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

Latest revision as of 18:33, 8 July 2025

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (coordinates and bashy algebra)
5 Solution 4 (Trigonometry)
6 Solution 5 (Circles and Trigonometry)
7 Solution 6 (Trig Identities; warning: bashy)
8 Solution 7 (analytic geometry with roots of unity)
9 Solution 8
10 Solution 9
11 Video Solution (Angle chasing and congruent triangles)
12 Remarks
13 See also

Problem

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

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 $\angle{CLK}=360^{\circ}-\angle{CLA}-\angle{ALK}=180^{\circ}-2\alpha=\angle{AKB}$ , 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 $[AKC]=[ALK]+\frac{14^2}{2}(\sin(60-2\alpha)+\sin 2\alpha)=98(\sin(60+2\alpha))+[ALK]=55\sqrt{3}+\frac{\sqrt{3}}{4}14^2=104\sqrt{3}$ , so our answer is $\boxed{104}$

~ Bluesoul

Solution 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 $\triangle LK'C$ 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 (coordinates and bashy algebra)

By drawing out 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 $\sqrt{38^{2} - (2b)^{2}}$ . We now have:

$a^{2} + b^{2} = 196 \hspace{1 cm} \textbf{(1)}$

$(a - c)^{2} + (b - d)^{2} = 196 \hspace{1 cm} \textbf{(2)}$

$c^{2} + d^{2} = 196 \hspace{1 cm} \textbf{(3)}$

$(c - \sqrt{38^{2} - (2b)^{2}})^{2} + d^{2} = 196 \hspace{1 cm} \textbf{(4)}$

Expanding equation (4) and simplifying, we end with $38^{2} - 4b^{2} = 2c \cdot \sqrt{38^{2} - 4b^{2}}$ . Next, squaring both sides and canceling terms, we have $4 \cdot 19^{2} - 4b^{2} = 4c^{2}$ which tells us that $c^{2} = 19^{2} - b^{2}$ . Now, plugging this value in into equation (3) tells us that $d^{2} = 196 - 19^{2} + b^{2}$ . We expand equation (2) to get $a^{2} - 2ac + c^{2} + b^{2} - 2bd + d^{2} = 196$ . Using equation (1), we can cancel terms and shift things over to get $196 = 2ac + 2bd$ which means $98 = ac + bd$ . From equation (1), we have $a^{2} = 196 - b^{2}$ . Now, plugging in all of our variables in terms of $b$ to this new equation, we have $98 = \sqrt{(361 - b^{2})(196 - b^{2}} + b\sqrt{b^{2} - 165}$ . We now move things over to get $98 - b\sqrt{b^{2} - 165} = \sqrt{(361 - b^{2})(196 - b^{2}}$ . Squaring both sides and canceling, we have $98^{2} - 196b\sqrt{b^{2} - 165} = (361)(196) - 392b^{2}$ . We can now divide both sides by $196$ to get $49 - b\sqrt{b^{2} - 165} = 361 - 2b^{2}$ . Rearranging and simplifying, we now have $2b^{2} - 312 = b\sqrt{b^{2} - 165}$ . Squaring both sides and combining like terms, we have $3b^{4} - 1083b^{2} + 312^{2} = 0$ . This part will be a bit of a bash. Quadratic Formula tells us that $b^{2} = \frac{1083 \pm \sqrt{1083^{2} - 12 \cdot 312^{2}}}{6}$ . The discriminant nicely simplifies to $\sqrt{4761} = 69$ (this will be an extremely long bash but it's worth it). In fact, after computing, we end up with $b^{2} = 192, 169$ . This leads us to solutions of $b = 8\sqrt{3}, 13$ . If we choose $b = 8\sqrt{3}$ , then $c^{2} = 361 - b^{2} = 361 - 192 = 169$ which tells us that $c = 13$ .(In fact if you choose $b = 13$ , then $c = 8\sqrt{3}$ so it's symmetrical and doesn't matter which one you choose). Next, $a^{2} = 196 - b^{2} = 196 - 192 = 4$ which tells us that $a = 2$ . Finally, $d^{2} = 196 - c^{2} = 196 - 169 = 27$ which tells us that $d = 3\sqrt{3}$ . Therefore, after all the bashing, our solution quadruple is $a = 2, b = 8\sqrt{3}, c = 13, d = 3\sqrt{3}$ . Now plugging in all the points and using the Pythagorean Theorem, we get the coordinates of the quadrilateral which are $(0, 16\sqrt{3}), (2, 8\sqrt{3}), (13, 3\sqrt{3})$ , and $(26, 0)$ . By Shoelace, our area is $104\sqrt{3}$ . Thus, the answer is $\boxed{104}$ .

