Difference between revisions of "1989 AIME Problems/Problem 15"
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== Problem == | == Problem == | ||
− | + | Point <math>P</math> is inside <math>\triangle ABC</math>. Line segments <math>APD</math>, <math>BPE</math>, and <math>CPF</math> are drawn with <math>D</math> on <math>BC</math>, <math>E</math> on <math>AC</math>, and <math>F</math> on <math>AB</math> (see the figure below). Given that <math>AP=6</math>, <math>BP=9</math>, <math>PD=6</math>, <math>PE=3</math>, and <math>CF=20</math>, find the area of <math>\triangle ABC</math>. | |
+ | [[Image:AIME_1989_Problem_15.png|center]] | ||
− | { | + | == Solutions == |
− | + | ||
− | + | === Solution 1 (Ceva's Theorem, Stewart's Theorem) === | |
− | + | ||
− | + | 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> | |
− | == | + | |
− | === Solution | + | <center><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></center> | ||
+ | |||
+ | This is true because triangles <math>XPY</math> and <math>YPL</math> have their areas in ratio <math>XP:PL</math> (as they share a common height from <math>Y</math>), and the same is true of triangles <math>ZPY</math> and <math>LPZ</math>. | ||
+ | |||
+ | We'll also use the related fact that <math>\dfrac{[XPY]}{[ZPX]} = \dfrac{YL}{LZ}</math>. 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. | ||
+ | |||
+ | <center><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></center> | ||
+ | |||
+ | Since <math>AP = PD = 6</math>, this means that <math>[APB] + [APC] = [BPC]</math>; thus <math>\triangle BPC</math> has half the area of <math>\triangle ABC</math>. And since <math>PE = 3 = \dfrac{1}{3}BP</math>, we can conclude that <math>\triangle APC</math> has one third of the combined areas of triangle <math>BPC</math> and <math>APB</math>, and thus <math>\dfrac{1}{4}</math> of the area of <math>\triangle ABC</math>. This means that <math>\triangle APB</math> is left with <math>\dfrac{1}{4}</math> of the area of triangle <math>ABC</math>: | ||
+ | <cmath> [BPC]: [APC]: [APB] = 2:1:1.</cmath> | ||
+ | Since <math>[APC] = [APB]</math>, and since <math>\dfrac{[APC]}{[APB]} = \dfrac{CD}{DB}</math>, this means that <math>D</math> is the midpoint of <math>BC</math>. | ||
+ | |||
+ | Furthermore, we know that <math>\dfrac{CP}{PF} = \dfrac{[APC] + [BPC]}{[APB]} = 3</math>, so <math>CP = \dfrac{3}{4} \cdot CF = 15</math>. | ||
+ | |||
+ | We now apply [[Stewart's theorem]] to segment <math>PD</math> in <math>\triangle BPC</math>—or rather, the simplified version for a median. This tells us that | ||
+ | <cmath> 2 BD^2 + 2 PD^2 = BP^2+ CP^2. </cmath> Plugging in we know, we learn that | ||
+ | <cmath> \begin{align*} | ||
+ | 2 BD^2 + 2 \cdot 36 &= 81 + 225 = 306, \\ | ||
+ | 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 <math>\boxed{108}</math>. | ||
+ | |||
+ | === Solution 2 (Mass Points, Stewart's Theorem, Heron's Formula) === | ||
Because we're given three concurrent [[cevian]]s and their lengths, it seems very tempting to apply [[Mass points]]. We immediately see that <math>w_E = 3</math>, <math>w_B = 1</math>, and <math>w_A = w_D = 2</math>. Now, we recall that the masses on the three sides of the triangle must be balanced out, so <math>w_C = 1</math> and <math>w_F = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>. | Because we're given three concurrent [[cevian]]s and their lengths, it seems very tempting to apply [[Mass points]]. We immediately see that <math>w_E = 3</math>, <math>w_B = 1</math>, and <math>w_A = w_D = 2</math>. Now, we recall that the masses on the three sides of the triangle must be balanced out, so <math>w_C = 1</math> and <math>w_F = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>. | ||
