Difference between revisions of "2020 AIME I Problems/Problem 13"
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D=extension(A,I,B,C); | D=extension(A,I,B,C); | ||
M=(A+D)/2; | M=(A+D)/2; | ||
+ | F=(0,9sqrt(7)/7); | ||
+ | E=(5,sqrt(28)); | ||
draw(B--EE,gray+dashed); | draw(B--EE,gray+dashed); |
Revision as of 22:38, 1 June 2020
Contents
Problem
Point lies on side of so that bisects The perpendicular bisector of intersects the bisectors of and in points and respectively. Given that and the area of can be written as where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. Find
Solution 1
Points are defined as shown. It is pretty easy to show that by spiral similarity at by some short angle chasing. Now, note that is the altitude of , as the altitude of . We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that , the altitude of . Similarly, the altitude of is the altitude of , or . However, it's not too hard to see that , and therefore . From here, we get that the area of is , by similarity. ~awang11
Solution 2(coord bash + basic geometry)
Let lie on the x-axis and be the origin. is . Use Heron's formula to compute the area of triangle . We have . and . We now find the altitude, which is , which is the y-coordinate of . We now find the x-coordinate of , which satisfies , which gives since the triangle is acute. Now using the Angle Bisector Theorem, we have and to get . The coordinates of D are . Since we want the area of triangle , we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is and the slope of AD is . The slope of the perpendicular bisector is . The equation is(in point slope form) . The slope of AB, or in trig words, the tangent of is . Finding and . Plugging this in to half angle tangent, it gives as the slope of the angle bisector, since it passes through , the equation is . Similarly, the equation for the angle bisector of will be . For use the B-angle bisector and the perpendicular bisector of AD equations to intersect at . For use the C-angle bisector and the perpendicular bisector of AD equations to intersect at . The area of AEF is equal to since AD is the altitude of that triangle with EF as the base, with being the height. and , so which gives . NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo
Solution 3 (Coordinate Bash + Trig)
Let and be the line . We compute that , so . Thus, lies on the line . The length of at a point is , so .
We now have the coordinates , and . We also have by the angle-bisector theorem and by taking the midpoint. We have that because , by half angle formula.
We also compute , so .
Now, has slope , so it's perpendicular bisector has slope and goes through .
We find that this line has equation .
As , we have that line has form . Solving for the intersection point of these two lines, we get and thus
We also have that because , has form .
Intersecting the line and the perpendicular bisector of yields .
Solving this, we get and so .
We now compute . We also have .
As , we have .
The desired answer is ~Imayormaynotknowcalculus
Solution 4 (Barycentric Coordinates)
As usual, we will use homogenized barycentric coordinates.
We have that will have form . Similarly, has form and has form . Since and , we also have . It remains to determine the equation of the line formed by the perpendicular bisector of .
This can be found using EFFT. Let a point on have coordinates . We then have that the displacement vector and that the displacement vector has form . Now, by EFFT, we have . This equates to .
Now, intersecting this with , we have , , and . This yields , , and , or .
Similarly, intersecting this with , we have , , and . Solving this, we obtain , , and , or .
We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being . We then have , thus .
Our second displacement vector is . As a result, , so .
As , the desired area is . ~Imayormaynotknowcalculus
Remark: The area of can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.
Solution 5 (geometry+trig)
size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; F=(0,9sqrt(7)/7); E=(5,sqrt(28)); draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(A--E); draw(A--F); draw(D--F); draw(D--E); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,SE); dot("$D$",D,S); dot("$E$",(5,sqrt(28)),N); dot("$M$",M,dir(70)); dot("$F$",(0,9sqrt(7)/7),N); label("$2$",B--D,S); label("$3$",D--C,S); label("$6$",A--C,N); label("$4$",A--B,W); (Error compiling LaTeX. e511ca2b14ce97b2971580a7c49206dadbed6bd7.asy: 13.1: modifying non-public field outside of structure)
To get the area of , we try to find and .
Since is the angle bisector, we can get that and . By applying Stewart's Theorem, we can get that . Therefore .
Since is the perpendicular bisector of , we know that . Since is the angle bisector of , we know that . By applying the Law of Sines to and , we know that . Since is not equal to and therefor these two triangles are not congruent, we know that and are supplement. Then we know that and are also supplement. Given that , we can get that is half of . Similarly, we have is half of .
By applying the Law of Cosines, we get , and then . Similarly, we can get and . Based on some trig identities, we can compute that , and .
Finally, the area of equals . Therefore, the final answer is . ~xamydad
Remark: I didn't figure out how to add segments , , and . Can someone please help add these segments?
Solution 6
First and foremost as is the perpendicular bisector of . Now note that quadrilateral is cyclic, because and . Similarly quadrilateral is cyclic, Let ,, be the ,, and excenters of respectively. Then it follows that . By angle bisector theorem we have . Now let the feet of the perpendiculars from and to be and resptively. Then by tangents we have From the previous ratios, Similarly we can find that and and thus -tkhalid
See Also
2020 AIME I (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.