~ilikemath247365

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: $\begin{aligned} 38^{2} & = 28^{2} (\cos^{2} x + \cos^{2} (30 - x)) \\ (\frac{19}{14})^{2} & = \cos^{2} x + (\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x)^{2} \\ = \cos^{2} x + \frac{3}{4} \cos^{2} x - \frac{\sqrt{3}}{2} \sin x \cos x + \frac{1}{4} \sin^{2} x \\ = \frac{3}{2} \cos^{2} x - \frac{\sqrt{3}}{2} \sin x \cos x + \frac{1}{4} \\ = \frac{3}{4} (2 \cos^{2} x - 1) - \frac{\sqrt{3}}{4} (2 \sin x \cos x) + 1 \\ (\frac{33}{14}) (\frac{5}{14}) & = \frac{\sqrt{3}}{2} (\frac{\sqrt{3}}{2} (\cos 2 x) - \frac{1}{2} (\sin 2 x)) \\ \frac{55 \sqrt{3}}{98} & = \cos (30 - 2 x) \end{aligned}$

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 $AC$ .

$\begin{array}{r} \frac{55 \sqrt{3}}{98} = \cos (30 - 2 x) \\ \sin (30 - 2 x) = \sqrt{1 - \cos^{2} (30 - 2 x)} = \frac{23}{98} \\ \cos (180 - 2 x) = (- \frac{\sqrt{3}}{2}) (\frac{55 \sqrt{3}}{98}) - (\frac{1}{2}) (\frac{23}{98}) \\ \cos (180 - 2 x) = - \frac{47}{49} \\ A C^{2} = 14^{2} + 14^{2} + 2 \cdot 14 \cdot 14 \cdot (\frac{47}{49}) \\ A C = \sqrt{768} = 16 \sqrt{3} \end{array}$

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

$\begin{array}{r} \frac{1}{2} (14) (14) (\sin (180 - 2 x)) + \frac{1}{2} (14) (28 \cos x) (\sin (120 + x)) \end{array}$

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:

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

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

~ lisztepos

~ Edited by Aoum

Solution 5 (Circles and Trigonometry)

Since $KB=KL=KA=14$ and $LK=LA=LC=14$ , we can construct 2 circles of radus 14 with $K$ and $L$ as the center of the two circles. Let the intersection of the 2 circles other than $A$ be point $M$ . Connect $BM$ , $CM$ , $KM$ , and $LM$ . Connect $AM$ , which is the radical axis of the 2 circles.

From the figure, we know that $[KLCB] = [KLCMB] - [BMC]$ $[KLCB] = [BKM] + [CLM] + [KLM] - [BMC]$

Let $\angle{BAM} = \theta$ , which means that $\angle{CAM} = \frac{\pi}{2} - \theta$ . For easier calculation, we temporarily define the radius of the 2 circles (which is 14) to be $R$ . $\angle{BAM}$ is an inscribed angle and $\angle{BKM}$ is a central angle, so $\angle{BKM} = 2\angle{BAM} = 2\theta$ . Similar with the other side, $\angle{CLM} = \pi-2\theta$ . $KM = KL = LM = R$ , so $\triangle{BKM}$ is an equilateral triangle.

