High School Olympiads Forum 9

by franzliszt, Jul 23, 2024, 8:23 PM

High School Olympiads Forum 9 wrote:

Let $ABC$ be an acute triangle with circumcircle $\Gamma$ and intouch triangle $DEF$ . A circle is drawn tangent to $\overline{BC}$ at $D$ and to minor arc $BC$ of $\Gamma$ at $X$ . Define $Y$ and $Z$ similarly. Prove that lines $\overline{DX}$ , $\overline{EY}$ , $\overline{FZ}$ are concurrent.

AoPS Link

Proof. Let $D'$ be the midpoint of $\widehat{BAC}$ containing $A$ and define $E',F'$ similarly. By Shooting Lemma, $XD\cap \Gamma=D'$ . In Barycentric Coordinates with reference triangle $ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ we have $D=(0:s-c:s-b)$ and $D'=(a^2:-b^2+bc:-c^2+bc)$ where we computed the coordinates of $D'=\frac{I_B+I_C}2$ using the midpoint formula on normalized $I_B=(a:-b:c),I_C=(a:b:-c)$ . We can compute $E',F'$ similarly. We claim that the concurrency point is $X_{55} =( a^2(s-a) : b^2(s-b) : c^2(s-c))$ .

Verify that $\begin{align*}\begin{vmatrix}0&s-c&s-b\\a^2(s-a) & b^2(s-b) & c^2(s-c)\\ a^2&-b^2+bc&-c^2+bc\end{vmatrix}&=\frac{a^2}{4}\begin{vmatrix}0&a+b-c&a-b+c\\-a+b+c & b^2(a-b+c) & c^2(a+b-c)\\ 1&-b^2+bc&-c^2+bc\end{vmatrix}\\&=0+(a+b-c)[c^2(a+b-c)-(-a+b+c)(-c^2+bc)]+(a-b+c)[(-a+b+c)(-b^2+bc)-b^2(a-b+c)]\\&=0\end{align*}$ where we simplified the last part with a little help.

We can use the exact same process to show that $X_{55}$ lies on $EE'$ and $FF'$ completing the proof.

USAMO 2015/2

by franzliszt, Jul 11, 2024, 6:24 PM

USAMO 2015/2 wrote:

Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$ . Let $X$ be a variable point on segment $\overline{PQ}$ . Line $AX$ meets $\omega$ again at $S$ (other than $A$ ). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$ . Let $M$ denote the midpoint of chord $\overline{ST}$ . As $X$ varies on segment $\overline{PQ}$ , show that $M$ moves along a circle.

AoPS Link

Proof. Let $O$ be the center of $\omega$ and $C$ be the midpoint of $\overline{AO}$ . Let $\triangle MDE$ be the medial triangle of $\triangle AST$ (so $D,E$ be the midpoints of $\overline{AT},\overline{AS}$ ) and let $\Omega$ be the circle centered at $C$ which passes through all of $A,U,V,O$ by a homothety centered at $A$ with scale factor $2$ . By checking extreme cases ( $X=P,Q,AB\cap PQ$ ), we can guess the locus circle of $M$ , motivating us to prove the following claim, which will actually finish the problem.

Claim: As $X$ varies on $\overline{PQ}$ , $M$ moves along a circle centered at $C$ .
Proof. To show this, it suffices to prove that $CM^2$ does not depend on $X$ . We will do this by Barycentrics on $\triangle AST$ . Set $A=(1,0,0),S=(0,1,0),T=(0,0,1)$ and $ST=a,TA=s,AS=t$ . By the midpoint formula, we have $M=\left(0,\frac12,\frac12\right), D=\left(\frac12,0,\frac12\right),E=\left(\frac12,\frac12,0\right)$ .

Note that points $A,D,E$ are on $\Omega$ . The general equation of a circle is $-a^2yz-s^2zx-t^2xy+(x+y+z)(ux+vy+wz)=0$ for some constants $u,v,w$ . To find the constants $u,v,w$ which determine $\Omega$ , we can plug in each of $A,D,E$ in the general form and solve for $u,v,w$ . Note that the equation of a circle is homogenous so we can use homogenized coordinates here.

Plugging in $A=(1,0,0)$ gives $u=0$ .
Plugging in $D=(1:0:1)$ gives $u+v=\frac{s^2}2$ .
Plugging in $E=(1:1:0)$ gives $u+w=\frac{t^2}2$ .