− | Recalling that <math>w_C = w_B = 1</math>, we see that <math>DC = DB</math> and <math>DP</math> is a [[median]] to <math>BC</math> in <math>\triangle BCP</math>. Applying [[Stewart's Theorem]], < | + | Recalling that <math>w_C = w_B = 1</math>, we see that <math>DC = DB</math> and <math>DP</math> is a [[median]] to <math>BC</math> in <math>\triangle BCP</math>. Applying [[Stewart's Theorem]], we have the following: |
+ | <cmath>\frac{BC}{2}(9^2+15^2)=BC(6^2+ \left(\frac{BC}{2} \right)^2).</cmath> | ||
+ | Eliminating <math>BC</math> on both sides, we have: | ||
+ | <cmath>\frac 12(9^2+15^2)=6^2+ \left(\frac{BC}{2} \right)^2.</cmath> | ||
+ | Combining terms and simplifying numbers, we have: | ||
+ | <cmath>153=36+\left(\frac{BC}{2} \right)^2.</cmath> | ||
+ | Subtracting 36 to the other side yields: | ||
+ | <cmath>117= \left(\frac{BC}{2} \right)^2.</cmath> | ||
+ | Finishing it off from there, we find that <math>BC=2 \sqrt{117}.</math> Now, notice that <math>2[BCP] = [ABC]</math>, because both triangles share the same base, <math>BC</math> and <math>h_{\triangle ABC} = 2h_{\triangle BCP}</math>. Applying [[Heron's formula]] on triangle <math>BCP</math> with sides <math>15</math>, <math>9</math>, and <math>2\sqrt{117}</math>, we have: | ||
+ | <cmath>\sqrt{(\sqrt{117}+12)(\sqrt{117}+12-9)(\sqrt{117}+12-15)(\sqrt{117}+12-2\sqrt{117})}.</cmath> | ||
+ | Combining terms results in: | ||
+ | <cmath>\sqrt{(\sqrt{117}+12)(\sqrt{117}+3)(\sqrt{117}-3)(-\sqrt{117}+12)}.</cmath> | ||
+ | Notice that these factors can be grouped into a difference of squares: | ||
+ | <cmath>\sqrt{(144-\sqrt{117}^2)(\sqrt{117}^2-9)}.</cmath> | ||
+ | Since <math>\sqrt{117}^2=117</math>, we have: | ||
+ | <cmath>\sqrt{(27)(108)}.</cmath> | ||
+ | After simplifying this radical, we find that it equals <math>54.</math> Therefore, <math>[BCP] = 54</math>, and hence <math>[ABC]=2 \cdot 54= \boxed{108}</math>. | ||
− | === Solution | + | (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 <math>\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}</math> | Using a different form of [[Ceva's Theorem]], we have <math>\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}</math> | ||
− | Solving <math>4y = x + y</math> and <math>x + y = 20</math>, we obtain <math>x = | + | Solving <math>4y = x + y</math> and <math>x + y = 20</math>, we obtain <math>x = CP = 15</math> and <math>y = FP = 5</math>. |
− | Let <math>Q</math> be the point on <math>AB</math> such that <math>FC\parallel QD</math>. | + | Let <math>Q</math> be the point on <math>AB</math> such that <math>FC \parallel QD</math>. |
− | Since <math>AP = PD</math> and <math>FP\parallel QD</math>, <math>QD = 2FP = | + | Since <math>AP = PD</math> and <math>FP\parallel QD</math>, <math>QD = 2FP = 10</math>. (Stewart's Theorem) |
− | Also, since <math>FC\parallel QD</math> and <math> | + | Also, since <math>FC\parallel QD</math> and <math>QD = \frac{FC}{2}</math>, we see that <math>FQ = QB</math>, <math>BD = DC</math>, etc. ([[Stewart's Theorem]]) |
Similarly, we have <math>PR = RB</math> (<math>= \frac12PB = 7.5</math>) and thus <math>RD = \frac12PC = 4.5</math>. | Similarly, we have <math>PR = RB</math> (<math>= \frac12PB = 7.5</math>) and thus <math>RD = \frac12PC = 4.5</math>. | ||
<math>PDR</math> is a <math>3-4-5</math> [[right triangle]], so <math>\angle PDR</math> (<math>\angle ADQ</math>) is <math>90^\circ</math>. | <math>PDR</math> is a <math>3-4-5</math> [[right triangle]], so <math>\angle PDR</math> (<math>\angle ADQ</math>) is <math>90^\circ</math>. | ||
Therefore, the area of <math>\triangle ADQ = \frac12\cdot 12\cdot 6 = 36</math>. | Therefore, the area of <math>\triangle ADQ = \frac12\cdot 12\cdot 6 = 36</math>. | ||
− | Using area ratio, <math>\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = 108</math>. | + | Using area ratio, <math>\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = \boxed{108}</math>. |
