Difference between revisions of "2020 AIME II Problems/Problem 15"
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Since <math>AXTY</math> is cyclical, <math>\angle XTY=\angle XHY</math> and this implies that <math>XHYT</math> is a paralelogram. | Since <math>AXTY</math> is cyclical, <math>\angle XTY=\angle XHY</math> and this implies that <math>XHYT</math> is a paralelogram. | ||
By the Law of Cosines: | By the Law of Cosines: | ||
− | <cmath>XY^2=XT^2+TY^2+2(XT)(TY)\cdot cos(\angle A)</cmath> | + | <cmath>XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)</cmath> |
− | <cmath>XY^2=XH^2+HY^2+2(XH)(HY) \cdot cos(\angle A)</cmath> | + | <cmath>XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)</cmath> |
− | <cmath>HT^2=HX^2+XT^2-2(HX)(XT) \cdot cos(\angle A)</cmath> | + | <cmath>HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)</cmath> |
− | <cmath>HT^2=HY^2+YT^2-2(HY)(YT) \cdot cos(\angle A).</cmath> | + | <cmath>HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).</cmath> |
We add all these equations to get: | We add all these equations to get: | ||
<cmath>HT^2+XY^2=2(HT^2+TY^2).</cmath> | <cmath>HT^2+XY^2=2(HT^2+TY^2).</cmath> |
Revision as of 12:32, 15 April 2021
Contents
[hide]Problem
Let be an acute scalene triangle with circumcircle
. The tangents to
at
and
intersect at
. Let
and
be the projections of
onto lines
and
, respectively. Suppose
,
, and
. Find
.
Solution
Assume to be the center of triangle
,
cross
at
, link
,
. Let
be the middle point of
and
be the middle point of
, so we have
. Since
, we have
. Notice that
, so
, and this gives us
. Since
is perpendicular to
,
and
cocycle (respectively), so
and
. So
, so
, which yields
So same we have
. Apply Ptolemy theorem in
we have
, and use Pythagoras theorem we have
. Same in
and triangle
we have
and
. Solve this for
and
and submit into the equation about
, we can obtain the result
.
(Notice that is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)
-Fanyuchen20020715
Solution 2 (Official MAA)
Let denote the midpoint of
. The critical claim is that
is the orthocenter of
, which has the circle with diameter
as its circumcircle. To see this, note that because
, the quadrilateral
is cyclic, it follows that
implying that
. Similarly,
. In particular,
is a parallelogram.
Hence, by the Parallelogram Law,
But
. Therefore
Solution 3 (No parallelogram Law)
Let be the orthocenter of
.
Lemma 1: is the midpoint of
.
Proof: Let be the midpoint of
, and observe that
and
are cyclical. Define
and
, then note that:
That implies that
,
, and
. Thus
and
;
is indeed the same as
, and we have proved lemma 1.
Since is cyclical,
and this implies that
is a paralelogram.
By the Law of Cosines:
We add all these equations to get:
We now use the conditions of the problem:
and
. Note that
, so by the Pythagorean Theorem, it follows that
. We were also given that
, into which we substitute to get
Therefore,
. ~ MathLuis
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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