Plugging in these constants in the general equation, we find that $\Omega$ has equation $-a^2yz-s^2zx-t^2xy+(x+y+z)\left(\frac{s^2}2\cdot y+\frac{t^2}2\cdot z\right)=0.$ Hence, using Barycentric Power of a Point, we can compute $\text{Pow}_{\Omega}(M)=-a^2\left(\frac12\right)\left(\frac12\right)-s^2\left(\frac12\right)(0)-t^2(0)\left(\frac12\right)+\left(0+\frac12+\frac12\right)\left(\frac{s^2}2\cdot \left(\frac12\right)+\frac{t^2}2\cdot \left(\frac12\right)\right)=\frac12\left(\frac{-a^2+s^2+t^2}2\right).$ By the Law of Cosines, $\frac{-a^2+s^2+t^2}2=s\cdot t\cdot\cos \angle SAT$ but note that we have $s\cdot\cos \angle SAT=s\cdot\cos \angle XAT=AX$ since $\triangle AXT$ is right. Hence $\begin{align*}\text{Pow}_{\Omega}(M)&=\frac12\left(\frac{-a^2+s^2+t^2}2\right)\\&= \frac12\left(s\cdot t\cdot\cos \angle SAT\right)\\&= \frac12(AS\cdot AX)\end{align*}$ , but note that $\triangle PAX\sim \triangle SAP$ in a negative orientation by AA similarity since $\measuredangle PAX=-\measuredangle SAP$ and $\measuredangle XPA=\measuredangle QPA=\measuredangle AQP=\measuredangle ASP=-\measuredangle PSA$ . Thus, by the similarity ratios, $AS\cdot AX=AP^2$ and we have $\text{Pow}_{\Omega}(M)=\frac12(AP^2)$ which does not depend on $X$ . Hence, $CM^2$ is fixed and we can conclude that $M$ lies on a fixed circle centered at $C$ , as desired. $\square$ .

Fortunately, that claim finishes the problem, so we are done. $\blacksquare$ .

https://cdn.artofproblemsolving.com/attachments/0/7/bbe4125d011f8fb775e0b45ad9b95851cb4123.png

USAJMO 2011/5

by franzliszt, Jul 11, 2024, 6:04 PM

USAJMO 2011/5 wrote:

Points $A,B,C,D,E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$ , (ii) $P, A, C$ are collinear, and (iii) $DE \parallel AC$ . Prove that $BE$ bisects $AC$ .

AoPS Link

Proof. Rename some points and use phantom points. We will solve the following equivalent problem:

Quote:

In $\triangle ABC$ with circumcircle $\Omega$ , let $D$ be a point on $\Omega$ such that $AD$ is the $A$ -symmedian. Let $T$ be the midpoint of $AD$ and let $BT$ meet the line through $C$ parallel to $AD$ at $E$ . Show that $E$ lies on $\Omega$ .

We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . Since $D$ is on the $A$ -symmedian, it can be parameterized by $D=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (D)=0$ . Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$ , so plugging in $D$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$ . So $D=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$ Thus, the point at infinity along the line $A$ -symmedian has coordinates $P_\infty=(b^2+c^2:-b^2:-c^2).$

Normalize $D$ by multiplying each component by $\frac{1}{-a^2+2b^2+2c^2}$ . The midpoint formula on normalized $D$ and $A$ gives $T=(-a^2+b^2+c^2:b^2:c^2)$ . Note that $E$ is the intersection of cevians $BT$ and $CP_\infty$ . Points on $BT$ can be parameterized by $(-a^2+b^2+c^2:s:c^2)$ while points on $CP_\infty$ can be parameterized by $(b^2+c^2:-b^2:t)$ . Hence, their intersection must have coordinates $E=[(b^2+c^2)(-a^2+b^2+c^2):-b^2(-a^2+b^2+c^2):c^2(b^2+c^2)].$ It remains to check that $\text{Pow}_\Omega (E)=0$ . Plugging $E$ into the circumcircle equation yeilds $a^2[-b^2(-a^2+b^2+c^2)c^2(b^2+c^2)]+b^2[(b^2+c^2)(-a^2+b^2+c^2)c^2(b^2+c^2)]+c^2[(b^2+c^2)(-a^2+b^2+c^2)(-b^2)(-a^2+b^2+c^2)]$ which is clearly true after factoring.