+ | |||
+ | === Solution 4 (Stewart's Theorem) === | ||
+ | |||
+ | First, let <math>[AEP]=a, [AFP]=b,</math> and <math>[ECP]=c.</math> Thus, we can easily find that <math>\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.</math> Now, <math>\frac{[ABP]}{[BPD]}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.</math> In the same manner, we find that <math>[CPD]=a+c.</math> Now, we can find that <math>\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.</math> We can now use this to find that <math>\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.</math> Plugging this value in, we find that <math>\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.</math> Now, since <math>\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},</math> we can find that <math>2AE=EC.</math> Setting <math>AC=b,</math> we can apply [[Stewart's Theorem]] on triangle <math>APC</math> to find that <math>(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).</math> Solving, we find that <math>b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.</math> But, <math>3^2+6^2=45,</math> meaning that <math>\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.</math> Since <math>[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,</math> we conclude that the answer is <math>\boxed{108}</math>. | ||
+ | |||
+ | === 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 <math>M(A)=6;M(D)=6;M(B)=3;M(E)=9</math>; we can get that <math>M(P)=12;M(F)=9;M(C)=3</math>; which leads to the ratio between segments, | ||
+ | <cmath>\frac{CE}{AE}=2;\frac{BF}{AF}=2;\frac{BD}{CD}=1.</cmath> Denoting that <math>CE=2x;AE=x; AF=y; BF=2y; CD=z; DB=z.</math> | ||
+ | |||
+ | Now we know three cevians' length, Applying Stewart theorem to them, getting three different equations: | ||
+ | <cmath>\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}</cmath> | ||
+ | After solving the system of equation, we get that <math>x=3\sqrt{5};y=\sqrt{13};z=3\sqrt{13}</math>; | ||
+ | |||
+ | pulling <math>x,y,z</math> back to get the length of <math>AC=9\sqrt{5};AB=3\sqrt{13};BC=6\sqrt{13}</math>; now we can apply Heron's formula here, which is <cmath>\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.</cmath> | ||
+ | |||
+ | Our answer is <math>\boxed{108}</math>. | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ====Note (how to find x and y without the system of equations)==== | ||
+ | To ease computation, we can apply Stewart's Theorem to find <math>x</math>, <math>y</math>, and <math>z</math> directly. Since <math>M(C)=3</math> and <math>M(F)=9</math>, <math>\overline{PC}=15</math> and <math>\overline{PF}=5</math>. We can apply Stewart's Theorem on <math>\triangle CPE</math> to get <math>(2x+x)(2x \cdot x) + 3^2 \cdot 3x = 15^2 \cdot x + 6^2\cdot 2x</math>. Solving, we find that <math>x=3\sqrt{5}</math>. We can do the same on <math>\triangle APB</math> and <math>\triangle BPC</math> to obtain <math>y</math> and <math>z</math>. We proceed with Heron's Formula as the solution states. | ||
+ | |||
+ | ~kn07 | ||
+ | |||
+ | === Solution 6 (easier version of Solution 5)=== | ||
+ | |||
+ | In Solution 5, instead of finding all of <math>x, y, z</math>, we only need <math>y, z</math>. This is because after we solve for <math>y, z</math>, we can notice that <math>\triangle BAD</math> is isosceles with <math>AB = BD</math>. Because <math>P</math> is the midpoint of the base, <math>BP</math> is an altitude of <math>\triangle BAD</math>. Therefore, <math>[BAD] = \frac{(AD)(BP)}{2} = \frac{12 \cdot 9}{2} = 54</math>. Using the same altitude property, we can find that <math>[ABC] = 2[BAD] = 2 \cdot 54 = \boxed{108}</math>. | ||
+ | |||
+ | -NL008 | ||
+ | |||
+ | === Solution 7 (Mass Points, Stewart's Theorem, Simple Version) === | ||
