Difference between revisions of "2015 AMC 12B Problems/Problem 19"
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==Problem== | ==Problem== | ||
+ | In <math>\triangle ABC</math>, <math>\angle C = 90^\circ</math> and <math>AB = 12</math>. Squares <math>ABXY</math> and <math>CBWZ</math> are constructed outside of the triangle. The points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle. What is the perimeter of the triangle? | ||
+ | <math>\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32</math> | ||
+ | ==Solution 1== | ||
+ | <asy> | ||
+ | pair A,B,C,M,E,W,Z,X,Y; | ||
+ | A=(2,0); | ||
+ | B=(0,2); | ||
+ | C=(0,0); | ||
+ | M=(A+B)/2; | ||
+ | W=(-2,2); | ||
+ | Z=(-2,-0); | ||
+ | X=(2,4); | ||
+ | Y=(4,2); | ||
+ | E=(W+Z)/2; | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(W--B--C--Z--cycle); | ||
+ | draw(A--B--X--Y--cycle); | ||
+ | dot(M); | ||
+ | dot(E); | ||
+ | label("W",W,NW); | ||
+ | label("Z",Z,SW); | ||
+ | label("C",C,S); | ||
+ | label("A",A,S); | ||
+ | label("B",B,N); | ||
+ | label("X",X,NE); | ||
+ | label("Y",Y,SE); | ||
+ | label("E",E,1.5*plain.W); | ||
+ | label("M",M,NE); | ||
+ | draw(circle(M,sqrt(10))); | ||
+ | </asy> | ||
+ | First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of <math>WZ</math> and <math>XY</math> and finding their intersection point. This point happens to be the midpoint of <math>AB</math>, the hypotenuse. Let this point be <math>M</math>. To find the radius, determine <math>MY</math>, where <math>MY^{2} = MA^2 + AY^2</math>, <math>MA = \frac{12}{2} = 6</math>, and <math>AY = AB = 12</math>. Thus, the radius <math>=r =MY = 6\sqrt5</math>. | ||
− | ==Solution== | + | Next we let <math>AC = b</math> and <math>BC = a</math>. Consider the right triangle <math>ACB</math> first. Using the Pythagorean theorem, we find that <math>a^2 + b^2 = 12^2 = 144</math>. |
+ | <asy> | ||
+ | pair A,B,C,M,E,W,Z,X,Y; | ||
+ | A=(2,0); | ||
+ | B=(0,2); | ||
+ | C=(0,0); | ||
+ | M=(A+B)/2; | ||
+ | W=(-2,2); | ||
+ | Z=(-2,-0); | ||
+ | X=(2,4); | ||
+ | Y=(4,2); | ||
+ | E=(W+Z)/2; | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(W--B--C--Z--cycle); | ||
+ | draw(A--B--X--Y--cycle); | ||
+ | dot(M); | ||
+ | dot(E); | ||
+ | label("W",W,NW); | ||
+ | label("Z",Z,SW); | ||
+ | label("C",C,S); | ||
+ | label("A",A,S); | ||
+ | label("B",B,N); | ||
+ | label("X",X,NE); | ||
+ | label("Y",Y,SE); | ||
+ | label("E",E,1.5*plain.W); | ||
+ | label("M",M,NE); | ||
+ | draw(circle(M,sqrt(10))); | ||
+ | draw(E--Z--M--cycle,dashed); | ||
+ | </asy> | ||
+ | Now, we let <math>E</math> be the midpoint of <math>WZ</math>, and we consider right triangle <math>ZEM</math>. By the Pythagorean theorem, we have that <math>\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180</math>. Expanding this equation, we get that | ||
+ | |||
+ | <cmath>\frac{1}{4}(a^2+b^2) + a^2 + ab = 180</cmath> | ||
+ | <cmath>\frac{144}{4} + a^2 + ab = 180</cmath> | ||
+ | <cmath>a^2 + ab = 144 = a^2 + b^2</cmath> | ||
+ | <cmath>ab = b^2</cmath> | ||
+ | <cmath>b = a</cmath> | ||
+ | |||
+ | This means that <math>ABC</math> is a 45-45-90 triangle, so <math>a = b = \frac{12}{\sqrt2} = 6\sqrt2</math>. Thus the perimeter is <math>a + b + AB = 12\sqrt2 + 12</math> which is answer <math>\boxed{\textbf{(C)}\; 12 + 12\sqrt2}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | The center of the circle on which <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie must be equidistant from each of these four points. Draw the perpendicular bisectors of <math>\overline{XY}</math> and of <math>\overline{WZ}</math>. Note that the perpendicular bisector of <math>\overline{WZ}</math> is parallel to <math>\overline{CW}</math> and passes through the midpoint of <math>\overline{AC}</math>. Therefore, the triangle that is formed by <math>A</math>, the midpoint of <math>\overline{AC}</math>, and the point at which this perpendicular bisector intersects <math>\overline{AB}</math> must be similar to <math>\triangle ABC</math>, and the ratio of a side of the smaller triangle to a side of <math>\triangle ABC</math> is 1:2. Consequently, the perpendicular bisector of <math>\overline{XY}</math> passes through the midpoint of <math>\overline{AB}</math>. The perpendicular bisector of <math>\overline{WZ}</math> must include the midpoint of <math>\overline{AB}</math> as well. Since all points on a perpendicular bisector of any two points <math>M</math> and <math>N</math> are equidistant from <math>M</math> and <math>N</math>, the center of the circle must be the midpoint of <math>\overline{AB}</math>. | ||
