2020 AIME I Problems/Problem 13
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3(coord bash + basic geometry)
- 6 Solution 4 (Coordinate Bash + Trig)
- 7 Solution 5 (Barycentric Coordinates)
- 8 Solution 6 (geometry+trig)
- 9 Solution 7
- 10 Solution 8 -Trigonometry(only)
- 11 Solution 9 (Official MAA 1)
- 12 Solution 10 (Official MAA 2)
- 13 Solution 11 Bash for life
- 14 Solution 12 (Simple geometry)
- 15 Video Solution
- 16 See Also
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
Diagram
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
Let , , be the midpoints of arcs , , . By Fact 5, we know that , and so by Ptolmey's theorem, we deduce that In particular, we have . Now the key claim is that:
Claim: and are homothetic at with ratio .
Proof. First, we show that is the midpoint of . Indeed, we have by Ratio lemma and Law of Sines.
Now observe that:
- is the perpendicular bisector of ,
- is the perpendicular bisector of , and
- .
Combining these facts gives that is a midline in , which proves the claim.
To finish, we compute , noting that .
By Heron's, we can calculate the circumradius , and by Law of Cosines, we get Then using , we can compute Thus , which gives a final answer of .
~pinetree1
Solution 3(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 4 (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 5 (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 6 (geometry+trig)
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 therefore these two triangles are not congruent, we know that and are supplementary. Then we know that and are also supplementary. 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?
(Added :) ~Math_Genius_164)
Solution 7
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
Solution 8 -Trigonometry(only)
Trig values we use here:
First let the incenter be . Let be the midpoint of minor arc on and let be the foot of to .
We can find using Stewart's Theorem: from Angle Bisector Theorem and . Then it is easy to find that .
Now we trig bash for . Notice that from the Incenter Excenter Lemma. We obtain that . To get we angle chase to get . Then gives . This means .
Now let . It is easy to angle chase and . Since , we compute that which implies which gives an answer of . ~Leonard_my_dude~
Solution 9 (Official MAA 1)
Let , , and . Notice that .
In , segment is the bisector of , and lies on the perpendicular bisector of side . Therefore is the midpoint of arc on the circumcircle of . It follows that and . Likewise, is cyclic, , and . Because is the perpendicular bisector of , triangles and are congruent, implying that
Applying the Law of Sines to and givesBy the Angle Bisector Theorem, and . Combining the above information yields Applying the Law of Cosines to gives , , and . By the Half Angle Formulas,Therefore The requested sum is .
Solution 10 (Official MAA 2)
Let the point be the midpoint of , let be the incenter of which is the common point of lines , , and , and let be the inradius of . The semiperimeter of isand Heron's Formula gives the area of asThis area is also implying that . Stewart's Theorem gives . Because the ratio of the areas of and is it follows thatThus .
Note that . Thus . The height of to is , and the height of to is , so . The needed area of is , as above.
Solution 11 Bash for life
Firstly, it is easy to find with angle bisector theorem.
Using LOC and some trig formulas we get all those values: Now we find the coordinates of points and we apply shoelace theorem later. Point A's coordinates is , Let is perpendicular to , , which means the slope of is . Find the coordinate of , it is easy, , the function is . Now find the intersection of , , getting that Now we look at line segment , since . Since the line passes , we can set the equation to get the , find the intersection of ,, getting that and in the end we use shoelace theorem with coordinates of getting the area leads to the final answer
~bluesoul
Solution 12 (Simple geometry)
Using the Claim (below) we get is orthocenter of So area of is
Semiperimeter of so the bisector
We get the inradius by applying Heron's formula
We use formulas for inradius and get The area
Claim
Let be incenter of Then bisector perpendicular bisector of and perpendicular dropped to bisector from point are concurrent.
Proof
Denote Then Denote the intersection point of and the tangent line to the circumcircle at point
WLOC, (case is trivial).
(this angles are measured by half the arc of the circumcircle).
Therefore bisector of angle P coincite with the perpendicular bisector of .
By applying the Law of Sines to we get
Let be crosspoint of and bisector By applying the Law of Sines to we get
Let be crosspoint of and the perpendicular dropped to bisector from point coincide with
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~MathProblemSolvingSkills.com
See Also
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