− | === | + | Set <math>AF=x,</math> and use [[mass points]] to find that <math>PF=5</math> and <math>BF=2x.</math> Using [[Stewart's Theorem]] on <math>APB,</math> we find that <math>AB=3\sqrt{13}.</math> Then we notice that <math>APB</math> is right, which means the area of <math>APB</math> is <math>27.</math> Because <math>CF=4\cdot PF,</math> the area of <math>ABC</math> is <math>4</math> times the area of <math>APB,</math> which means the area of <math>ABC=4\cdot 27=\boxed{108}.</math> |
− | |||
− | + | === Solution 8 (Ratios, Auxiliary Lines and 3-4-5 triangle) === | |
− | + | We try to solve this using only elementary concepts. Let the areas of triangles <math>BCP</math>, <math>ACP</math> and <math>ABP</math> be <math>X</math>, <math>Y</math> and <math>Z</math> respectively. Then <math>\frac{X}{Y+Z}=\frac{6}{6}=1</math> and <math>\frac{Y}{X+Z}=\frac{3}{9}=\frac{1}{3}</math>. Hence <math>\frac{X}{2}=Y=Z</math>. Similarly <math>\frac{FP}{PC}=\frac{Z}{X+Y}=\frac{1}{3}</math> and since <math>CF=20</math> we then have <math>FP=5</math>. Additionally we now see that triangles <math>FPE</math> and <math>CPB</math> are similar, so <math>FE \parallel BC</math> and <math>\frac{FE}{BC} = \frac{1}{3}</math>. Hence <math>\frac{AF}{FB}=\frac{1}{2}</math>. Now construct a point <math>K</math> on segment <math>BP</math> such that <math>BK=6</math> and <math>KP=3</math>, we will have <math>FK \parallel AP</math>, and hence <math>\frac{FK}{AP} = \frac{FK}{6} = \frac{2}{3}</math>, giving <math>FK=4</math>. Triangle <math>FKP</math> is therefore a 3-4-5 triangle! So <math>FK \perp BE</math> and so <math>AP \perp BE</math>. Then it is easy to calculate that <math>Z = \frac{1}{2} \times 6 \times 9 = 27</math> and the area of triangle <math>ABC = X+Y+Z = 4Z = 4 \times 27 = \boxed{108}</math>. | |
+ | ~Leole | ||
− | |||
− | + | === Solution 9 (Just Trig Bash) === | |
− | |||
− | + | We start with mass points as in Solution 2, and receive <math>BF:AF = 2</math>, <math>BD:CD = 1</math>, <math>CE:AE = 2</math>. [[Law of Cosines]] on triangles <math>ADB</math> and <math>ADC</math> with <math>\theta = \angle ADB</math> and <math>BD=DC=x</math> gives | |
+ | <cmath>36+x^2-12x\cos \theta = 81</cmath> | ||
+ | <cmath>36+x^2-12x\cos (180-\theta) = 36+x^2+12x\cos \theta = 225</cmath> | ||
+ | Adding them: <math>72+2x^2=306 \implies x=3\sqrt{13}</math>, so <math>BC = 6\sqrt{13}</math>. Similarly, <math>AB = 3\sqrt{13}</math> and <math>AC = 9\sqrt{5}</math>. Using Heron's, | ||
+ | <cmath>[ \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}.</cmath> | ||
− | + | ~sml1809 | |
== See also == | == See also == |
Latest revision as of 23:46, 30 June 2024
Contents
- 1 Problem
- 2 Solutions
- 2.1 Solution 1 (Ceva's Theorem, Stewart's Theorem)
- 2.2 Solution 2 (Mass Points, Stewart's Theorem, Heron's Formula)
- 2.3 Solution 3 (Ceva's Theorem, Stewart's Theorem)
- 2.4 Solution 4 (Stewart's Theorem)
- 2.5 Solution 5 (Mass of a Point, Stewart's Theorem, Heron's Formula)
- 2.6 Solution 6 (easier version of Solution 5)
- 2.7 Solution 7 (Mass Points, Stewart's Theorem, Simple Version)
- 2.8 Solution 8 (Ratios, Auxiliary Lines and 3-4-5 triangle)
- 2.9 Solution 9 (Just Trig Bash)
- 3 See also
Problem
Point is inside
. Line segments
,
, and
are drawn with
on
,
on
, and
on
(see the figure below). Given that
,
,
,
, and
, find the area of
.
Solutions
Solution 1 (Ceva's Theorem, Stewart's Theorem)
Let be the area of polygon
. We'll make use of the following fact: if
is a point in the interior of triangle
, and line
intersects line
at point
, then
![[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]](http://latex.artofproblemsolving.com/3/c/9/3c978a1fdb0a6c3a79b9ee477956a440b734f13b.png)
This is true because triangles and
have their areas in ratio
(as they share a common height from
), and the same is true of triangles
and
.