AIME I 2019/15

by franzliszt, Jul 11, 2024, 6:02 PM

AIME I 2019/15 wrote:

Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ , $PB=3$ , $XY=11$ , and $PQ^2 = \tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

AoPS Link

Proof. Note by Radical Axis that the tangents from $A$ and $B$ and the line $XY$ concur at a point we will call $Z$ . Let $AQ$ meet $\omega$ again at $D$ .

Lemma: In $\triangle ABC$ with circumcircle $\Omega$ , let $D$ be a point on $\Omega$ such that $AD$ is the $A$ -symmedian. Let $T$ be the midpoint of $AD$ and let $BT$ meet the line through $C$ parallel to $AD$ at $E$ . Show that $E$ lies on $\Omega$ . (USAJMO 2011/5)

Proof. We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . Since $D$ is on the $A$ -symmedian, it can be parameterized by $D=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (D)=0$ . Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$ , so plugging in $D$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$ . So $D=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$ Thus, the point at infinity along the line $A$ -symmedian has coordinates $P_\infty=(b^2+c^2:-b^2:-c^2).$
Normalize $D$ by multiplying each component by $\frac{1}{-a^2+2b^2+2c^2}$ . The midpoint formula on normalized $D$ and $A$ gives $T=(-a^2+b^2+c^2:b^2:c^2)$ . Note that $E$ is the intersection of cevians $BT$ and $CP_\infty$ . Points on $BT$ can be parameterized by $(-a^2+b^2+c^2:s:c^2)$ while points on $CP_\infty$ can be parameterized by $(b^2+c^2:-b^2:t)$ . Hence, their intersection must have coordinates $E=[(b^2+c^2)(-a^2+b^2+c^2):-b^2(-a^2+b^2+c^2):c^2(b^2+c^2)].$ It remains to check that $\text{Pow}_\Omega (E)=0$ . Plugging $E$ into the circumcircle equation yields $a^2[-b^2(-a^2+b^2+c^2)c^2(b^2+c^2)]+b^2[(b^2+c^2)(-a^2+b^2+c^2)c^2(b^2+c^2)]+c^2[(b^2+c^2)(-a^2+b^2+c^2)(-b^2)(-a^2+b^2+c^2)]$ which is clearly true after factoring. $\square$

Consider the homothety centered at $A$ that maps $\triangle AQP\mapsto ADB$ . By construction, $QP\parallel DB$ so we can conclude that $Q$ is the midpoint of $\overline{XY}$ by our lemma.

Unfortunately, we cannot do the answer extraction with bary since we are not given the side lengths of $\triangle XAB$ , but Power of a Point on $P$ with respect to $\omega$ and minimal computations give away an answer of $PQ^2=\frac{61}{4}$ or $\boxed{65}.$

2017 AMC 12B Problem 15

by franzliszt, Jul 11, 2024, 6:00 PM

2017 AMC 12B Problem 15 wrote:

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB' = 3AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC' = 3BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA' = 3CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ ?

$\textbf{(A) }9:1\qquad\textbf{(B) }16:1\qquad\textbf{(C) }25:1\qquad\textbf{(D) }36:1\qquad\textbf{(E) }37:1$

AoPS Link

Proof. Use barycentrics on $\triangle ABC$ . Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ so that $A'=(4,0,-3),B'=(-3,4,0),C'=(0,-3,4)$ . Then by the area formula, $\frac{[A'B'C']}{[ABC]}=\begin{vmatrix}4&0&-3\\ -3&4&0\\ 0&-3&4\end{vmatrix}=\frac{37}{1}$ or $\textbf{(E)} \text{ }37:1$ .

This post has been edited 1 time. Last edited by franzliszt, Jul 11, 2024, 6:00 PM

AIME II 2022/11

by franzliszt, Jul 11, 2024, 5:58 PM

AIME II 2022/11 wrote:

Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

AoPS Link

Proof. Let $AB\cap DC=P$ , let $M$ be the midpoint of $BC$ , and employ barycentrics on $\triangle PAD$ with $P=(1,0,0),A=(0,1,0),D=(0,0,1)$ . Let $PB=b-2$ and $PC=c-3$ and $AD=a=7$ . Then we can crank out $B=(2:b-2:0)$ and $C=(3:0:c-3)$ . Normalizing, we have $B=\left(\frac2b,\frac{b-2}b,0\right)$ and $C=\left(\frac3c,0,\frac{c-3}c\right)$ . Since $M$ is the midpoint of $BC$ , it has coordinates $M=\frac{B+C}2=\left(\frac{3b+2c}{2bc},\frac{bc-2c}{2bc},\frac{bc-3b}{2bc} \right).$ However, $M$ is also the incenter, so it has coordinates $M=(a:c:b)=\left(\frac{7}{7+b+c},\frac{c}{7+b+c},\frac{b}{7+b+c} \right).$ Hence, we have $M=\left(\frac{3b+2c}{2bc},\frac{bc-2c}{2bc},\frac{bc-3b}{2bc} \right)=\left(\frac{7}{7+b+c},\frac{c}{7+b+c},\frac{b}{7+b+c} \right)$ and we can compare components to get the vector-component equation $\left(\frac{3b+2c}{2bc},\frac{bc-2c}{2bc},\frac{bc-3b}{2bc} \right)-\left(\frac{7}{7+b+c},\frac{c}{7+b+c},\frac{b}{7+b+c} \right)=(3 b^2 - 9 b c + 21 b + 2 c^2 + 14 c,c(b^2 - b c + 5 b - 2 c - 14),b(-b c - 3 b + c^2 + 4 c - 21))=(0,0,0).$ where we used unhomogenized coordinates. We can solved this to get $(b,c)=(8,9)$ . So $B=(2:6:0)$ and $C=(3:0:6)$ and we can normalize these to $B=(1/4,3/4,0)$ and $C=(1/3,0,2/3)$ .

Note that the semiperimiter of $\triangle ABC$ is $\frac{7+8+9}2=12$ . By Heron's, we have $[ABC]=\sqrt{12(12-7)(12-8)(12-9)}=12\sqrt{5}$ . To finish, note that $[ABCD]=[BAD]+[DCB]$ . We have $[BAD]=\begin{vmatrix}1/4&3/4&0\\ 0&1&0\\ 0&0&1\end{vmatrix}\cdot[ABC]=\frac14\cdot 12\sqrt{5}=3\sqrt5$ and $[DCB]=\begin{vmatrix}0&0&1\\ 1/3&0&2/3\\ 1/4&3/4&0\end{vmatrix}\cdot[ABC]=\frac14\cdot 12\sqrt{5}=3\sqrt5$ so $[ABCD]=6\sqrt5$ , so the answer is $\boxed{180}$ .

AIME I 2016/6

by franzliszt, Jul 11, 2024, 5:57 PM

AIME I 2016/6 wrote:

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 $IC = \tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ .

AoPS Link

Proof. Switch to $A$ -indexing so that $C\mapsto A$ and $A\mapsto C$ in the original problem statement. We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . By the angle bisector theorem, $\begin{align*}L&=(0:b:c)\\I&=(a:b:c)=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)\\I_a&=(-a:b:c)=\left(\frac{-a}{-a+b+c},\frac{b}{-a+b+c},\frac{c}{-a+b+c}\right).\end{align*}$ By Fact 5, $D$ is the midpoint of $II_a$ so $D=\frac{I+I_a}2=\frac12\cdot\frac{1}{(a+b+c)(-a+b+c)}[(-a^2+ab+ac,-ab+b^2+bc,-ac+bc+c^2)+(-a^2-ab-ac,ab+b^2+bc,ac+bc+c^2)].$ Thus, $D=(-a^2:b^2+bc:c^2+bc).$
Note that $\overrightarrow{LI}=\vec{I}-\vec{L}=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)-\left(0,\frac{b}{b+c},\frac{c}{b+c}\right)=\left(\frac{a}{a+b+c},\frac{b}{a+b+c}-\frac{b}{b+c},\frac{c}{a+b+c}-\frac{c}{b+c}\right)=\frac{1}{(b+c)(a+b+c)}\cdot(ab+ac,-ab,-ac).$ By the barycentric distance formula, $|\overrightarrow{LI}|^2=-a^2yz-b^2zx-c^2xy=\left(\frac{1}{(b+c)(a+b+c)}\right)^2[-a^2(-ab)(-ac)-b^2(-ac)(ab+ac)-c^2(ab+ac)(-ab)]=\left(\frac{a\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{\left(b+c\right)\left(b+c+a\right)}\right)^2$ so $LI=\frac{a\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{\left(b+c\right)\left(b+c+a\right)}=\frac{a}{b+c}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right).$
By more barycentric distance formula computations, $AI=\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}$ and similar calculations give $AI_A=\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c-a}=\frac{b+c+a}{b+c-a}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right).$ Furthermore, $DI_A=ID=\frac{II_A}2=\frac{AI_A-AI}2=\frac12\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c-a}-\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)=\frac{a\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{(b+c-a)(b+c+a)}=\frac{a}{b+c-a}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)$ and $LD=ID-IL=\frac{a}{b+c-a}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)-\frac{a}{b+c}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)=\frac{a^2}{(b + c) (b + c-a)}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right).$
Summarizing,

