=BH$ , so $BH=1$ .

Note that triangles $MBI$ and $NBI$ are congruent, and $HBE$ and $NBE$ are also congruent. Denoting the measure of angles $MBI$ and $NBI$ measure $\alpha$ and the measure of angles $HBE$ and $NBE$ measure $\beta$ , straight angle $MBH=2\alpha+2\beta$ , so $\alpha + \beta=90^\circ$ . This means that angle $IBE$ is a right angle, so it forms a right triangle.

Setting the base of the right triangle to $IE$ , the height is $BN=1$ and the base consists of $IN=r$ and $EN=r_N$ . Triangles $INB$ and $BNE$ are similar to $IBE$ , so $\frac{IN}{BN}=\frac{BN}{EN}$ , or $\frac{r}{1}=\frac{1}{r_N}$ . This makes $r_N$ the reciprocal of $r$ , so $r_N=\sqrt{\frac{x+1}{x-1}}$ .

Circle $\omega$ 's radius can be expressed by the distance from the incenter $I$ to the bottom of the excircle with center $E_N$ . This length is equal to $r+2r_N$ , or $\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}$ . Denote this value $r_\omega$ .

Finally, we calculate the distance from the incenter $I$ to the closest point on the excircle tangent to $AB$ , which forms another radius of circle $\omega$ and is equal to $r_\omega$ . We denote the center of the excircle $E_M$ and the radius $r_M$ . We also denote the points where the excircle intersects $AB$ and the extension of $BC$ using $J$ and $K$ , respectively. In order to calculate the distance, we must find the distance between $I$ and $E_M$ and subtract off the radius $r_M$ .

We first must calculate the radius of the excircle. Because the excircle is tangent to both $AB$ and the extension of $AC$ , its center must lie on the angle bisector formed by the two lines, which is parallel to $BC$ . This means that the distance from $E_M$ to $K$ is equal to the length of $AN$ , so the radius is also $\sqrt{x^2-1}$ .

Next, we find the length of $IE_M$ . We can do this by forming the right triangle $IAE_M$ . The length of leg $AI$ is equal to $AN$ minus $r$ , or $\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}$ . In order to calculate the length of leg $AE_M$ , note that right triangles $AJE_M$ and $BNA$ are congruent, as $JE_M$ and $NA$ share a length of $\sqrt{x^2-1}$ , and angles $E_MAJ$ and $NAB$ add up to the right angle $NAE_M$ . This means that $AE_M=BA=x$ .

Using Pythagorean Theorem, we get $IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.$ Bringing back $r_\omega=IE_M-r_M$ and substituting in some values, the equation becomes $r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.$ Rearranging and squaring both sides gets $\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.$ Distributing both sides yields $r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.$ Canceling terms results in $r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.$ Since $-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),$ We can further simplify to $r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.$ Substituting out $r_\omega$ gets $\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2$ which when distributed yields $\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.$ After some canceling, distributing, and rearranging, we obtain $4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.$ Multiplying both sides by $x-1$ results in $4x+4=x^3-x^2-8x^2+8x-4x+4$ which can be rearranged into $x^3-9x^2=0$ and factored into $x^2(x-9)=0.$ This means that $x$ equals $0$ or $9$ , and since a side length of $0$ cannot exist, $x=9$ .

As a result, the triangle must have sides in the ratio of $9:2:9$ . Since the triangle must have integer side lengths, and these values share no common factors greater than 1, the triangle with the smallest possible perimeter under these restrictions has a perimeter of $9+2+9=\boxed{020}$ . ~emerald_block

