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

Revision as of 03:01, 14 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 $\boxed{104}$ .

~ethanzhang1001

Solution 3

From the given condition, we could get $\angle{LAK}=60^{\circ}$ and $\triangle{LCA}, \triangle{BAK}$ are isosceles. Denote $\angle{BAK}=\alpha, \angle{CAL}=30^{\circ}-\alpha$ . From the isosceles condition, we have $\angle{BKA}=180^{\circ}-2\alpha, \angle{CLA}=120^{\circ}-2\alpha$

Since $\angle{CAB}$ is right, then $AB^2+AC^2=BC^2$ , we could use law of cosines to express $AC^2, AB^2, AC^2+AB^2=2\cdot 14^2(2-\cos \angle{BKA}-\angle {CLA})=2\cdot 14^2(2+\cos(2\alpha)+\cos(60^{\circ}-2\alpha))=38^2$

Which simplifies to $\cos(2\alpha)+\cos(60^{\circ}-2\alpha)=\frac{165}{98}$ , expand the expression by angle subtraction formula, we could get $\sqrt{3}\sin(2\alpha+60^{\circ})=\frac{165}{98}, \sin(2\alpha+60^{\circ})=\frac{55\sqrt{3}}{98}$

Conenct $CK$ we could notice $\angle{CLA}=360^{\circ}-\angle{CLA}-\angle{ALK}=180^{\circ}-2\alpha=\angle{AKB}$ , since $CL=LK=AK=KB$ we have $\triangle{CLK}\cong \triangle{AKB}$ . Moreover, since $K$ lies on the perpendicular bisector of $AB$ , the distance from $K$ to $AC$ is half of the length of $AB$ , which means $[ACK]=\frac{[ABC]}{2}$ , and we could have $[ACK]=[ACL]+[ALK]+[ABK]=[ABC]-[BKLC]$ , so $[BKLC]=[AKC]$ . We have $[AKC]=[ALK]+\frac{14^2}{2}(\sin(60-2\alpha)+\sin \alpha)=98(\sin(60+2\alpha))+[ALK]=55\sqrt{3}+\frac{\sqrt{3}}{4}14^2=104\sqrt{3}$ , so our answer is $\boxed{104}$

~Bluesoul

Solution 4 (Trigonometry)

$[asy] import math; import geometry; import olympiad; point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE); markscalefactor=1; draw(anglemark(L,C,A)); draw(anglemark(A,B,K)); [/asy]$ Immediately we should see that $\triangle{AKL}$ is equilateral, so $\angle{KAL}=60$ .

We assume $\angle{LCA}=x$ , and it is easily derived that $\angle{KBA}=30-x$ . Using trigonometry, we can say that $AC=28\cos{x}$ and $AB=28\cos{30-x}$ . Pythagoras tells us that $BC^2=AC^2+AB^2$ so now we evaluate as follows: \begin{align*} 38^2 &=28^2(\cos^2{x}+\cos^2{(30-x)}) \\ (\frac{19}{14})^2 &=\cos^2{x}+(\frac{\sqrt{3}}{2} \cos{x} - \frac{1}{2} \sin{x})^2 \\ &=\cos^2{x}+\frac{3}{4} \cos^2{x}-\frac{\sqrt{3}}{2}\sin{x} \cos{x}+\frac{1}{4}\sin^2{x} \\ &=\frac{3}{2} \cos^2{x}-\frac{\sqrt{3}}{2}\sin{x} \cos{x}+\frac{1}{4} \\ &=\frac{3}{4}(2\cos^2{x}-1)-\frac{\sqrt{3}}{4} (2\sin{x} \cos{x})+1 \\ (\frac{33}{14})(\frac{5}{14})&=\frac{\sqrt{3}}{2}(\frac{\sqrt{3}}{2}(\cos{2x})-\frac{1}{2} (\sin{2x})) \\ \frac{55\sqrt{3}}{98}&=\cos{(30-2x)} \\ \end{align*}

It is obvious that $\angle{ALC}=180-2x$ . We can easily derive $\cos{150+(30-2x)}$ using angle addition we know, and then using cosine rule to find side $AC$ .

\begin{align*} \frac{55\sqrt{3}}{98}=\cos{(30-2x)} \\ \sin{(30-2x)}=\sqrt{1-\cos^2{(30-2x)}}=\frac{23}{98} \\ \cos{(180-2x)}=(-\frac{\sqrt{3}}{2})(\frac{55\sqrt{3}}{98})-(\frac{1}{2})(\frac{23}{98}) \\ \cos{(180-2x)}=-\frac{47}{49} \\ AC^2=14^2+14^2+2\cdot 14\cdot 14\cdot (\frac{47}{49}) \\ AC=\sqrt{768}=16\sqrt3 \\ \end{align*}

We easily find $\cos{x}=\frac{4\sqrt{3}}{7}$ and $\sin{x}=\frac{1}{7}$ (draw a perpendicular down from $L$ to $AC$ ). What we are trying to find is the area of $BKLC$ , which can be found by adding the areas of $\triangle{BKL}$ and $\triangle{BLC}$ . It is trivial that $\triangle{BKL}$ and $\triangle{ACL}$ are congruent, so we know that $BL=28\cos{x}$ . What we require is

\begin{align*} \frac{1}{2}(14)(14)(\sin{(180-2x)})+\frac{1}{2}(14)(28\cos{x})(\sin{(120+x)}) \\ \end{align*}

We do similar calculations to obtain that $\sin{(120+x)}=\frac{11}{14}$ and $\cos{(180-2x)}=-\frac{47}{49}$ implies $\sin{(180-2x)}=\frac{8\sqrt{3}}{49}$ , so now we plug in everything we know to calculate the area of the quadrilateral:

\begin{align*} & \frac{1}{2}(14)(14)(\sin{(180-2x)})+\frac{1}{2}(14)(28\cos{x})(\sin{(120+x)}) \\ &=\frac{1}{2}(14)(14)(\frac{8\sqrt{3}}{49})+\frac{1}{2}(14)(16\sqrt{3})(\frac{11}{14}) \\ &=16\sqrt{3}+88\sqrt{3} \\ &=104\sqrt{3} \\ \end{align*}

We see that $n=\boxed{104}$ .

-lisztepos

Remarks

This problem can be approached either by analytic geometry or by trigonometric manipulation. The characteristics of this problem make it highly similar to 2017 AIME I Problem 15 (Link).

@@ Line 92: / Line 92: @@
 -lisztepos
+==Remarks==
+This problem can be approached either by analytic geometry or by trigonometric manipulation. The characteristics of this problem make it highly similar to 2017 AIME I Problem 15 ([[2017_AIME_I_Problems/Problem_15|Link]]).
 == See also ==

2025 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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