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2025 AIME II Problems/Problem 14

Revision as of 23:27, 13 February 2025 by Iamgreatperson (talk | contribs) (→‎Solution 2)

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 $\boxed{104}$.

~ethanzhang1001

See also

2025 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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