$AI=\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}$
$IL=\frac{a}{b+c}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)$
$LD=\frac{a^2}{(b + c) (b + c-a)}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)$
$DI_A=\frac{a}{b+c-a}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)$ .

After noting that $IL+LD=ID=DI_A$ and substituting $AI=\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}$ , we have $IL+LD=\frac{a}{b+c-a}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)=\frac{a}{b+c-a}AI$ but we also have $\frac{LD}{IL}=\frac{\frac{a^2}{(b + c) (b + c-a)}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)}{\frac{a}{b+c}\left(\frac{\sqrt{bc\left(\left(b+c\right)^2-a^2\right)}}{b+c+a}\right)}=\frac{a}{b+c-a}.$ Combining, $AI=\frac{IL}{LD}(IL+LD)=\frac23(2+3)=\frac{10}3\Rightarrow\boxed{13}.$
$[asy] import cse5; size(7cm); pathpen = black; pointpen = black; pair A = dir(200), B = dir(340), C = dir(120), D = dir(degrees(A)/2 + degrees(B)/2), I = incenter(A,B,C), L = extension(A,B,C,D), O = origin; Drawing(A--B--C--cycle); Drawing(unitcircle,rgb(0,0.6,1)); Drawing(C--D); // Show AD and BD. //Drawing(A--D--B,rgb(0,0.4,0)); // Show circumcircle of AIB //Drawing(CirclebyPoint(D,I,degrees(B-D)-15,degrees(A-D)+15),rgb(0.7,0.4,1)); // Show IA, IB //Drawing(A--I--B,rgb(1,0.4,0.1)); // IX, where X is foot from I to AB //Drawing(I--foot(I,A,B),rgb(1,0.2,0.4)); //Drawing("X",foot(I,A,B),dir(250)); // Show OI (be sure to show O) //Drawing(O--I,rgb(0,0.4,0)); // OD (passing through M) (be sure to show O and M) //Drawing(O--D,rgb(1,0.2,0.4)); // Show M, midpoint of AB //Drawing("M",(A+B)/2,dir(45)); // Show circumcenter //Drawing("O",O,dir(90)); Drawing("B",A,A); Drawing("C",B,B); Drawing("A",C,C); Drawing("D",D,D); Drawing("L",L,dir(50)); Drawing("I",I,dir(50)); [/asy]$

AIME II 2012/15

by franzliszt, Jul 11, 2024, 5:54 PM

AIME II 2012/15 wrote:

Triangle $ABC$ is inscribed in circle $\omega$ with $AB = 5$ , $BC = 7$ , and $AC = 3$ . The bisector of angle $A$ meets side $BC$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $DE$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where m and n are relatively prime positive integers. Find $m + n$ .

AoPS Link

Proof. Employ Barycentric Coordinates on $\triangle ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . My solution to Russia 2009 tells us that $F=(-a^2:2b^2:2c^2)$ . Of course, $\overrightarrow{AF}=\left(\frac{-a^2}{-a^2+2b^2+2c^2}-1,\frac{2b^2}{-a^2+2b^2+2c^2},\frac{2c^2}{-a^2+2b^2+2c^2}\right)$ so $|\overrightarrow{AF}|^2=-a^2\left(\frac{2b^2}{-a^2+2b^2+2c^2}\right)\left(\frac{2c^2}{-a^2+2b^2+2c^2}\right)-b^2\left(\frac{2c^2}{-a^2+2b^2+2c^2}\right)\left(\frac{-a^2}{-a^2+2b^2+2c^2}-1\right)-c^2\left(\frac{-a^2}{-a^2+2b^2+2c^2}-1\right)\left(\frac{2b^2}{-a^2+2b^2+2c^2}\right)$ by the barycentric distance formula. Fortunately, the sides of this triangle are small integers and after a bit of computation, we find $AF^2 = \frac{900}{19} \rightarrow \boxed{919}$ .

