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Because the trapezoid is isosceles, by symmetry <math>PQ</math> is parallel to <math>AB</math> and <math>BC</math>. Therefore, <math>\angle PAB \cong \angle APP'</math> by interior angles and <math>\angle PAB \cong \angle PAD</math> by the problem statement. Thus, <math>\triangle P'AP</math> is isosceles with <math>P'P = P'A</math>. By symmetry, <math>P'DP</math> is also isosceles, and thus <math>P'A = \frac{AD}{2}</math>. Similarly, the same thing is happening on the right side of the trapezoid, and thus <math>P'Q'</math> is the midline of the trapezoid. Then, <math>PQ = P'Q' - (P'P + Q'Q)</math>. | Because the trapezoid is isosceles, by symmetry <math>PQ</math> is parallel to <math>AB</math> and <math>BC</math>. Therefore, <math>\angle PAB \cong \angle APP'</math> by interior angles and <math>\angle PAB \cong \angle PAD</math> by the problem statement. Thus, <math>\triangle P'AP</math> is isosceles with <math>P'P = P'A</math>. By symmetry, <math>P'DP</math> is also isosceles, and thus <math>P'A = \frac{AD}{2}</math>. Similarly, the same thing is happening on the right side of the trapezoid, and thus <math>P'Q'</math> is the midline of the trapezoid. Then, <math>PQ = P'Q' - (P'P + Q'Q)</math>. | ||
− | Since <math>P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}</math> and <math>AD = BC = 333</math>, we have <math>P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333</math>. The length of the midline of a trapezoid is the average of their bases, so <math>P'Q' = \frac{500+650}{2} = 575</math>. Finally, <math>PQ = 575 - 333 = \boxed{242}</math> | + | Since <math>P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}</math> and <math>AD = BC = 333</math>, we have <math>P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333</math>. The length of the midline of a trapezoid is the average of their bases, so <math>P'Q' = \frac{500+650}{2} = 575</math>. Finally, <math>PQ = 575 - 333 = \boxed{242}</math>. |
~KingRavi | ~KingRavi |
Revision as of 03:08, 26 February 2022
Contents
Problem
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
.
Diagram
~MRENTHUSIASM ~ihatemath123
Solution 1
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
,
~Cytronical
Solution 2
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,
.
~KingRavi
Solution 3
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,
.
~ihatemath123
Solution 4
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
.
~Orange_Quail_9
Solution 5
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
~regular
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=fNAvxXnvAxs
Video Solution
https://www.youtube.com/watch?v=h_LOT-rwt08
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.