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Difference between revisions of "2025 AIME II Problems/Problem 14"

(→‎Solution 1(Coordinates and Bashy Algebra))
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~ilikemath247365
 
~ilikemath247365
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==Solution 2==
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<asy>
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import math; import geometry; import olympiad;
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point A,C,B,L,K,D,F,G,O; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); D=(16sqrt(3),26); F=(13sqrt(3),13); G=(8sqrt(3),24); O=(8sqrt(3),13);
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draw(A--B--D--C--A--L--C--F--L--K--A--D); draw(K--B--G--D--F--G--K--F); draw(B--O--L); draw(C--O--G);
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label("A",A,SW); label("B",B,NW); label("C",C,SE); label("A'",D,NE); label("K",K,W); label("L",L,NW); label("L'",G,SE); label("K'",F,E); label("O",O,NNW);
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</asy>
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Let <math>O</math> be the midpoint of <math>BC</math>. Take the diagram and rotate it <math>180^{\circ}</math> around <math>O</math> to get the diagram shown. Notice that we have <math>\angle ABC+\angle ACB=90^{\circ}</math>. Because <math>\triangle AKL</math> is equilateral, then <math>\angle KAL=60^{\circ}</math>, so <math>\angle BAK+\angle CAL=30^{\circ}</math>. Because of isosceles triangles <math>\triangle BAK</math> and <math>\triangle CAL</math>, we get that <math>\angle ABK+\angle ACL=30^{\circ}</math> too, implying that <math>\angle KBC+\angle LCB=60^{\circ}</math>. But by our rotation, we have <math>\angle LCO=\angle L'BO</math>, so this implies that <math>\angle KBL'=60^{\circ}</math>, or that <math>\triangle KBL'</math> is equilateral. We can similarly derive that <math>\angle KBO=\angle K'CO</math> implies <math>\angle LCK'=60^{\circ}</math> so that <math>\triangle LK'O</math> is also equilateral. At this point, notice that quadrilateral <math>KL'K'L</math> is a rhombus. The area of our desired region is now <math>[BKLC]=\frac{1}{2}[BL'K'CLK]</math>. We can easily find the areas of <math>\triangle KBL'</math> and <math>\triangle LK'C</math> to be <math>\frac{\sqrt{3}}{4}\cdot 14^2=49\sqrt{3}</math>. Now it remains to find the area of rhombus <math>KL'K'L</math>.
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<asy>
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import math; import geometry; import olympiad;
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point A,K,O,L,M; A=(-7sqrt(3),0); K=(0,7); O=(55sqrt(3)/14,23/14); L=(0,-7); M=(0,0);
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draw(A--K--O--L--A--O--M--A); draw(K--L);
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label("A",A,W); label("K",K,N); label("O",O,E); label("L",L,S); label("M",M,SE);
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</asy>
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Focus on the quadrilateral <math>AKOL</math>. Restate the configuration in another way - we have equilateral triangle <math>\triangle AKL</math> with side length 14, and a point <math>O</math> such that <math>AO=19</math> and <math>\angle KOL=90^{\circ}</math>. We are trying to find the area of <math>\triangle KOL</math>. Let <math>M</math> be the midpoint of <math>KL</math>. We see that <math>AM=7\sqrt{3}</math>, and since <math>M</math> is the circumcenter of <math>\triangle KOL</math>, it follows that <math>MO=7</math>. Let <math>\angle KMO=\theta</math>. From the Law of Cosines in <math>\triangle AMO</math>, we can see that <cmath>(7\sqrt{3})^2+7^2-2(7\sqrt{3})(7)\cos (\angle AMO)=361,</cmath> so after simplification we get that <math>\cos (\theta +90)=-\frac{55\sqrt{3}}{98}</math>. Then by trigonometric identities this simplifies to <math>\sin \theta =\frac{55\sqrt{3}}{98}</math>. Applying the definition <math>\cos^2\theta +\sin^2\theta =1</math> gives us that <math>\cos \theta =\frac{23}{98}</math>. Applying the Law of Cosines again in <math>\triangle KMO</math>, we get that <cmath>49+49-2\cdot 7\cdot 7\cdot \cos \theta =98-98\cdot \frac{23}{98}=98-23-75=KO^2,</cmath> which tells us that <math>KO=5\sqrt{3}</math>. The Pythagorean Theorem in <math>\triangle KOL</math> gives that <math>OL=11</math>, so the area of <math>\triangle KOL</math> is <math>\frac{55\sqrt{3}}{2}</math>. The rhombus <math>KL'K'L</math> consists of four of these triangles, so its area is <math>4\cdot \frac{55\sqrt{3}}{2}=110\sqrt{3}</math>.
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Finally, the area of hexagon <math>BL'K'CLK</math> is <math>49\sqrt{3}+110\sqrt{3}+49\sqrt{3}=208\sqrt{3}</math>, and since this consists of quadrilaterals <math>BKLC</math> and <math>CK'L'B</math> which must be congruent by that rotation, the area of <math>BKLC</math> is <math>104\sqrt{3}</math>. Therefore the answer is <math>\boxed{104}</math>.

