Difference between revisions of "1987 AIME Problems/Problem 15"
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== Problem == | == Problem == | ||
+ | Squares <math>S_1</math> and <math>S_2</math> are [[inscribe]]d in [[right triangle]] <math>ABC</math>, as shown in the figures below. Find <math>AC + CB</math> if area <math>(S_1) = 441</math> and area <math>(S_2) = 440</math>. | ||
+ | |||
+ | [[Image:AIME_1987_Problem_15.png]] | ||
== Solution == | == Solution == | ||
+ | [[Image:1987 AIME-15a.png|360px]] | ||
+ | |||
+ | Because all the [[triangle]]s in the figure are [[similar]] to triangle <math>ABC</math>, it's a good idea to use [[area ratios]]. In the diagram above, <math>\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.</math> Hence, <math>T_3 = \frac {440}{441}T_1</math> and <math>T_4 = \frac {440}{441}T_2</math>. Additionally, the area of triangle <math>ABC</math> is equal to both <math>T_1 + T_2 + 441</math> and <math>T_3 + T_4 + T_5 + 440.</math> | ||
+ | |||
+ | Setting the equations equal and solving for <math>T_5</math>, <math>T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}</math>. Therefore, <math>441T_5 = 441 + T_1 + T_2</math>. However, <math>441 + T_1 + T_2</math> is equal to the area of triangle <math>ABC</math>! This means that the ratio between the areas <math>T_5</math> and <math>ABC</math> is <math>441</math>, and the ratio between the sides is <math>\sqrt {441} = 21</math>. As a result, <math>AB = 21\sqrt {440} = \sqrt {AC^2 + BC^2}</math>. We now need <math>(AC)(BC)</math> to find the value of <math>AC + BC</math>, because <math>AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2</math>. | ||
+ | |||
+ | Let <math>h</math> denote the height to the [[hypotenuse]] of triangle <math>ABC</math>. Notice that <math>h - \frac {1}{21}h = \sqrt {440}</math>. (The height of <math>ABC</math> decreased by the corresponding height of <math>T_5</math>) Thus, <math>(AB)(h) = (AC)(BC) = 22\cdot 21^2</math>. Because <math>AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2</math>, <math>AC + BC = (21)(22) = \boxed{462}</math>. | ||
+ | |||
+ | == Easy Trig Solution == | ||
+ | |||
+ | Let <math>\tan\angle ABC = x</math>. Now using the 1st square, <math>AC=21(1+x)</math> and <math>CB=21(1+x^{-1})</math>. Using the second square, <math>AB=\sqrt{440}(1+x+x^{-1})</math>. We have <math>AC^2+CB^2=AB^2</math>, or <cmath>441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).</cmath> Rearranging and letting <math>u=x+x^{-1} \Rightarrow u^2 - 2 = x^2 + x^{-2}</math> gives us <math>u^2+2u-440=0.</math> We take the positive root, so <math>u=20</math>, which means <math>AC+CB=21(2+x+x^{-1})=21(2+u)=\boxed{462}</math>. | ||
+ | |||
+ | == Messy Trig Solution == | ||
+ | |||
+ | Let <math>\theta</math> be the smaller angle in the triangle. Then the sum of shorter and longer leg is <math>\sqrt{441}(2+\tan{\theta}+\cot{\theta})</math>. We observe that the short leg has length <math>\sqrt{441}(1+\tan{\theta}) = \sqrt{440}(\sec{\theta}+\sin{\theta})</math>. Grouping and squaring, we get <math>\sqrt{\frac{440}{441}} = \frac{\sin{\theta}+\cos{\theta}}{1+\sin{\theta}\cos{\theta}}</math>. Squaring and using the double angle identity for sine, we get, <math>110(\sin{2\theta})^2 + \sin{2\theta} - 1 = 0</math>. Solving, we get <math>\sin{2\theta} = \frac{1}{10}</math>. Now to find <math>\tan{\theta}</math>, we find <math>\cos{2\theta}</math> using the Pythagorean | ||
+ | Identity, and then use the tangent double angle identity. Thus, <math>\tan{\theta} = 10-3\sqrt{11}</math>. Substituting into the original sum, | ||
+ | we get <math>\boxed{462}</math>. | ||
+ | |||
+ | == Solution 4 (Algebra) == | ||
+ | <asy> | ||
+ | size(200); | ||
+ | pair A, B, C, D, E, F; | ||
+ | A = (0, 5); | ||
+ | B = (12, 0); | ||
+ | C = (0, 0); | ||
+ | D = (0, 60/17); | ||
+ | E = (60/17, 60/17); | ||
+ | F = (60/17, 0); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(D--E--F); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,dir(0)); | ||