Using the Law of Cosines, we get the area of each little triangle. $[BKM] = \frac{1}{2}\cdot R^2\cdot\sin(2\theta)$ $[CLM] = \frac{1}{2}\cdot R^2\cdot\sin(\pi-2\theta) = \frac{1}{2}\cdot R^2\cdot\sin(2\theta)$ $[KLM] = \frac{1}{2}\cdot\sin({\frac{\pi}{3}})=\frac{\sqrt3}{4}R^2$ $\begin{aligned} [B M C] & = \frac{1}{2} \cdot | B M | \cdot | M C | \cdot \sin (\frac{5 π}{6}) \\ = \frac{1}{2} \cdot \frac{1}{2} \cdot 2 R \sin (θ) \cdot 2 R \sin (\frac{π}{2} - θ) \\ = R^{2} \cdot \sin (θ) \cos (θ) \\ = \frac{1}{2} \cdot R^{2} \sin (2 θ) \end{aligned}$

We can conclude that $[KLCB] = \frac{1}{2}\cdot R^2\cdot\sin(2\theta)+\frac{1}{2}\cdot R^2\cdot\sin(2\theta)+\frac{\sqrt3}{4}R^2-\frac{1}{2}\cdot R^2\sin(2\theta)$ $[KLCB] = {14}^2\cdot(\frac{\sin(2\theta)}{2}+\frac{\sqrt3}{4})$

Now, we just needed to find the value of $\sin(2\theta)$ . We analyze the $\triangle{BMC}$ . We already know that $\angle{BMC} = {150}^{\circ}$ and $BM = 2R\sin(\theta)$ and $BM = 2R\cos(\theta)$ . Using the Laws of Cosines (again!) and the given condition of $BC = 38$ , we can create a formula on $\theta$ .

${BC}^2 = {BM}^2+{CM}^2-2\cdot BM\cdot MC\cdot\cos(\angle{BMC})$ ${BC}^2 = (2R\sin(\theta))^2+(2R\cos(\theta))^2-2\cdot\cos({150}^{\circ})\cdot(2R\cos(\theta))\cdot(2R\cos(\theta)) = {38}^2$ $4R^2(\sin^2(\theta)+\cos^2(\theta)+\sqrt3\sin(\theta)\cos(\theta)) = {38}^2$ $4R^2(1+\frac{\sqrt3}{2}\cdot\sin(2\theta)) = {38}^2$ $4R^2+\frac{4R^2\sqrt3}{2}\cdot\sin(2\theta)) = {38}^2$ $\sin(2\theta) = \frac{2}{\sqrt3}\cdot(\frac{{38}^2}{4\cdot{38}^2}-1)$ $\sin(2\theta) = \frac{2\cdot165}{\sqrt3\cdot{14}^2} = \frac{165}{98\sqrt3}$

We put the calculated value of $\sin(2\theta)$ back into $[KLCB]$ : $[KLCB] = {14}^2\cdot(\frac{165}{2\cdot98\sqrt3}+\frac{\sqrt3}{4})$ $[KLCB] = 55\sqrt3+49\sqrt3 = 104\sqrt3$

Therefore, $n=\boxed{104}$ .

~cassphe

Solution 6 (Trig Identities; warning: bashy)

Consider a diagram to the original problem (credit to solution 4):

$[asy] import math; import geometry; import olympiad; point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE); markscalefactor=1; [/asy]$

Now, let us simplify the problem further. We know that $K$ and $L$ must lie on the perpendicular bisectors of $AB$ and $AC$ , respectively. The real problem here is the equilateral triangle in the middle, inscribed in a rectangle with diagonal length 19.

We create a further simplified problem: given that the inscribed equilateral triangle of a certain rectangle with diagonal length $19$ has side length $14$ , find the sides and intersection points on this rectangle. For reference, here is a diagram:

$[asy] import math; import geometry; import olympiad; point A,B,C,D,L,K; A=(0,0); D=(13,0); B=(0,8sqrt(3)); C=(13,8sqrt(3)); L=(13,3sqrt(3)); K=(2,8sqrt(3)); draw(A--B--C--D--cycle); draw(A--K--L--cycle); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("K",K,N); label("L",L,E); label("D",D,SE); markscalefactor=1; [/asy]$

Note the angles $\angle{LAD}$ and $\angle{BAK}$ . Since $\angle{LAD} + \angle{BAK} + 60^{\circ} = 90^{\circ}$ , $\angle{LAD} + \angle{BAK} = 30^{\circ}$ , and $\angle{BAK} = 30^{\circ} - \angle{LAD}$ . Thus, let $\angle{LAD} = \alpha$ and $\angle{BAK} = 30 - \alpha$ .

