2017 USAJMO Problems/Problem 3

Revision as of 15:14, 10 May 2021 by Sakshamsethi (talk | contribs) (Solution 5)

Problem

($*$) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$; let lines $PB$ and $CA$ intersect at $E$; and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of triangle $DEF$ is twice that of triangle $ABC$.

[asy]     size(3inch);     pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3));     draw(Circle(O, 2sqrt(3)), black);     draw(A--B--C--cycle);     draw(B--E1--C);     draw(C--F--B);     draw(A--P);     draw(D--E1--F--cycle, dashed);     label("A", A, N);     label("B", B, W);     label("C", C, E);     label("P", P, S);     label("D", D, NW);     label("E", E1, SE);     label("F", F, SW); [/asy]

Solution (No Trig/Bash)

Extend $DP$ to hit $EF$ at $K$. Then note that $[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].$ Letting $BF=x$ and $PF=y$, we have that $\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.$ Solving and simplifying using LoC on $\triangle BPC$ gives $\frac{AB}{AF}=\frac{PC}{PB+PC}.$ Similarly, $\frac{AC}{AE}=\frac{PB}{PB+PC}.$


Now we find $\frac{AK}{DK}.$ Note that $\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.$ Now let $E'=DE\cap AF$ and $F'=DF\cap AE$. Then by an area/concurrence theorem, we have that $\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,$ or $\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.$ Thus we have that $\frac{DK}{AK}=2\cdot\frac{DP}{AP}.$


Manipulating these gives $\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.$ Thus $\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,$ and we are done.

~cocohearts

Solution 1

WLOG, let $AB = 1$. Let $[ABD] = X, [ACD] = Y$, and $\angle BAD = \theta$. After some angle chasing, we find that $\angle BCF \cong \angle BEC \cong \theta$ and $\angle FBC \cong \angle BCE \cong 120^{\circ}$. Therefore, $\triangle FBC$ ~ $\triangle BCE$.


Lemma 1: If $BF = k$, then $CE = \frac 1k$. This lemma results directly from the fact that $\triangle FBC$ ~ $\triangle BCE$; $\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}$, or $CE = \frac{1}{BF}$.


Lemma 2: $[AEF] = (k+\frac 1k + 2)(X+Y)$. We see that $[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)$, as desired.


Lemma 3: $\frac{X}{Y} = k$. We see that \[\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.\] However, after some angle chasing and by the Law of Sines in $\triangle BCF$, we have $\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}$, or $k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}$, which implies the result.


By the area lemma, we have $[BDF] = kX$ and $[CDF] = \frac{Y}{k}$.

We see that $[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk$. Thus, it suffices to show that $X + Y + \frac Xk + Yk = 2X + 2Y$, or $\frac Xk + Yk = X + Y$. Rearranging, we find this to be equivalent to $\frac XY = k$, which is Lemma 3, so the result has been proven.

Solution 2

We will use barycentric coordinates and vectors. Let $\vec{X}$ be the position vector of a point $X.$ The point $(x, y, z)$ in barycentric coordinates denotes the point $x\vec{A} + y\vec{B} + z\vec{C}.$ For all points in the plane of $\triangle{ABC},$ we have $x + y + z = 1.$ It is clear that $A = (1, 0, 0)$; $B = (0, 1, 0)$; and $C = (0, 0, 1).$

Define the point $P$ as $P = \left(x_P, y_P, z_P\right).$ The fact that $P$ lies on the circumcircle of $\triangle{ABC}$ gives us $x^2_P + y^2_P + z^2_P = 1.$ This, along with the condition $x_P + y_P + z_P = 1$ inherent to barycentric coordinates, gives us $x_Py_P + y_Pz_P + z_Px_P = 0.$

We can write the equations of the following lines: \[BC: x = 0\] \[CA: y = 0\] \[AB: z = 0\] \[PA: \frac{y}{y_P} = \frac{z}{z_P}\] \[PB: \frac{x}{x_P} = \frac{z}{z_P}\] \[PC: \frac{x}{x_P} = \frac{y}{y_P}.\]

We can then solve for the points $D, E, F$: \[D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)\] \[E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)\] \[F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).\]

The area of an arbitrary triangle $XYZ$ is: \[[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|\] \[[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.\]

To calculate $[DEF],$ we wish to compute $(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).$ After a lot of computation, we obtain the following: \[(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].\]

Evaluating the denominator, \[(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)\] \[(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.\]

Since $x_P + y_P + z_P = 1$ and $x_Py_P + y_Pz_P + z_Px_P = 0,$ it follows that: \[(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.\]

We thus conclude that: \[(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].\]

From this, it follows that $[DEF] = 2[ABC],$ and we are done.


Solution 3

[asy]    import cse5;    import graph;    import olympiad;      size(3inch);           pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3));     pair P = (-1, -sqrt(11)+sqrt(3));     path circle = Circle(O, 2sqrt(3));     pair D = extension(A,P,B,C);     pair E1 = extension(A,C,B,P);     pair F=extension(A,B,C,P);     draw(circle, black);     draw(A--B--C--cycle);     draw(B--E1--C);     draw(C--F--B);     draw(A--P);     draw(D--E1--F--cycle, dashed);     pair G = extension(O,D,F,E1);     draw(O--G,dashed);     label("A", A, N);     label("B", B, W);     label("C", C, E);     label("P", P, S);     label("D", D, NW);     label("E", E1, SE);     label("F", F, SW);     dot("O", O, SE); [/asy]

We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\sqrt{3})$. Using the facts $\triangle{CBP} \sim  \triangle{CFB}$ and $\triangle{BCP} \sim  \triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \frac{1}{2x}$, and $E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})$.