+ | label("$C$",C,SW); | ||
+ | label("$D$",D,W); | ||
+ | label("$E$",E,NE); | ||
+ | label("$F$",F,S); | ||
+ | label("$S_1$",(30/17,30/17)); | ||
+ | </asy> | ||
+ | |||
+ | <asy> | ||
+ | size(200); | ||
+ | pair A, B, C, W, X, Y, Z; | ||
+ | A = (0, 5); | ||
+ | B = (12, 0); | ||
+ | C = (0, 0); | ||
+ | real m = 1.31004366812; | ||
+ | real n = 3.1441048035; | ||
+ | W = (0,m); | ||
+ | X = (m,m+n); | ||
+ | Y = (m+n,n); | ||
+ | Z = (n,0); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(W--X--Y--Z--cycle); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,dir(0)); | ||
+ | label("$C$",C,SW); | ||
+ | label("$W$",W,dir(180)); | ||
+ | label("$X$",X,NE); | ||
+ | label("$Y$",Y,NE); | ||
+ | label("$Z$",Z,S); | ||
+ | label("$S_2$",(2.22707423581,2.22707423581)); | ||
+ | </asy> | ||
+ | |||
+ | Label points as above. Let <math>x=AC</math>, <math>y=BC</math>, <math>s_1 = 21</math> be the side length of <math>S_1</math>, and <math>s_2 = \sqrt{440}</math> be the side length of <math>S_2</math>. | ||
+ | |||
+ | Since <math>\triangle ABC\sim\triangle AED</math>, we have <math>\frac{x}{y} = \frac{x-s_1}{s_1}</math> | ||
+ | |||
+ | <math>\implies xs_1=xy-ys_1</math> | ||
+ | |||
+ | <math>\implies xy=s_1(x+y)</math> | ||
+ | |||
+ | <math>\implies xy=21(x+y) \qquad \qquad (*)</math>. | ||
+ | Since <math>\triangle ABC\sim\triangle AWX\sim\triangle ZBY</math>, we have <math>s_2 + \frac{s_2x}{y} + \frac{s_2y}{x}=\sqrt{x^2+y^2}</math> | ||
+ | |||
+ | <math>\implies s_2(x^2+xy+y^2)=xy\sqrt{x^2+y^2}</math> | ||
+ | |||
+ | <math>\implies s_2^2(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)</math> | ||
+ | |||
+ | <math>\implies 440(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)</math> | ||
+ | |||
+ | Let <math>t=x+y</math>. Repeatedly applying <math>(*)</math>, we get | ||
+ | <cmath>440(t^2-21t)^2 = 441t^2(t^2 - 42t)</cmath> | ||
+ | <cmath>440(t-21)^2 = 441(t^2-42t)</cmath> | ||
+ | <cmath>440t^2 - 42\cdot 440t + 440\cdot 441 = 441t^2 - 441\cdot 42t</cmath> | ||
+ | <cmath>t^2-42t-440\cdot 441=0</cmath> | ||
+ | <cmath>(t-21)^2 = 441^2</cmath> | ||
+ | <cmath>t-21=441</cmath> | ||
+ | <cmath>t=\boxed{462}</cmath> | ||
+ | |||
+ | ~rayfish | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1987|num-b=14|after=Last<br />Question}} | |
− | {{ | + | [[Category:Intermediate Geometry Problems]] |
+ | {{MAA Notice}} |
Latest revision as of 17:05, 19 February 2022
Contents
[hide]Problem
Squares and are inscribed in right triangle , as shown in the figures below. Find if area and area .
Solution
Because all the triangles in the figure are similar to triangle , it's a good idea to use area ratios. In the diagram above, Hence, and . Additionally, the area of triangle is equal to both and
Setting the equations equal and solving for , . Therefore, . However, is equal to the area of triangle ! This means that the ratio between the areas and is , and the ratio between the sides is . As a result, . We now need to find the value of , because .
Let denote the height to the hypotenuse of triangle . Notice that . (The height of decreased by the corresponding height of ) Thus, . Because , .
Easy Trig Solution
Let . Now using the 1st square, and . Using the second square, . We have , or Rearranging and letting gives us We take the positive root, so , which means .
Messy Trig Solution
Let be the smaller angle in the triangle. Then the sum of shorter and longer leg is . We observe that the short leg has length . Grouping and squaring, we get . Squaring and using the double angle identity for sine, we get, . Solving, we get . Now to find , we find using the Pythagorean Identity, and then use the tangent double angle identity. Thus, . Substituting into the original sum, we get .
Solution 4 (Algebra)
Label points as above. Let , , be the side length of , and be the side length of .
Since , we have
.
Since , we have
Let . Repeatedly applying , we get
~rayfish
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last 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.