Difference between revisions of "2015 AMC 10B Problems/Problem 19"
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<math>b^2+a^2=12^2</math>, so <math>ab=a^2 \Rightarrow a=b</math> | <math>b^2+a^2=12^2</math>, so <math>ab=a^2 \Rightarrow a=b</math> | ||
− | Therefore, <math>\triangle ABC</math> is an isosceles right triangle. <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, so the perimeter is < | + | Therefore, <math>\triangle ABC</math> is an isosceles right triangle. <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, so the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math> |
~BakedPotato66 | ~BakedPotato66 | ||
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Therefore, <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>. | Therefore, <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>. | ||
− | ~isabelchen | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=gDSIM9SAstk | ||
+ | |||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Fileagingassisitant fileagingassisitant] | ||
==See Also== | ==See Also== |
Latest revision as of 21:12, 25 August 2023
Problem
In ,
and
. Squares
and
are constructed outside of the triangle. The points
, and
lie on a circle. What is the perimeter of the triangle?
Solution 1
The center of the circle lies on the intersection between the perpendicular bisectors of chords and
. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be
. Draw perpendiculars to
and
from
, and connect
and
.
. Let
and
. Then
. Simplifying this gives
. But by Pythagorean Theorem on
, we know
, because
. Thus
. So our equation simplifies further to
. However
, so
, which means
, or
. Aha! This means
is just an isosceles right triangle, so
, and thus the perimeter is
.
Solution 2
Let and
(and we're given that
). Draw line segments
and
. Now we have cyclic quadrilateral
This means that opposite angles sum to . Therefore,
. Simplifying carefully, we get
. Similarly,
=
.
That means .
Setting up proportions,
Cross-multiplying we get:
But also, by Pythagoras,
, so
Therefore, is an isosceles right triangle.
, so the perimeter is
~BakedPotato66
~LegionOfAvatars
Solution 3
Both solution 1 and 2 uses Pythagorean Theorem to prove is isosceles right triangle. I'm going to prove
is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation.
Let be the the midpoint of
. The perpendicular bisector of line
and
will meet at
. Thus
is the center of the circle points
,
,
, and
lies on.
,
,
, and
,
,
by
,
. Because
,
is a
rotation about point
of
. So,
.
Because and
is the radius of
,
. Because
is the midpoint of hypotenuse
,
,
,
by
. Because
,
is a
rotation about point
of
. So,
.
,
,
is isosceles right triangle,
. So,
is isosceles right triangle.
Therefore, , the perimeter is
.
Video Solution
https://www.youtube.com/watch?v=gDSIM9SAstk
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.