Difference between revisions of "2020 AIME II Problems/Problem 15"
(→Solution 4 (Similarity and median)) |
(→Solution 4 (Similarity and median)) |
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==Solution 4 (Similarity and median)== | ==Solution 4 (Similarity and median)== | ||
+ | [[File:2020 AIME II 15.png|300px|right]] | ||
<i><b>Lemma</b></i> | <i><b>Lemma</b></i> | ||
Line 83: | Line 84: | ||
<math>BM \perp MT, TX \perp AB \implies \angle MTX = \angle MBA.</math> | <math>BM \perp MT, TX \perp AB \implies \angle MTX = \angle MBA.</math> | ||
− | <math>\angle CBT = \angle BAC = \frac {\overset{\Large\frown} {BC}}{ 2} | + | <math>\angle CBT = \angle BAC = \frac {\overset{\Large\frown} {BC}}{ 2} \implies \triangle ABC \sim \triangle XTM.</math> |
− | |||
<i><b>Solution</b></i> | <i><b>Solution</b></i> | ||
− | + | [[File:2020 AIME II 15a.png|300px|right]] | |
<math>\triangle ABC \sim \triangle XTM \sim \triangle YMT.</math> | <math>\triangle ABC \sim \triangle XTM \sim \triangle YMT.</math> | ||
Corresponding sides of similar <math>\triangle XTM \sim \triangle YMT</math> is <math>MT,</math> so | Corresponding sides of similar <math>\triangle XTM \sim \triangle YMT</math> is <math>MT,</math> so | ||
− | < | + | |
+ | <math>\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT</math> – parallelogram. | ||
<cmath>4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.</cmath> | <cmath>4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.</cmath> | ||
The formula for median <math>DT</math> of triangle <math>XYT</math> is | The formula for median <math>DT</math> of triangle <math>XYT</math> is | ||
− | <cmath>2 DT^2 = XT^2 + TY^2 – XY^2 | + | <cmath>2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},</cmath> |
<cmath>3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,</cmath> | <cmath>3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,</cmath> | ||
<cmath>3 \cdot XY^2 = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}. </cmath> | <cmath>3 \cdot XY^2 = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}. </cmath> |
Revision as of 08:16, 3 June 2022
Contents
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 (Law of Cosines)
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 have that
and
using our midpoints. Note that
, so by the Pythagorean Theorem, it follows that
. We were also given that
, which we multiply by
to use equation
.
Since
, we have
Therefore,
. ~ MathLuis
Solution 4 (Similarity and median)
Lemma
Let be an acute scalene triangle with circumcircle
. The tangents to
at
and
intersect at
. Let
be the projections of
onto line
. Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.
Proof
the quadrilateral
is cyclic.
Solution
Corresponding sides of similar is
so
– parallelogram.
The formula for median
of triangle
is
~vvsss, www.deoma-cmd.ru
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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