1989 AIME Problems/Problem 15


Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure below). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$.

AIME 1989 Problem 15.png


Solution 1 (Ceva's Theorem, Stewart's Theorem)

Let $[RST]$ be the area of polygon $RST$. We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$, and line $XP$ intersects line $YZ$ at point $L$, then $\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.$

[asy] size(170); pair X = (1,2), Y = (0,0), Z = (3,0); real x = 0.4, y = 0.2, z = 1-x-y; pair P = x*X + y*Y + z*Z; pair L = y/(y+z)*Y + z/(y+z)*Z; draw(X--Y--Z--cycle); draw(X--P); draw(P--L, dotted); draw(Y--P--Z); label("$X$", X, N); label("$Y$", Y, S); label("$Z$", Z, S); label("$P$", P, NE); label("$L$", L, S);[/asy]

This is true because triangles $XPY$ and $YPL$ have their areas in ratio $XP:PL$ (as they share a common height from $Y$), and the same is true of triangles $ZPY$ and $LPZ$.

We'll also use the related fact that $\dfrac{[XPY]}{[ZPX]} = \dfrac{YL}{LZ}$. This is slightly more well known, as it is used in the standard proof of Ceva's theorem.

Now we'll apply these results to the problem at hand.

[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label("$A$", A, S); label("$B$", B, S); label("$C$", C, N); label("$D$", D, NE); label("$E$", EE, NW); label("$F$", F, S); label("$P$", P, E); [/asy]

Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\triangle BPC$ has half the area of $\triangle ABC$. And since $PE = 3 = \dfrac{1}{3}BP$, we can conclude that $\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\dfrac{1}{4}$ of the area of $\triangle ABC$. This means that $\triangle APB$ is left with $\dfrac{1}{4}$ of the area of triangle $ABC$: \[[BPC]: [APC]: [APB] = 2:1:1.\] Since $[APC] = [APB]$, and since $\dfrac{[APC]}{[APB]} = \dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$.

Furthermore, we know that $\dfrac{CP}{PF} = \dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \dfrac{3}{4} \cdot CF = 15$.

We now apply Stewart's theorem to segment $PD$ in $\triangle BPC$—or rather, the simplified version for a median. This tells us that \[2 BD^2 + 2 PD^2 = BP^2+ CP^2.\] Plugging in we know, we learn that \begin{align*} 2 BD^2 + 2 \cdot 36 &= 81 + 225 = 306, \\ BD^2 &= 117. \end{align*} Happily, $BP^2 + PD^2 = 81 + 36$ is also equal to 117. Therefore $\triangle BPD$ is a right triangle with a right angle at $B$; its area is thus $\dfrac{1}{2} \cdot 9 \cdot 6 = 27$. As $PD$ is a median of $\triangle BPC$, the area of $BPC$ is twice this, or 54. And we already know that $\triangle BPC$ has half the area of $\triangle ABC$, which must therefore be $\boxed{108}$.

Solution 2 (Mass Points, Stewart's Theorem, Heron's Formula)

Because we're given three concurrent cevians and their lengths, it seems very tempting to apply Mass points. We immediately see that $w_E = 3$, $w_B = 1$, and $w_A = w_D = 2$. Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C = 1$ and $w_F = 3$. Thus, $CP = 15$ and $PF = 5$.

Recalling that $w_C = w_B = 1$, we see that $DC = DB$ and $DP$ is a median to $BC$ in $\triangle BCP$. Applying Stewart's Theorem, we have the following: \[\frac{BC}{2}(9^2+15^2)=BC(6^2+ \left(\frac{BC}{2} \right)^2).\] Eliminating $BC$ on both sides, we have: \[\frac 12(9^2+15^2)=6^2+ \left(\frac{BC}{2} \right)^2.\] Combining terms and simplifying numbers, we have: \[153=36+\left(\frac{BC}{2} \right)^2.\] Subtracting 36 to the other side yields: \[117= \left(\frac{BC}{2} \right)^2.\] Finishing it off from there, we find that $BC=2 \sqrt{117}.$ Now, notice that $2[BCP] = [ABC]$, because both triangles share the same base, $BC$ and $h_{\triangle ABC} = 2h_{\triangle BCP}$. Applying Heron's formula on triangle $BCP$ with sides $15$, $9$, and $2\sqrt{117}$, we have: \[\sqrt{(\sqrt{117}+12)(\sqrt{117}+12-9)(\sqrt{117}+12-15)(\sqrt{117}+12-2\sqrt{117})}.\] Combining terms results in: \[\sqrt{(\sqrt{117}+12)(\sqrt{117}+3)(\sqrt{117}-3)(-\sqrt{117}+12)}.\] Notice that these factors can be grouped into a difference of squares: \[\sqrt{(144-\sqrt{117}^2)(\sqrt{117}^2-9)}.\] Since $\sqrt{117}^2=117$, we have: \[\sqrt{(27)(108)}.\] After simplifying this radical, we find that it equals $54.$ Therefore, $[BCP] = 54$, and hence $[ABC]=2 \cdot 54= \boxed{108}$.

