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

Revision as of 01:29, 10 April 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 $\triangle ABC$ .

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 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

Because the length of cevian $BE$ is unknown, we can examine what happens when we extend it or decrease its length and see that it simply changes the angles between the cevians. Wouldn't it be great if it the length of $BE$ was such that $\angle APC = 90^\circ$ ? Let's first assume it's a right angle and hope that everything works out.

Extend $AD$ to $Q$ so that $PD = DQ = 6$ . The result is that $BQ = 9$ , $PQ = 12$ , and $BP = 15$ because $\triangle CDP\cong \triangle BDQ$ . Now we see that if we are able to show that $BE = 20$ , that is $PE = 5$ , then our right angle assumption will be true.

Apply the Pythagorean Theorem on $\triangle APC$ to get $AC = 3\sqrt {13}$ , so $AE = \sqrt {13}$ and $CE = 2\sqrt {13}$ . Now, we apply the Law of Cosines on triangles $CEP$ and $AEP$ .

Let $PE = x$ . Notice that $\angle CEB = 180^\circ - \angle AEB$ and $\cos CEB = - \cos AEB$ , so we get two nice equations.

$81 = 52 + y^2 - 2y \sqrt {13}\cos CEF$ $36 = 13 + y^2 + y \sqrt {13} \cos CEF$

Solving, $y = 5$ (yay!).

Now, the area is easy to find. $[ABC] = [AQB] + [APC] = \frac12(9)(18) + \frac12(6)(9) = \boxed{108}$ .

[however, I think this solution is wrong, the A, B, and Cs do not match with the picture]

Solution 5

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+heron)

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:

$(1): (3x)^2 * 2y+(2z)^2 * y=(3y)(2y^2+400)\\ (2): (3y)^2 * z+(3x)^2 * z=(2z)(z^2+144)\\ (3): (2z)^2 * x+(3y)^2 * x=(3x)(2x^2+144)$

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

@@ Line 5: / Line 5: @@
 == Solutions ==
-== Solution 1 ==
+=== Solution 1 ===
 Let <math>[RST]</math> be the area of polygon <math>RST</math>.  We'll make use of the following fact: if <math>P</math> is a point in the interior of triangle <math>XYZ</math>, and line <math>XP</math> intersects line <math>YZ</math> at point <math>L</math>, then <math>\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.</math>

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

Page

Toolbox

Search