Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AIME Problems/Problem 15"

m (→‎Solution 5 (Mass of a Point, Stewart's Theorem, Heron's Formula))
(→‎Solution 1)
Line 64: Line 64:
 
2 BD^2 + 2 \cdot 36 &= 81 + 225 = 306, \\
 
2 BD^2 + 2 \cdot 36 &= 81 + 225 = 306, \\
 
BD^2 &= 117. \end{align*} </cmath>
 
BD^2 &= 117. \end{align*} </cmath>
−
Happily, <math>BP^2 + PD^2 = 81 + 36</math> is also equal to 117.  Therefore <math>\triangle BPD</math> is a right triangle with a right angle at <math>B</math>; its area is thus <math>\dfrac{1}{2} \cdot 9 \cdot 6 = 27</math>.  As <math>PD</math> is a median of <math>\triangle BPC</math>, the area of <math>BPC</math> is twice this, or 54.  And we already know that <math>\triangle BPC</math> has half the area of <math>\triangle ABC</math>, which must therefore be 108.
+
Happily, <math>BP^2 + PD^2 = 81 + 36</math> is also equal to 117.  Therefore <math>\triangle BPD</math> is a right triangle with a right angle at <math>B</math>; its area is thus <math>\dfrac{1}{2} \cdot 9 \cdot 6 = 27</math>.  As <math>PD</math> is a median of <math>\triangle BPC</math>, the area of <math>BPC</math> is twice this, or 54.  And we already know that <math>\triangle BPC</math> has half the area of <math>\triangle ABC</math>, which must therefore be <math>\boxed{108}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 21:56, 18 December 2022

Problem

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

Solutions

Solution 1

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

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, $BC^2 + 12^2 = 2(15^2 + 9^2)$, and $BC = 6\sqrt {13}$. Now notice that $2[BCP] = [ABC]$, because both triangles share the same base and the $h_{\triangle ABC} = 2h_{\triangle BCP}$. Applying Heron's formula on triangle $BCP$ with sides $15$, $9$, and $6\sqrt{13}$, $[BCP] = 54$ and $[ABC] = \boxed{108}$.

Solution 3

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

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 $\boxed{108}$.

~bluesoul

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Final Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_15&oldid=184244"