We'll also use the related fact that . 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]](http://latex.artofproblemsolving.com/7/c/9/7c9e14d181ecf013675ff1bfbe1aee4cc51d0c7e.png)
Since , this means that
; thus
has half the area of
. And since
, we can conclude that
has one third of the combined areas of triangle
and
, and thus
of the area of
. This means that
is left with
of the area of triangle
:
Since
, and since
, this means that
is the midpoint of
.
Furthermore, we know that , so
.
We now apply Stewart's theorem to segment in
—or rather, the simplified version for a median. This tells us that
Plugging in we know, we learn that
Happily,
is also equal to 117. Therefore
is a right triangle with a right angle at
; its area is thus
. As
is a median of
, the area of
is twice this, or 54. And we already know that
has half the area of
, which must therefore be
.
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 ,
, and
. Now, we recall that the masses on the three sides of the triangle must be balanced out, so
and
. Thus,
and
.
Recalling that , we see that
and
is a median to
in
. Applying Stewart's Theorem, we have the following:
Eliminating
on both sides, we have:
Combining terms and simplifying numbers, we have:
Subtracting 36 to the other side yields:
Finishing it off from there, we find that
Now, notice that
, because both triangles share the same base,
and
. Applying Heron's formula on triangle
with sides
,
, and
, we have:
Combining terms results in:
Notice that these factors can be grouped into a difference of squares:
Since
, we have:
After simplifying this radical, we find that it equals
Therefore,
, and hence
.
(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
Solving and
, we obtain
and
.
Let be the point on
such that
.
Since
and
,
. (Stewart's Theorem)
Also, since and
, we see that
,
, etc. (Stewart's Theorem)
Similarly, we have
(
) and thus
.
is a
right triangle, so
(
) is
.
Therefore, the area of
.
Using area ratio,
.
Solution 4 (Stewart's Theorem)
First, let and
Thus, we can easily find that
Now,
In the same manner, we find that
Now, we can find that
We can now use this to find that
Plugging this value in, we find that
Now, since
we can find that
Setting
we can apply Stewart's Theorem on triangle
to find that
Solving, we find that
But,
meaning that
Since
we conclude that the answer is
.
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 ; we can get that
; which leads to the ratio between segments,
Denoting that
Now we know three cevians' length, Applying Stewart theorem to them, getting three different equations:
After solving the system of equation, we get that
;
pulling back to get the length of
; now we can apply Heron's formula here, which is
Our answer is .
~bluesoul
Note (how to find x and y without the system of equations)
To ease computation, we can apply Stewart's Theorem to find ,
, and
directly. Since
and
,
and
. We can apply Stewart's Theorem on
to get
. Solving, we find that
. We can do the same on
and
to obtain
and
. We proceed with Heron's Formula as the solution states.
~kn07
Solution 6 (easier version of Solution 5)
In Solution 5, instead of finding all of , we only need
. This is because after we solve for
, we can notice that
is isosceles with
. Because
is the midpoint of the base,
is an altitude of
. Therefore,
. Using the same altitude property, we can find that
.
-NL008
Solution 7 (Mass Points, Stewart's Theorem, Simple Version)
Set and use mass points to find that
and
Using Stewart's Theorem on
we find that
Then we notice that
is right, which means the area of
is
Because
the area of
is
times the area of
which means the area of
Solution 8 (Ratios, Auxiliary Lines and 3-4-5 triangle)
We try to solve this using only elementary concepts. Let the areas of triangles ,
and
be
,
and
respectively. Then
and
. Hence
. Similarly
and since
we then have
. Additionally we now see that triangles
and
are similar, so
and
. Hence
. Now construct a point
on segment
such that
and
, we will have
, and hence
, giving
. Triangle
is therefore a 3-4-5 triangle! So
and so
. Then it is easy to calculate that
and the area of triangle
.
~Leole
Solution 9 (Just Trig Bash)
We start with mass points as in Solution 2, and receive ,
,
. Law of Cosines on triangles
and
with
and
gives
Adding them:
, so
. Similarly,
and
. Using Heron's,
~sml1809
See also
1989 AIME (Problems • Answer Key • Resources) | ||
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