+ | |||
+ | Now the distance between the midpoint of <math>\overline{AB}</math> and <math>Z</math>, which is equal to the radius of this circle, is <math>\sqrt{12^2 + 6^2} = \sqrt{180}</math>. Let <math>a=AC</math>. Then the distance between the midpoint of <math>\overline{AB}</math> and <math>Y</math>, also equal to the radius of the circle, is given by <math>\sqrt{\left(\frac{a}{2}\right)^2 + \left(a + \frac{\sqrt{144 - a^2}}{2}\right)^2}</math> (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have | ||
+ | |||
+ | <cmath>\left(\frac{a}{2}\right)^2+\left(a+\frac{\sqrt{144-a^2}}{2}\right)^{2} = 180</cmath> | ||
+ | <cmath>144 - a^2 = a\sqrt{144-a^2}</cmath> | ||
+ | <cmath>(144-a^2)^2 = a^2(144-a^2)</cmath> | ||
+ | |||
+ | Since <math>a</math> cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by <math>(144-a^2)</math>, and arrive at <math>a = 6\sqrt{2}</math>. The length of other leg of the triangle must be <math>\sqrt{144-72} = 6\sqrt{2}</math>. Thus, the perimeter of the triangle is <math>12+2(6\sqrt{2}) = \boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | In order to solve this problem, we can search for similar triangles. Begin by drawing triangle <math>ABC</math> and squares <math>ABXY</math> and <math>ACWZ</math>. Draw segments <math>\overline{YZ}</math> and <math>\overline{WX}</math>. Because we are given points <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle, we can conclude that <math>WXYZ</math> forms a cyclic quadrilateral. Take <math>\overline{AC}</math> and extend it through a point <math>P</math> on <math>\overline{YZ}</math>. Now, we must do some angle chasing to prove that <math>\triangle WBX</math> is similar to <math>\triangle YAZ</math>. | ||
+ | |||
+ | Let <math>\alpha</math> denote the measure of <math>\angle ABC</math>. Following this, <math>\angle BAC</math> measures <math>90 - \alpha</math>. By our construction, <math>\overline{CAP}</math> is a straight line, and we know <math>\angle YAB</math> is a right angle. Therefore, <math>\angle PAY</math> measures <math>\alpha</math>. Also, <math>\angle CAZ</math> is a right angle and thus, <math>\angle ZAP</math> is a right angle. Sum <math>\angle ZAP</math> and <math>\angle PAY</math> to find <math>\angle ZAY</math>, which measures <math>90 + \alpha</math>. We also know that <math>\angle WBY</math> measures <math>90 + \alpha</math>. Therefore, <math>\angle ZAY = \angle WBX</math>. | ||
+ | |||
+ | Let <math>\beta</math> denote the measure of <math>\angle AZY</math>. It follows that <math>\angle WZY</math> measures <math>90 + \beta^\circ</math>. Because <math>WXYZ</math> is a cyclic quadrilateral, <math>\angle WZY + \angle YXW = 180^\circ</math>. Therefore, <math>\angle YXW</math> must measure <math>90 - \beta</math>, and <math>\angle BXW</math> must measure <math>\beta</math>. Therefore, <math>\angle AZY = \angle BXW</math>. | ||
+ | |||
+ | <math>\angle ZAY = \angle WBX</math> and <math>\angle AZY = \angle BXW</math>, so <math>\triangle AZY \sim \triangle BXW</math>! Let <math>x = AC = WC</math>. By Pythagorean theorem, <math>BC = \sqrt{144-x^2}</math>. Now we have <math>WB = WC + BC = x + \sqrt{144-x^2}</math>, <math>BX = 12</math>, <math>YA = 12</math>, and <math>AZ = x</math>. | ||
+ | We can set up an equation: | ||
+ | |||
+ | <cmath>\frac{YA}{AZ} = \frac{WB}{BX}</cmath> | ||
+ | <cmath>\frac{12}{x} = \frac{x+\sqrt{144-x^2}}{12}</cmath> | ||
+ | <cmath>144 = x^2 + x\sqrt{144-x^2}</cmath> | ||
+ | <cmath>12^2 - x^2 = x\sqrt{144-x^2}</cmath> | ||
+ | <cmath>12^4 - 2*12^2*x^2 + x^4 = 144x^2 - x^4</cmath> | ||
+ | <cmath>2x^4 - 3(12^2)x^2 + 12^4 = 0</cmath> | ||
+ | <cmath>(2x^2 - 144)(x^2 - 144) = 0</cmath> | ||
+ | |||
+ | Solving for <math>x</math>, we find that <math>x = 6\sqrt{2}</math> or <math>x = 12</math>, which we omit. The perimeter of the triangle is <math>12 + x + \sqrt{144-x^2}</math>. Plugging in <math>x = 6\sqrt{2}</math>, we get <math>\boxed{\textbf{(C)}\; 12+12\sqrt{2}}</math>. | ||