(The original author made a mistake in their solution. Corrected and further explained by dbnl.)

Solution 3 (Ceva's Theorem, Stewart's Theorem)

Using a different form of Ceva's Theorem, we have $\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}$

Solving $4y = x + y$ and $x + y = 20$, we obtain $x = CP = 15$ and $y = FP = 5$.

Let $Q$ be the point on $AB$ such that $FC \parallel QD$. Since $AP = PD$ and $FP\parallel QD$, $QD = 2FP = 10$. (Stewart's Theorem)

Also, since $FC\parallel QD$ and $QD = \frac{FC}{2}$, we see that $FQ = QB$, $BD = DC$, etc. (Stewart's Theorem) Similarly, we have $PR = RB$ ($= \frac12PB = 7.5$) and thus $RD = \frac12PC = 4.5$.

$PDR$ is a $3-4-5$ right triangle, so $\angle PDR$ ($\angle ADQ$) is $90^\circ$. Therefore, the area of $\triangle ADQ = \frac12\cdot 12\cdot 6 = 36$. Using area ratio, $\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = \boxed{108}$.

Solution 4 (Stewart's Theorem)

First, let $[AEP]=a, [AFP]=b,$ and $[ECP]=c.$ Thus, we can easily find that $\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.$ Now, $\frac{[ABP]}{[BPD]}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.$ In the same manner, we find that $[CPD]=a+c.$ Now, we can find that $\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.$ We can now use this to find that $\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.$ Plugging this value in, we find that $\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.$ Now, since $\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},$ we can find that $2AE=EC.$ Setting $AC=b,$ we can apply Stewart's Theorem on triangle $APC$ to find that $(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).$ Solving, we find that $b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.$ But, $3^2+6^2=45,$ meaning that $\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.$ Since $[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,$ we conclude that the answer is $\boxed{108}$.

Solution 5 (Mass of a Point, Stewart's Theorem, Heron's Formula)

Firstly, since they all meet at one single point, denoting the mass of them separately. Assuming $M(A)=6;M(D)=6;M(B)=3;M(E)=9$; we can get that $M(P)=12;M(F)=9;M(C)=3$; which leads to the ratio between segments, \[\frac{CE}{AE}=2;\frac{BF}{AF}=2;\frac{BD}{CD}=1.\] Denoting that $CE=2x;AE=x; AF=y; BF=2y; CD=z; DB=z.$

Now we know three cevians' length, Applying Stewart theorem to them, getting three different equations: \begin{align} (3x)^2 \cdot 2y+(2z)^2 \cdot y&=(3y)(2y^2+400), \\ (3y)^2 \cdot z+(3x)^2 \cdot z&=(2z)(z^2+144), \\ (2z)^2 \cdot x+(3y)^2 \cdot x&=(3x)(2x^2+144). \end{align} After solving the system of equation, we get that $x=3\sqrt{5};y=\sqrt{13};z=3\sqrt{13}$;

pulling $x,y,z$ back to get the length of $AC=9\sqrt{5};AB=3\sqrt{13};BC=6\sqrt{13}$; now we can apply Heron's formula here, which is \[\sqrt\frac{(9\sqrt{5}+9\sqrt{13})(9\sqrt{13}-9\sqrt{5})(9\sqrt{5}+3\sqrt{13})(9\sqrt{5}-3\sqrt{13})}{16}=108.\]

Our answer is $\boxed{108}$.