The slope of $FE$ is \[k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}\] It is well-known that $\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\frac{1}{k}$. Since $O=(0,\frac{1}{\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \frac{k}{\sqrt{3}}$, i.e., $D = (\frac{k}{\sqrt{3}},0)$. Using shoelace, \[2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})-   (-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}}\] \[= 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)\] \[= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2} {2+\frac{1}{x}+x})\] \[= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})\] \[= 4\sqrt{3}\] So $[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]$. Q.E.D

By Mathdummy.

Solution 4 Without the nasty computations

Note that $\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths.

Let $BP = x$ and $CP = y$. Then, From Law of Cosine, $BC^2 = x^2 + y^2 + xy$.

From Ptolemy's theorem, $AP BC = x AC + y AB$, so $AP = x + y$.

Lemma 1: In Triangle ABC with side lengths $a,b,c$ and $\angle A =120^o$, the length of the angle bisector of $A$ is \[d = \frac{bc}{b+c}\] This can be easily proved with Stewart's and Law of Cosine.

Using Lemma 1, we have \[PD = \frac{xy}{x+y}\] \[x = \frac{FP AP}{FP + AP}\] \[y = \frac{EP AP}{EP + AP}\] Plug in $AP=x+y$, we get: \[PD = \frac{xy}{x+y}\] \[FP = \frac{x(x+y)}{y}\] \[EP = \frac{y(x+y)}{x}\] Then \[[\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)\] \[= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2)\] \[= \frac{\sqrt{3}}{2}(x^2 + xy + y^2)\] \[= \frac{\sqrt{3}}{2} BC^2\] \[= 2 [\triangle{ABC}]\]

By Mathdummy.

Solution 5

Extend AP to meet FE at point X. Note that there is symmetry of triangle $\triangle{AFE}$ across line AX. Therefore, if we can prove that the area of $\triangle{FDX}$ is twice the area of $\triangle{BAD}$, then we will be done. For simplicity, let the side length of equilateral triangle $\triangle{ABC}$ be $1$.

We can easily find the area of $\triangle{BAD}$. Since the area formula for an equilateral triangle is $\frac{s^2 \sqrt{3}}{4}$ , where $s$ is the side length, the area of $\triangle{ABC}$ is $\frac{\sqrt{3}}{4}$. By symmetry, the area of $\triangle{BAD}$ is exactly half the area of $\triangle{ABC}$. Thus the area of $\triangle{BAD} = \frac{\frac{\sqrt{3}}{4}}{2} = \boxed{\frac{\sqrt{3}}{8}}$.

Now we just need to find the area of $\triangle{FDX}$. We have to find $FX$ and $DX$ for the area. Note that $FX = \frac{FE}{2}$. It looks by eye that $\triangle{AFE}$ is equilateral. Let's try to prove it.



$\textbf{\textit{Lemma 1:$ (Error compiling LaTeX. ! File ended while scanning use of \textbf .)\triangle{AFE}$is equilateral.}}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

$\textbf{\textit{Proof of Lemma 1:}}$ By symmetry, across lines $AX, FC, BE$, we arrive to some conclusions: \begin{center}

       $AF=AE$
       
       $AF = FE$
       
       $FE = AE$. 
       \end{center}
       
       Substituting the first equation into the second, we see that 					      $AF=AE=FE$, which is necessary and sufficient to prove that $\triangle{AFE}$ is equilateral. $\square$ 
       


Since $\triangle{AFE}$ is equilateral, and so is $\triangle{ABC}$, we find that $\triangle{ABC} \sim \triangle{AFE}$. Because $AB = \frac{AF}{2}$ by symmetry across $BE$, we know that the similarity ratio is $\frac{1}{2}$.
Thus the side length of $\triangle{AFE}$ is $2$, and $FX = \frac{FE}{2} = 1$.



Now we need to find $DX$. We can do $AX-AD$. Well, that's easy! Since $\triangle{BAD}$ is a 30-60-90 triangle by symmetry across line $AX$, and $BD = \frac{BC}{2} = \frac{1}{2}$, we know that $AD = \frac{\sqrt{3}}{2}$. To find $AX$, we do pretty much the same thing: $FAX$ is a 30-60-90 triangle, and $FX = \frac{2}{2} = 1$, so $AX = \sqrt{3}$. We can now find $DX$: 
\begin{center}
$DX = AX - AD = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$. 
\end{center}




Recalling why we were doing all this, we find the area of $\triangle{FDX}$. It is $\frac{FX \times DX}{2} = \frac{1 \times \frac{\sqrt{3}}{2}}{2} = \boxed{\frac{\sqrt{3}}{4}}$. 



Let's summarize: 

\begin{center}
The area of $\triangle{ABC}$ is $\boxed{\frac{\sqrt{3}}{8}}$

The area of $\triangle{FDX}$ is $\boxed{\frac{\sqrt{3}}{4}}$, 

which proves our initial claim. 



\end{center}

$\square{}$

See also

2017 USAJMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAJMO Problems and Solutions
Invalid username
Login to AoPS