+ | |||
+ | |||
+ | Alternatively, let <math>BC = S_1, AB = S_2</math> and <math>AC = x</math>. Because <math>\frac{WB}{BX} = \frac{AY}{AZ}</math>, we get that <math>S_2^2 = S_1^2 + S_1x</math>. <math>S_1 = x</math> satisfies the equation because of Pythagorean theorem, so <math>\triangle ABC</math> is right isosceles. | ||
+ | |||
+ | ==Solution 4== | ||
+ | We claim that <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles right triangle. | ||
+ | |||
+ | ''Proof:'' If <math>\triangle ACB</math> is an isosceles right triangle, then <math>\angle WAY=180º</math>. Therefore, <math>W</math>, <math>A</math>, and <math>Y</math> are collinear. Since <math>WY</math> and <math>YX</math> form a right angle, <math>WX</math> is the diameter of the circumcircle of <math>\triangle WYX</math>. Similarly, <math>Z</math>, <math>A</math>, and <math>X</math> are collinear, and <math>ZX</math> forms a right angle with <math>ZW</math>. Thus, <math>WX</math> is also the diameter of the circumcircle of <math>\triangle WZX</math>. Therefore, since <math>\triangle WYX</math> and <math>\triangle WZX</math> share a circumcircle, <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> lie on a circle if <math>\triangle ACB</math> is an isosceles triangle. | ||
+ | |||
+ | If <math>\triangle ACB</math> is isosceles, then its legs have length <math>6\sqrt{2}</math>. The perimeter of <math>\triangle ACB</math> is <math>\boxed{\textbf{(C) }12+12\sqrt{2}}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Note that because <math>\angle WZA=\angle AYX=90^\circ</math>, these two angles inscribe the semicircle defined by diameter <math>WX</math>. Since <math>A</math> lies on line <math>ZA</math>, and <math>\angle XAY=45^\circ</math> (because <math>ABXY</math> is a square), we can find that <math>\angle ZAY= 180^\circ - 45^\circ = 135^\circ</math>. | ||
+ | |||
+ | Now, we can see that <math>\angle BAC=45^\circ</math> to complete the full <math>360^\circ</math>. Therefore, <math>\triangle ABC</math> is and isosceles right triangle, and <math>AC=BC=6\sqrt2</math>. So Our answer is <math>6\sqrt2 + 6\sqrt 2 + 12 = \boxed{\textbf{(C) }12+12\sqrt{2}}</math>. | ||
+ | . | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | We see a circle and little information of position of the shapes inside the triangle, so we think of things associated with circles and think of cyclic quads. We then notice that quadrilateral <math>WXYZ</math> is cyclic. Then we see that <math>\angle W = \angle X</math> making extensions of <math>\overline{WA}</math> and <math>\overline{XZ}</math> diagonals of <math>WXYZ</math>. Let <math>x = \overline{AC}=\overline{AW}</math>. We can see that <math>\overline{AY} = 12\sqrt{2}</math> and <math>\overline{AZ} = x\sqrt{2}</math>. Thus <math>\overline{YZ}^2 = x^2 + (x+12\sqrt{2})^2 = 2x^2 + 24x\sqrt{2} + 288</math>. Let <math>\angle ZAY = \alpha</math>. Then we can see that <math>\angle ZAC</math> and <math>\angle BAY</math> are <math>45</math> degrees, making <math>\angle ZAY</math> <math>\alpha + 90</math> degrees. | ||
+ | |||
+ | We can verify with the cosine addition identity that <math>-\cos{\alpha + 90 degree} = \sin{\alpha}</math> (knowing that <math>\sin{\theta + 90 degree} = \cos{\theta}</math> motivates this). By law of cosine's, <math>\overline{YZ}^2 = (12\sqrt{2})^2 + x\sqrt{2})^2 - 2\cdot 12\sqrt{2} \cdot x\sqrt{2}cos{\alpha + 90 degree} = 288 + x^2 + 48cos{\alpha + 90 degree} = 288 + 2x^2 + 48x\sin{\alpha}</math>. Since <math>\angle{A}</math> is opposite <math>\overline{BC}</math>, <math>\sin{\alpha} = \frac{\sqrt{144-x^2}}{12}</math>. Thus <math>\overline{YZ}^2 = 288 + 2x^2 + 4x\sqrt{144-x^2}</math>. | ||
+ | |||
+ | Setting the first paragraph's <math>\overline{YZ}^2</math> equal to the second: | ||
+ | <cmath>288 + x^2 + 4x\sqrt{144-x^2}2x^2 + 24x\sqrt{2} + 288</cmath> | ||
Latest revision as of 20:05, 7 November 2021
Contents
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
First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of and and finding their intersection point. This point happens to be the midpoint of , the hypotenuse. Let this point be . To find the radius, determine , where , , and . Thus, the radius .