Solution 3

Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM

@@ Line 14: / Line 14: @@
 -franchester
-==Solution 2==
+==Solution 2 (Lots of Pythagorean Theorem)==
+First, assume <math>BC=2</math> and <math>AB=AC=x</math>. The triangle can be scaled later if necessary. Let <math>I</math> be the incenter and let <math>r</math> be the inradius. Let the points at which the incircle intersects <math>AB</math>, <math>BC</math>, and <math>CA</math> be denoted <math>M</math>, <math>N</math>, and <math>O</math>, respectively.
+Next, we calculate <math>r</math> in terms of <math>x</math>. Note the right triangle formed by <math>A</math>, <math>I</math>, and <math>M</math>. The length <math>IM</math> is equal to <math>r</math>. Using the Pythagorean Theorem, the length <math>AN</math> is <math>\sqrt{x^2-1}</math>, so the length <math>AI</math> is <math>\sqrt{x^2-1}-r</math>. Note that <math>BN</math> is half of <math>BC=2</math>, and by symmetry caused by the incircle, <math>BN=BM</math> and <math>BM=1</math>, so <math>MA=x-1</math>. Applying the Pythagorean Theorem to <math>AIM</math>, we get
+<cmath>r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.</cmath>
+Expanding yields
+<cmath>r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,</cmath>
+which can be simplified to
+<cmath>2r\sqrt{x^2-1}=2x-2.</cmath>
+Dividing by <math>2</math> and then squaring results in
+<cmath>r^2(x^2-1)=(x-1)^2,</cmath>
+and isolating <math>r^2</math> gets us
+<cmath>r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},</cmath>
+so <math>r=\sqrt{\frac{x-1}{x+1}}</math>.
+We then calculate the radius of the excircle tangent to <math>BC</math>. We denote the center of the excircle <math>E_N</math> and the radius <math>r_N</math>.
+Consider the quadrilateral formed by <math>M</math>, <math>I</math>, <math>E_N</math>, and the point at which the excircle intersects the extension of <math>AB</math>, which we denote <math>H</math>. By symmetry caused by the excircle, <math>BN=BH</math>, so <math>BH=1</math>.
+Note that triangles <math>MBI</math> and <math>NBI</math> are congruent, and <math>HBE</math> and <math>NBE</math> are also congruent. Denoting the measure of angles <math>MBI</math> and <math>NBI</math> measure <math>\alpha</math> and the measure of angles <math>HBE</math> and <math>NBE</math> measure <math>\beta</math>, straight angle <math>MBH=2\alpha+2\beta</math>, so <math>\alpha + \beta=90^\circ</math>. This means that angle <math>IBE</math> is a right angle, so it forms a right triangle.
+Setting the base of the right triangle to <math>IE</math>, the height is <math>BN=1</math> and the base consists of <math>IN=r</math> and <math>EN=r_N</math>. Triangles <math>INB</math> and <math>BNE</math> are similar to <math>IBE</math>, so <math>\frac{IN}{BN}=\frac{BN}{EN}</math>, or <math>\frac{r}{1}=\frac{1}{r_N}</math>. This makes <math>r_N</math> the reciprocal of <math>r</math>, so <math>r_N=\sqrt{\frac{x+1}{x-1}}</math>.
+Circle <math>\omega</math>'s radius can be expressed by the distance from the incenter <math>I</math> to the bottom of the excircle with center <math>E_N</math>. This length is equal to <math>r+2r_N</math>, or <math>\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}</math>. Denote this value <math>r_\omega</math>.
+Finally, we calculate the distance from the incenter <math>I</math> to the closest point on the excircle tangent to <math>AB</math>, which forms another radius of circle <math>\omega</math> and is equal to <math>r_\omega</math>. We denote the center of the excircle <math>E_M</math> and the radius <math>r_M</math>. We also denote the points where the excircle intersects <math>AB</math> and the extension of <math>BC</math> using <math>J</math> and <math>K</math>, respectively. In order to calculate the distance, we must find the distance between <math>I</math> and <math>E_M</math> and subtract off the radius <math>r_M</math>.
+We first must calculate the radius of the excircle. Because the excircle is tangent to both <math>AB</math> and the extension of <math>AC</math>, its center must lie on the angle bisector formed by the two lines, which is parallel to <math>BC</math>. This means that the distance from <math>E_M</math> to <math>K</math> is equal to the length of <math>AN</math>, so the radius is also <math>\sqrt{x^2-1}</math>.
+Next, we find the length of <math>IE_M</math>. We can do this by forming the right triangle <math>IAE_M</math>. The length of leg <math>AI</math> is equal to <math>AN</math> minus <math>r</math>, or <math>\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}</math>. In order to calculate the length of leg <math>AE_M</math>, note that right triangles <math>AJE_M</math> and <math>BNA</math> are congruent, as <math>JE_M</math> and <math>NA</math> share a length of <math>\sqrt{x^2-1}</math>, and angles <math>E_MAJ</math> and <math>NAB</math> add up to the right angle <math>NAE_M</math>. This means that <math>AE_M=BA=x</math>.
+Using Pythagorean Theorem, we get
+<cmath>IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.</cmath>
+Bringing back
+<cmath>r_\omega=IE_M-r_M</cmath>
+and substituting in some values, the equation becomes
+<cmath>r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.</cmath>
+Rearranging and squaring both sides gets
+<cmath>\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.</cmath>
+Distributing both sides yields
+<cmath>r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.</cmath>
+Canceling terms results in
+<cmath>r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.</cmath>
+Since
+<cmath>-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),</cmath>
+We can further simplify to
+<cmath>r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.</cmath>
+Substituting out <math>r_\omega</math> gets
+<cmath>\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2</cmath>
+which when distributed yields
+<cmath>\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.</cmath>
+After some canceling, distributing, and rearranging, we obtain
+<cmath>4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.</cmath>
+Multiplying both sides by <math>x-1</math> results in
+<cmath>4x+4=x^3-x^2-8x^2+8x-4x+4</cmath>
+which can be rearranged into
+<cmath>x^3-9x^2=0</cmath>
+and factored into
+<cmath>x^2(x-9)=0.</cmath>
+This means that <math>x</math> equals <math>0</math> or <math>9</math>, and since a side length of <math>0</math> cannot exist, <math>x=9</math>.
+As a result, the triangle must have sides in the ratio of <math>9:2:9</math>. Since the triangle must have integer side lengths, and these values share no common factors greater than 1, the triangle with the smallest possible perimeter under these restrictions has a perimeter of <math>9+2+9=\boxed{020}</math>.
+~[[User:emerald_block|emerald_block]]
+==Solution 3==
 Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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