Difference between revisions of "1987 AIME Problems/Problem 15"
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we get <math>\boxed{462}</math>. | 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}} | {{AIME box|year=1987|num-b=14|after=Last<br />Question}} |
Revision as of 00:45, 27 December 2021
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.