Next we let and . Consider the right triangle first. Using the Pythagorean theorem, we find that . Now, we let be the midpoint of , and we consider right triangle . By the Pythagorean theorem, we have that . Expanding this equation, we get that
This means that is a 45-45-90 triangle, so . Thus the perimeter is which is answer .
Solution 2
The center of the circle on which , , , and lie must be equidistant from each of these four points. Draw the perpendicular bisectors of and of . Note that the perpendicular bisector of is parallel to and passes through the midpoint of . Therefore, the triangle that is formed by , the midpoint of , and the point at which this perpendicular bisector intersects must be similar to , and the ratio of a side of the smaller triangle to a side of is 1:2. Consequently, the perpendicular bisector of passes through the midpoint of . The perpendicular bisector of must include the midpoint of as well. Since all points on a perpendicular bisector of any two points and are equidistant from and , the center of the circle must be the midpoint of .
Now the distance between the midpoint of and , which is equal to the radius of this circle, is . Let . Then the distance between the midpoint of and , also equal to the radius of the circle, is given by (the ratio of the similar triangles is involved here). Squaring these two expressions for the radius and equating the results, we have
Since cannot be equal to 12, the length of the hypotenuse of the right triangle, we can divide by , and arrive at . The length of other leg of the triangle must be . Thus, the perimeter of the triangle is .
Solution 3
In order to solve this problem, we can search for similar triangles. Begin by drawing triangle and squares and . Draw segments and . Because we are given points , , , and lie on a circle, we can conclude that forms a cyclic quadrilateral. Take and extend it through a point on . Now, we must do some angle chasing to prove that is similar to .
Let denote the measure of . Following this, measures . By our construction, is a straight line, and we know is a right angle. Therefore, measures . Also, is a right angle and thus, is a right angle. Sum and to find , which measures . We also know that measures . Therefore, .
Let denote the measure of . It follows that measures . Because is a cyclic quadrilateral, . Therefore, must measure , and must measure . Therefore, .
and , so ! Let . By Pythagorean theorem, . Now we have , , , and . We can set up an equation:
Solving for , we find that or , which we omit. The perimeter of the triangle is . Plugging in , we get .
Alternatively, let and . Because , we get that . satisfies the equation because of Pythagorean theorem, so is right isosceles.
Solution 4
We claim that , , , and lie on a circle if is an isosceles right triangle.
Proof: If is an isosceles right triangle, then . Therefore, , , and are collinear. Since and form a right angle, is the diameter of the circumcircle of . Similarly, , , and are collinear, and forms a right angle with . Thus, is also the diameter of the circumcircle of . Therefore, since and share a circumcircle, , , , and lie on a circle if is an isosceles triangle.
If is isosceles, then its legs have length . The perimeter of is .
Solution 5
Note that because , these two angles inscribe the semicircle defined by diameter . Since lies on line , and (because is a square), we can find that .
Now, we can see that to complete the full . Therefore, is and isosceles right triangle, and . So Our answer is . .
Solution 5
We see a circle and little information of position of the shapes inside the triangle, so we think of things associated with circles and think of cyclic quads. We then notice that quadrilateral is cyclic. Then we see that making extensions of and diagonals of . Let . We can see that and . Thus . Let . Then we can see that and are degrees, making degrees.
We can verify with the cosine addition identity that (knowing that motivates this). By law of cosine's, . Since is opposite , . Thus .
Setting the first paragraph's equal to the second:
See Also
2015 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.