Difference between revisions of "2015 AMC 10B Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
− | The center of the circle lies on the perpendicular bisectors of | + | The center of the circle lies on the intersection between the perpendicular bisectors of chords <math>ZW</math> and <math>YX</math>. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be <math>O</math>. Draw perpendiculars to <math>ZW</math> and <math>YX</math> from <math>O</math>, and connect <math>OZ</math> and <math>OY</math>. <math>OY^2=6^2+12^2=180</math>. Let <math>AC=a</math> and <math>BC=b</math>. Then <math>\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180</math>. Simplifying this gives <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180</math>. But by Pythagorean Theorem on <math>\triangle ABC</math>, we know <math>a^2+b^2=144</math>, because <math>AB=12</math>. Thus <math>\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36</math>. So our equation simplifies further to <math>a^2+ab=144</math>. However <math>a^2+b^2=144</math>, so <math>a^2+ab=a^2+b^2</math>, which means <math>ab=b^2</math>, or <math>a=b</math>. <i>Aha</i>! This means <math>\triangle ABC</math> is just an isosceles right triangle, so <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, and thus the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>. |
<asy> | <asy> | ||
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
Line 69: | Line 69: | ||
<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 | ||
~LegionOfAvatars | ~LegionOfAvatars | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Both solution 1 and 2 uses Pythagorean Theorem to prove <math>\triangle ABC</math> is isosceles right triangle. I'm going to prove <math>\triangle ABC</math> is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation. | ||
+ | |||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
+ | import graph; size(11.5cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ | ||
+ | |||
+ | draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); | ||
+ | draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); | ||
+ | /* draw figures */ | ||
+ | draw((3.46,0.96)--(3.44,-3.36)); | ||
+ | draw((3.44,-3.36)--(8.02,-3.44)); | ||
+ | draw((8.02,-3.44)--(3.46,0.96)); | ||
+ | draw((3.46,0.96)--(-0.86,0.98)); | ||
+ | draw((-0.86,0.98)--(-0.88,-3.34)); | ||
+ | draw((-0.88,-3.34)--(3.44,-3.36)); | ||
+ | draw((3.46,0.96)--(8.02,-3.44)); | ||
+ | draw((8.02,-3.44)--(12.42,1.12)); | ||
+ | draw((12.42,1.12)--(7.86,5.52)); | ||
+ | draw((7.86,5.52)--(3.46,0.96)); | ||
+ | draw((5.74,-1.24)--(-0.86,0.98)); | ||
+ | draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4")); | ||
+ | draw((5.74,-1.24)--(7.86,5.52)); | ||
+ | draw((5.74,-1.24)--(10.14,3.32), linetype("4 4")); | ||
+ | draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2")); | ||
+ | draw((8.02,-3.44)--(-0.86,0.98)); | ||
+ | draw((3.44,-3.36)--(7.86,5.52)); | ||
+ | draw((3.44,-3.36)--(5.74,-1.24)); | ||
+ | /* dots and labels */ | ||
+ | dot((3.46,0.96),dotstyle); | ||
+ | label("$A$", (3.2,1.06), NE * labelscalefactor); | ||
+ | dot((3.44,-3.36),dotstyle); | ||
+ | label("$C$", (3.14,-3.86), NE * labelscalefactor); | ||
+ | dot((8.02,-3.44),dotstyle); | ||
+ | label("$B$", (8.06,-3.8), NE * labelscalefactor); | ||
+ | dot((-0.86,0.98),dotstyle); | ||
+ | label("$Z$", (-1.34,1.12), NE * labelscalefactor); | ||
+ | dot((-0.88,-3.34),dotstyle); | ||
+ | label("$W$", (-1.48,-3.54), NE * labelscalefactor); | ||
+ | dot((12.42,1.12),dotstyle); | ||
+ | label("$X$", (12.5,1.24), NE * labelscalefactor); | ||
+ | dot((7.86,5.52),dotstyle); | ||
+ | label("$Y$", (7.94,5.64), NE * labelscalefactor); | ||
+ | dot((5.74,-1.24),dotstyle); | ||
+ | label("$O$", (5.52,-1.82), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>O</math> be the the midpoint of <math>AB</math>. The perpendicular bisector of line <math>WZ</math> and <math>XY</math> will meet at <math>O</math>. Thus <math>O</math> is the center of the circle points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lies on. | ||
+ | |||
+ | <math>\angle ZAC=\angle OAY=90^{\circ}</math>, <math>\angle ZAC+\angle BAC=\angle OAY+\angle BAC</math>, <math>\angle ZAB=\angle CAY</math>, and <math>AZ=AC</math>, <math>AB=AY</math>, <math>\triangle AZB \cong \triangle ACY</math> by <math>SAS</math>, <math>BZ=YC</math>. Because <math>AZ \perp AC</math>, <math>\triangle ACY</math> is a <math>90^{\circ}</math> rotation about point <math>A</math> of <math>\triangle AZB</math>. So, <math>BZ \perp YC</math>. | ||
+ | |||
+ | Because <math>OZ</math> and <math>OY</math> is the radius of <math>\odot O</math>, <math>OZ=OY</math>. Because <math>O</math> is the midpoint of hypotenuse <math>AB</math>, <math>OB=OC</math>, <math>BZ=CY</math>, <math>\triangle OBZ \cong \triangle OCY</math> by <math>SSS</math>. Because <math>BZ \perp CY</math>, <math>\triangle OCY</math> is a <math>90^{\circ}</math> rotation about point <math>O</math> of <math>\triangle OBZ</math>. So, <math>OB \perp OC</math>. | ||
+ | |||
+ | <math>\angle COB = 90^{\circ}</math>, <math>OC=OB</math>, <math>\triangle OCB</math> is isosceles right triangle, <math>\angle ABC=\angle OBC=45^{\circ}</math>. So, <math>\triangle ABC</math> is isosceles right triangle. | ||
+ | |||
+ | Therefore, <math>AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}</math>, the perimeter is <math>\boxed{\textbf{(C)}\ 12+12\sqrt{2}}</math>. | ||
+ | |||
+ | ~[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.