Difference between revisions of "2020 AIME II Problems/Problem 13"
m (→Solution 2 (Complex Bash)) |
(→Solution 3 (Guess)) |
||
Line 24: | Line 24: | ||
~Lopkiloinm | ~Lopkiloinm | ||
+ | |||
+ | ==Solution 4 (Official MAA 1)== | ||
+ | Let <math>\omega</math> be the inscribed circle, <math>I</math> be its center, and <math>r</math> be its radius. The area of <math>ABCDE</math> is equal to its semiperimeter, <math>15,</math> times <math>r</math>, so the problem is reduced to finding <math>r</math>. Let <math>a</math> be the length of the tangent segment from <math>A</math> to <math>\omega</math>, and analogously define <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math>. Then <math>a+b=5</math>, <math>b+c= c+d=d+e=6</math>, and <math>e+a=7</math>, with a total of <math>a+b+c+d+e=15</math>. Hence <math>a=3</math>, <math>b=d=2</math>, and <math>c=e=4</math>. It follows that <math>\angle B= \angle D</math> and <math>\angle C= \angle E</math>. Let <math>Q</math> be the point where <math>\omega</math> is tangent to <math>\overline{CD}</math>. Then <math>\angle IAE = \angle IAB =\frac{1}{2}\angle A</math>. The sum of the internal angles in polygons <math>ABCQI</math> and <math>AIQDE</math> are equal, so <math>\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA</math>, which implies that <math>\angle AIQ</math> must be <math>180^\circ</math>. Therefore points <math>A</math>, <math>I</math>, and <math>Q</math> are collinear. | ||
+ | <asy> | ||
+ | defaultpen(fontsize(8pt)); | ||
+ | unitsize(0.025cm); | ||
+ | |||
+ | pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; | ||
+ | string[] labels = {"$A$", "$B$", "$C$", "$D$", "$E$"}; | ||
+ | pair[] dirs = {SW, SE,E, N, NW}; | ||
+ | |||
+ | |||
+ | string[] smallLabels = {"$a$", "$b$", "$c$", "$d$", "$e$"}; | ||
+ | |||
+ | pair I = (3,4); | ||
+ | real rad = 4; | ||
+ | pair Q = foot(I, vertices[2], vertices[3]); | ||
+ | |||
+ | pair[] interpoints = {}; | ||
+ | for(int i =0; i<vertices.length; ++i){ | ||
+ | interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length])); | ||
+ | } | ||
+ | |||
+ | |||
+ | for(int i = 0; i< vertices.length; ++i){ | ||
+ | draw(vertices[i]--vertices[(i+1)%vertices.length]); | ||
+ | dot(labels[i],vertices[i],dirs[i]); | ||
+ | draw(I--vertices[i]); | ||
+ | } | ||
+ | draw(Circle(I, rad)); | ||
+ | dot("$I$", I, dir(200)); | ||
+ | draw(I--Q); | ||
+ | dot("$Q$", Q, NE); | ||
+ | |||
+ | for(int i = 0; i < vertices.length; ++i){ | ||
+ | label(smallLabels[i], vertices[i] --interpoints[i]); | ||
+ | //dot(interpoints[i], blue); | ||
+ | label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]); | ||
+ | } | ||
+ | |||
+ | </asy> | ||
+ | Because <math>\overline{AQ} \perp \overline{CD}</math>, it follows that<cmath>AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.</cmath>Another expression for <math>AC^2-AD^2</math> can be found as follows. Note that <math>\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}</math> and <math>\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}</math>, so | ||
+ | <cmath>\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}</cmath>and | ||
+ | <cmath>\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.</cmath>Applying the Law of Cosines to <math>\triangle ABC</math> and <math>\triangle AED</math> gives | ||
+ | \[ | ||
+ | AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2} | ||
+ | \]and | ||
+ | \[ | ||
+ | AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}. | ||
+ | \]Hence | ||
+ | \[ | ||
+ | 12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}, | ||
+ | \]yielding | ||
+ | \[ | ||
+ | 2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36; | ||
+ | \]equivalently | ||
+ | \[ | ||
+ | 7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2). | ||
+ | \] | ||
+ | Substituting <math>x=r^2</math> gives the quadratic equation <math>5x^2-84x+64=0</math>, with solutions <math>\frac{42 - 38}{5}=\frac45</math>, and <math>\frac{42 + 38}{5}= 16</math>. The solution <math>r^2=\frac45</math> corresponds to a five-pointed star, which is not convex. Indeed, if <math>r<3</math>, then <math> \tan \left(\frac{\angle A}{2}\right)</math>, <math>\tan \left(\frac{\angle C}{2}\right)</math>, and <math>\tan \left(\frac{\angle E}{2}\right)</math> are less than <math>1,</math> implying that <math>\angle A</math>, <math>\angle C</math>, and <math>\angle E</math> are acute, which cannot happen in a convex pentagon. Thus <math>r^2=16</math> and <math>r=4</math>. The requested area is <math>15\cdot4 = 60</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 12:25, 16 June 2020
Contents
Problem
Convex pentagon has side lengths , , and . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of .
Solution 1 (Misplaced problem?)
Assume the incircle touches , , , , at respectively. Then let , , . So we have , and =7, solve it we have , , . Let the center of the incircle be , by SAS we can proof triangle is congruent to triangle , and triangle is congruent to triangle . Then we have , . Extend , cross ray at , ray at , then by AAS we have triangle is congruent to triangle . Thus . Let , then . So by law of cosine in triangle and triangle we can obtain , solved it gives us , which yield triangle to be a triangle with side length 15, 15, 24, draw a height from to divides it into two triangles with side lengths 9, 12, 15, so the area of triangle is 108. Triangle is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is .
-Fanyuchen20020715
Solution 2 (Complex Bash)
Suppose that the circle intersects , , , , and at , , , , and respectively. Then , , , , and . So , , , , and . Then , so . Then we can solve for each individually. , , , , and . To find the radius, we notice that , or . Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, is real. Expanding, we get: , then . On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0. . . must equal 4, as r cannot be negative or be approximately equal to 1. Thus, the area of is
-nihao4112
Solution 3 (Guess)
This pentagon is very close to a regular pentagon with side lengths . The area of a regular pentagon with side lengths is . is slightly greater than given that is slightly less than . is then slightly greater than . We will approximate that to be . The area is now roughly , but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the into thus turning the area into which is and since is a multiple of the semiperimeter , we can safely say that the answer is most likely .
~Lopkiloinm
Solution 4 (Official MAA 1)
Let be the inscribed circle, be its center, and be its radius. The area of is equal to its semiperimeter, times , so the problem is reduced to finding . Let be the length of the tangent segment from to , and analogously define , , , and . Then , , and , with a total of . Hence , , and . It follows that and . Let be the point where is tangent to . Then . The sum of the internal angles in polygons and are equal, so , which implies that must be . Therefore points , , and are collinear. Because , it follows thatAnother expression for can be found as follows. Note that and , so and Applying the Law of Cosines to and gives \[ AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2} \]and \[ AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}. \]Hence \[ 12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}, \]yielding \[ 2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36; \]equivalently \[ 7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2). \] Substituting gives the quadratic equation , with solutions , and . The solution corresponds to a five-pointed star, which is not convex. Indeed, if , then , , and are less than implying that , , and are acute, which cannot happen in a convex pentagon. Thus and . The requested area is .
Video Solution
https://youtu.be/bz5N-jI2e0U?t=327
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.