2022 AIME I Problems/Problem 3
In isosceles trapezoid , parallel bases and have lengths and , respectively, and . The angle bisectors of and meet at , and the angle bisectors of and meet at . Find .
We have the following diagram:
Let and be the points where and extend to meet , and be the height of . As proven in Solution 2, triangles and are congruent right triangles. Therefore, . We can apply this logic to triangles and as well, giving us . Since , .
Additionally, we can see that is similar to and . We know that . So, we can say that the height of the triangle is while the height of the triangle is . After that, we can figure out the distance from to and the height of triangle .
Finally, since the ratio between the height of to the height of is and is ,
Extend line to meet at and at . The diagram looks like this: Because the trapezoid is isosceles, by symmetry is parallel to and . Therefore, by interior angles and by the problem statement. Thus, is isosceles with . By symmetry, is also isosceles, and thus . Similarly, the same thing is happening on the right side of the trapezoid, and thus is the midline of the trapezoid. Then, .
Since and , we have . The length of the midline of a trapezoid is the average of their bases, so . Finally, .
We have the following diagram: Extend lines and to meet line at points and , respectively, and extend lines and to meet at points and , respectively.
Claim: quadrilaterals and are rhombuses.
Proof: Since , . Therefore, triangles , , and are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, is congruent to the other three. Therefore, , so is a rhombus. By symmetry, is also a rhombus.
Extend line to meet and at and , respectively. Because of rhombus properties, . Also, by rhombus properties, and are the midpoints of segments and , respectively; therefore, by trapezoid properties, . Finally, .
Let and be the feet of the altitudes from and , respectively, to , and let and be the feet of the altitudes from and , respectively, to . Side is parallel to side , so is a rectangle with width . Furthermore, because and trapezoid is isosceles, .
Also because is isosceles, is half the total sum of angles in , or . Since and bisect and , respectively, we have , so .
Letting , applying Pythagoras to yields . We then proceed using similar triangles: and , so by AA similarity . Likewise, and , so by AA similarity . Thus .
Adding our two equations for and gives . Therefore, the answer is .
This will be my first solution on AoPS. My apologies in advance for any errors.
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that is equidistant from and and is equidistant from and If we let the feet of the altitudes from to and be called and respectively, we can say that Analogously, we let the feet of the altitudes from to and be and respectively. Thus, Because is an isosceles trapezoid, we can say that all of the altitudes are equal to each other.
By SA as well as SS congruence for right triangles, we find that triangles and are congruent. Similarly, and by the same reasoning. Additionally, since and are congruent rectangles.
If we then let let and let we can create the following system of equations with the given side length information: Adding the first two equations, subtracting by twice the second, and dividing by yields
Extend line to meet at and at . The diagram looks like this: Since , it follows that . Thus, , implying that and are right triangles. Since and are medians, . Since , we have , or .
Solution 7 (Trigonometry)
Let . Note that since bisects and bisects , we have Let . We have that Now, drop an altitude from to at . Notice that . By the definition of cosine, we have Notice, however, that we can also apply this to ; we have By the Pythagorean Theorem, we get Then, drop an altitude from to at ; if , then . Because is an angle bisector, we see that . Again, by the definition of cosine, we have Finally, .
Video Solution (Mathematical Dexterity)
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