AIME I 2020/13

by franzliszt, Jul 11, 2024, 5:51 PM

AIME I 2020/13 wrote:

Point $D$ lies on side $BC$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC$ . The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F$ , respectively. Given that $AB=4$ , $BC=5$ , $CA=6$ , the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt n}p$ , where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p$ .

AoPS Link

Proof. Apply barycentrics on $\triangle ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . We can get the coordinates of $D=\left(0,\frac{3}{5},\frac{2}{5}\right)$ and $M=\left(\frac{5}{10},\frac{3}{10},\frac{2}{10}\right)$ by the Angle-Bisector Theorem and Midpoint Formula, respectively.

Now, let $T=(t,1-t,0)$ be a point on side $AB$ such that $MT\perp AD$ . We have $\overrightarrow{AD}=\left(-1,\frac{3}{5},\frac{2}{5}\right)=(-5,3,2)$ and $\overrightarrow{MT}=\left(\frac{5}{10}-t,\frac{3}{10}-(1-t),\frac{2}{10}-0\right)=(5-10t,-7+10t,2)$ . Observe that these vectors still have coefficient sum $0$ so we can apply EFFT. We find that $MT\perp AD$ iff $0=5^2(3(2)+2(-7+10t))+6^2(2(5-t)+-5(2))+4^2(-5(2+t)+3(5-t))$ so $t=\frac25$ . Thus, $T=(2:3:0)$ .

Let the incenter have coordinates $I=(a:b:c)=(5:6:4)$ . Since $E$ lies on cevian $BI$ , we can parameterize it with $E=(5:t:4)$ for some $t\in\mathbb{R}$ . However, $E$ also lies on line $MT$ so we must have $\begin{vmatrix}2&3&0\\ 5&3&2\\ 5&t&4\end{vmatrix}=0 \iff 2(12-2t)+3(10-20)=0\iff t=-\frac32.$ Thus, $E$ has coordinates $E=\left(5:-\frac32,4\right)$ . Homogenized, this becomes $E=\left(\frac{2}{3},-\frac{1}{5},\frac{8}{15}\right)$ .

Similar work shows that $F=\left(\frac{3}{7},\frac{18}{35},\frac{2}{35}\right)$ .

The barycentric area formula tells us $\frac{[AEF]}{[ABC]}=\begin{vmatrix}1&0&0\\\frac{2}{3}&-\frac{3}{15}&\frac{8}{15}\\ \frac{3}{7}&\frac{18}{35}&\frac{2}{35}\end{vmatrix}=\frac{2}{7}$ where we take the absolute value since areas are signed. However, by Heron's, $[ABC] = \frac{15\sqrt{7}}{4}$ , so $[AEF]=\frac27\cdot[ABC] = \frac{15\sqrt{7}}{14}\implies \boxed{036}.$

AIME I 2009/4

by franzliszt, Jul 11, 2024, 5:50 PM

AIME I 2009/4 wrote:

In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac{AM}{AB} = \frac{17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac{AN}{AD} = \frac{17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac{AC}{AP}$ .

AoPS Link

Proof. We employ barycentrics with reference triangle $ABD$ . Set $A=(1,0,0),B=(0,1,0),D=(0,0,1)$ . By the parallelogram lemma, $C=(-1,1,1)$ . Clearly we have $\frac{AM}{MB}=\frac{17}{1000-17}=\frac{17}{983}$ and $\frac{AN}{ND}=\frac{17}{2009-17}=\frac{17}{1992}$ so $M=(983:17:0)$ and $N=(1992:0:17)$ . The point $P$ is parameterized by $(1-2t,t,t)$ . Since $N,P,M$ are colinear, we must have $\begin{align*}0&=\begin{vmatrix} 983&17&0\\1992&0&17\\1-2t&t&t\end{vmatrix}\\ &= 983(-17t)+17[17(1-2t)-1992t]\\&=-3009t+17 \end{align*}$ so $t=\frac{17}{3009}=\frac{1}{177}$ . So $\frac{AC}{AP}=\frac{1}{\frac{1}{177}}=177$ .

This post has been edited 1 time. Last edited by franzliszt, Jul 11, 2024, 5:52 PM

Geometry through Barycentric Coordinates