Difference between revisions of "1989 AIME Problems/Problem 15"
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[[Image:AIME_1989_Problem_15.png|center]] | [[Image:AIME_1989_Problem_15.png|center]] | ||
− | == Solution == | + | == Solutions == |
− | === Solution | + | |
+ | == 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> | ||
+ | |||
+ | <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 108. | ||
+ | |||
+ | === Solution 2 === | ||
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]], <math>BC^2 + 12^2 = 2(15^2 + 9^2)</math>, and <math>BC = 6\sqrt {13}</math>. Now notice that <math>2[BCP] = [ABC]</math>, because both triangles share the same base and the <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>6\sqrt{13}</math>, <math>[BCP] = 54</math> and <math>[ABC] = \boxed{108}</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]], <math>BC^2 + 12^2 = 2(15^2 + 9^2)</math>, and <math>BC = 6\sqrt {13}</math>. Now notice that <math>2[BCP] = [ABC]</math>, because both triangles share the same base and the <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>6\sqrt{13}</math>, <math>[BCP] = 54</math> and <math>[ABC] = \boxed{108}</math>. | ||
− | === Solution | + | === Solution 3 === |
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> | ||
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Using area ratio, <math>\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = \boxed{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 | + | === Solution 4 === |
Because the length of cevian <math>BE</math> 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 <math>BE</math> was such that <math>\angle APC = 90^\circ</math>? Let's first assume it's a right angle and hope that everything works out. | Because the length of cevian <math>BE</math> 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 <math>BE</math> was such that <math>\angle APC = 90^\circ</math>? Let's first assume it's a right angle and hope that everything works out. | ||
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[however, I think this solution is wrong, the A, B, and Cs do not match with the picture] | [however, I think this solution is wrong, the A, B, and Cs do not match with the picture] | ||
− | === Solution | + | === Solution 5 === |
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>. | 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>. |
Revision as of 01:20, 10 April 2022
Contents
[hide]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
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
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.
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 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 , , 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, , and . Now notice that , because both triangles share the same base and the . Applying Heron's formula on triangle with sides , , and , and .
Solution 3
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
Because the length of cevian 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 was such that ? Let's first assume it's a right angle and hope that everything works out.
Extend to so that . The result is that , , and because . Now we see that if we are able to show that , that is , then our right angle assumption will be true.
Apply the Pythagorean Theorem on to get , so and . Now, we apply the Law of Cosines on triangles and .
Let . Notice that and , so we get two nice equations.
Solving, (yay!).
Now, the area is easy to find. .
[however, I think this solution is wrong, the A, B, and Cs do not match with the picture]
Solution 5
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+heron)
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
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.