Note (how to find x and y without the system of equations)

To ease computation, we can apply Stewart's Theorem to find $x$, $y$, and $z$ directly. Since $M(C)=3$ and $M(F)=9$, $\overline{PC}=15$ and $\overline{PF}=5$. We can apply Stewart's Theorem on $\triangle CPE$ to get $(2x+x)(2x \cdot x) + 3^2 \cdot 3x = 15^2 \cdot x + 6^2\cdot 2x$. Solving, we find that $x=3\sqrt{5}$. We can do the same on $\triangle APB$ and $\triangle BPC$ to obtain $y$ and $z$. We proceed with Heron's Formula as the solution states.


Solution 6 (easier version of Solution 5)

In Solution 5, instead of finding all of $x, y, z$, we only need $y, z$. This is because after we solve for $y, z$, we can notice that $\triangle BAD$ is isosceles with $AB = BD$. Because $P$ is the midpoint of the base, $BP$ is an altitude of $\triangle BAD$. Therefore, $[BAD] = \frac{(AD)(BP)}{2} = \frac{12 \cdot 9}{2} = 54$. Using the same altitude property, we can find that $[ABC] = 2[BAD] = 2 \cdot 54 = \boxed{108}$.


Solution 7 (Mass Points, Stewart's Theorem, Simple Version)

Set $AF=x,$ and use mass points to find that $PF=5$ and $BF=2x.$ Using Stewart's Theorem on $APB,$ we find that $AB=3\sqrt{13}.$ Then we notice that $APB$ is right, which means the area of $APB$ is $27.$ Because $CF=4\cdot PF,$ the area of $ABC$ is $4$ times the area of $APB,$ which means the area of $ABC=4\cdot 27=\boxed{108}.$

Solution 8 (Ratios, Auxiliary Lines and 3-4-5 triangle)

We try to solve this using only elementary concepts. Let the areas of triangles $BCP$, $ACP$ and $ABP$ be $X$, $Y$ and $Z$ respectively. Then $\frac{X}{Y+Z}=\frac{6}{6}=1$ and $\frac{Y}{X+Z}=\frac{3}{9}=\frac{1}{3}$. Hence $\frac{X}{2}=Y=Z$. Similarly $\frac{FP}{PC}=\frac{Z}{X+Y}=\frac{1}{3}$ and since $CF=20$ we then have $FP=5$. Additionally we now see that triangles $FPE$ and $CPB$ are similar, so $FE \parallel BC$ and $\frac{FE}{BC} = \frac{1}{3}$. Hence $\frac{AF}{FB}=\frac{1}{2}$. Now construct a point $K$ on segment $BP$ such that $BK=6$ and $KP=3$, we will have $FK \parallel AP$, and hence $\frac{FK}{AP} = \frac{FK}{6} = \frac{2}{3}$, giving $FK=4$. Triangle $FKP$ is therefore a 3-4-5 triangle! So $FK \perp BE$ and so $AP \perp BE$. Then it is easy to calculate that $Z = \frac{1}{2} \times 6 \times 9 = 27$ and the area of triangle $ABC = X+Y+Z = 4Z = 4 \times 27 = \boxed{108}$. ~Leole

Solution 9 (Just Trig Bash)

We start with mass points as in Solution 2, and receive $BF:AF = 2$, $BD:CD = 1$, $CE:AE = 2$. Law of Cosines on triangles $ADB$ and $ADC$ with $\theta = \angle ADB$ and $BD=DC=x$ gives \[36+x^2-12x\cos \theta = 81\] \[36+x^2-12x\cos (180-\theta) = 36+x^2+12x\cos \theta = 225\] Adding them: $72+2x^2=306 \implies x=3\sqrt{13}$, so $BC = 6\sqrt{13}$. Similarly, $AB = 3\sqrt{13}$ and $AC = 9\sqrt{5}$. Using Heron's, \[[ \triangle ABC ]= \sqrt{\left(\dfrac{9\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{9\sqrt{13}09\sqrt{5}}{2}\right)\left(\dfrac{3\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{-3\sqrt{13}+9\sqrt{5}}{2}\right)} = \boxed{108}.\]


See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Final Question
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