Revision as of 23:25, 13 February 2025

Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$

Solution 1(Coordinates and Bashy Algebra)

By drawing our the triangle, I set A to be (0, 0) in the coordinate plane. I set C to be (x, 0) and B to be (0, y). I set K to be (a, b) and L to be (c, d). Then, since all of these distances are 14, I used coordinate geometry to set up the following equations: $a^{2}$ + $b^{2}$ = 196; $a^{2}$ + $(b - y)^{2}$ = 196; $(a - c)^{2}$ + $(b - d)^{2}$ = 196; $c^{2}$ + $d^{2}$ = 196; $(c - x)^{2}$ + $d^{2}$. = 196. Notice by merging the first two equations, the only possible way for it to work is if $b - y$ = $-b$ which means $y = 2b$. Next, since the triangle is right, and we know one leg is $2b$ as $y = 2b$, the other leg, x, is $\sqrt{38^{2} - (2b)^{2}}$.Then, plugging these in, we get a system of equations with 4 variables and 4 equations and solving, we get a = 2, b = 8$\sqrt{3}$, c = 13, d = 3$\sqrt{3}$. Now plugging in all the points and using the Pythagorean Theorem, we get the coordinates of the quadrilateral. By Shoelace, our area is 104$\sqrt{3}$. Thus, the answer is $\boxed{104}$.

~ilikemath247365

Solution 2

[asy] import math; import geometry; import olympiad; point A,C,B,L,K,D,F,G,O; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); D=(16sqrt(3),26); F=(13sqrt(3),13); G=(8sqrt(3),24); O=(8sqrt(3),13); draw(A--B--D--C--A--L--C--F--L--K--A--D); draw(K--B--G--D--F--G--K--F); draw(B--O--L); draw(C--O--G); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("A'",D,NE); label("K",K,W); label("L",L,NW); label("L'",G,SE); label("K'",F,E); label("O",O,NNW); [/asy] Let $O$ be the midpoint of $BC$. Take the diagram and rotate it $180^{\circ}$ around $O$ to get the diagram shown. Notice that we have $\angle ABC+\angle ACB=90^{\circ}$. Because $\triangle AKL$ is equilateral, then $\angle KAL=60^{\circ}$, so $\angle BAK+\angle CAL=30^{\circ}$. Because of isosceles triangles $\triangle BAK$ and $\triangle CAL$, we get that $\angle ABK+\angle ACL=30^{\circ}$ too, implying that $\angle KBC+\angle LCB=60^{\circ}$. But by our rotation, we have $\angle LCO=\angle L'BO$, so this implies that $\angle KBL'=60^{\circ}$, or that $\triangle KBL'$ is equilateral. We can similarly derive that $\angle KBO=\angle K'CO$ implies $\angle LCK'=60^{\circ}$ so that $\triangle LK'O$ is also equilateral. At this point, notice that quadrilateral $KL'K'L$ is a rhombus. The area of our desired region is now $[BKLC]=\frac{1}{2}[BL'K'CLK]$. We can easily find the areas of $\triangle KBL'$ and $\triangle LK'C$ to be $\frac{\sqrt{3}}{4}\cdot 14^2=49\sqrt{3}$. Now it remains to find the area of rhombus $KL'K'L$. [asy] import math; import geometry; import olympiad; point A,K,O,L,M; A=(-7sqrt(3),0); K=(0,7); O=(55sqrt(3)/14,23/14); L=(0,-7); M=(0,0); draw(A--K--O--L--A--O--M--A); draw(K--L); label("A",A,W); label("K",K,N); label("O",O,E); label("L",L,S); label("M",M,SE); [/asy] Focus on the quadrilateral $AKOL$. Restate the configuration in another way - we have equilateral triangle $\triangle AKL$ with side length 14, and a point $O$ such that $AO=19$ and $\angle KOL=90^{\circ}$. We are trying to find the area of $\triangle KOL$. Let $M$ be the midpoint of $KL$. We see that $AM=7\sqrt{3}$, and since $M$ is the circumcenter of $\triangle KOL$, it follows that $MO=7$. Let $\angle KMO=\theta$. From the Law of Cosines in $\triangle AMO$, we can see that \[(7\sqrt{3})^2+7^2-2(7\sqrt{3})(7)\cos (\angle AMO)=361,\] so after simplification we get that $\cos (\theta +90)=-\frac{55\sqrt{3}}{98}$. Then by trigonometric identities this simplifies to $\sin \theta =\frac{55\sqrt{3}}{98}$. Applying the definition $\cos^2\theta +\sin^2\theta =1$ gives us that $\cos \theta =\frac{23}{98}$. Applying the Law of Cosines again in $\triangle KMO$, we get that \[49+49-2\cdot 7\cdot 7\cdot \cos \theta =98-98\cdot \frac{23}{98}=98-23-75=KO^2,\] which tells us that $KO=5\sqrt{3}$. The Pythagorean Theorem in $\triangle KOL$ gives that $OL=11$, so the area of $\triangle KOL$ is $\frac{55\sqrt{3}}{2}$. The rhombus $KL'K'L$ consists of four of these triangles, so its area is $4\cdot \frac{55\sqrt{3}}{2}=110\sqrt{3}$.

Finally, the area of hexagon $BL'K'CLK$ is $49\sqrt{3}+110\sqrt{3}+49\sqrt{3}=208\sqrt{3}$, and since this consists of quadrilaterals $BKLC$ and $CK'L'B$ which must be congruent by that rotation, the area of $BKLC$ is $104\sqrt{3}$. Therefore the answer is $\boxed{104}$.

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