Now, we know that $AB^2 + AD^2 = 19^2$ , as the hypotenuse of the larger right triangle is $38$ . However, we can also express AB and AB in terms of $\alpha$ : $AB = 14(\cos(30^{\circ}-\alpha))$ and $AD = 14(\cos(\alpha))$ . Thus, $\cos^2(\alpha) + \cos^2(30^{\circ}-\alpha) = 361/196$ . We expand this using the cosine difference identity:

$\cos^2(\alpha) + (\cos(30^{\circ})\cos(\alpha) + \sin(30^{\circ})\sin(\alpha))^2 = \frac{361}{196}$

$\frac{7}{4}\cos^2(\alpha) + \frac{1}{4}\sin^2(\alpha) + \frac{\sqrt3}{2}\sin(\alpha)\cos(\alpha) = \frac{361}{196}$

Using the fact that $\sin^2(\alpha) + \cos^2(\alpha) = 1$ , then multiplying the entire equation by $2$ ,

$3\cos^2(\alpha) + \sqrt3\sin(\alpha)\cos(\alpha) = \frac{156}{49}$

Now, to save some writing, let us denote $\sin(\alpha)$ with $x$ , and $\cos(\alpha)$ with $y$ .

We have the following equations:

$x^2 + y^2 = 1$

$3y^2 + \sqrt3xy = \frac{156}{49}$

Substituting $x$ for $y$ , moving $3y^2$ to the left side, squaring, and dividing by 9, we end up with the quartic:

$\frac{4}{3}y^4 - \frac{361}{147}y^2 + \frac{52^2}{49^2} = 0$

Using the quadratic formula, we end up with this:

$y^2 = \frac{\frac{361}{49} \pm \frac{1}{49}\cdot\sqrt{361^2 - 208^2\cdot3}}{8}$

Now, we could just compute $361^2 - 208^2\cdot3$ , but instead, we can do this:

$361^2 - 208^2\cdot3 = (129600 + 720 + 1) - (40000 + 3200 + 64)\cdot3$

$(129600 + 721) - (43200 + 64)\cdot3$

$(129600 + 721) - (129600 + 192) = 529 = 23^2$

Thus, we have two cases:

$1. \cos(\alpha) = \frac{13}{14}$

$2. \cos(\alpha) = \frac{4\sqrt3}{7}$

Both lead to the same side lengths of the rectangle: $8\sqrt3$ , and $13$ . Referring back to our original rectangle diagram and plugging in our trigonometric values, we get that $CK = 13 - 2 = 11$ , and $CL = 8\sqrt3 - 3\sqrt3 = 5\sqrt3$ . Thus, the area of the original quadrilateral is $\frac{88\sqrt3 + 55\sqrt3 + 65\sqrt3}{2}$ , or $\boxed{104}\sqrt3$ .

~Stead

Solution 7 (analytic geometry with roots of unity)

$[asy] import math; import geometry; import olympiad; point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); point M,N; M=(8sqrt(3), 0); N=(0,13); draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(N--K); draw(L--M); label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,ENE); label("L",L,NE); label(“N”,N,W); label(“M”,M,S); point am, ml, an, nk, mc, bn; am=(4sqrt(3), 0); mc=(12sqrt(3),0); ml=(8sqrt(3),1); an=(0,6.5); bn=(0,19.5); nk=(2.5981, 13); label(“a",am,S); label(“a",mc,S); label(“y”,ml,ESE); label(“b”,an,W); label(“b”,bn,W); label(“x”,nk,S); markscalefactor=1; [/asy]$ This diagram is modified from the solution 4 diagram. Let $M$ be the midpoint of $AC$ , and let $N$ be the midpoint of $AB$ .

We place the diagram onto the Cartesian coordinate grid. Let $A = (0, 0)$ , $M = (a, 0)$ , $C = (2a, 0)$ , $N = (0, b)$ , and $B = (0, 2b)$ . We are given $AL = CL$ , so $\triangle ACL$ is isosceles. Therefore, $LM$ is the perpendicular bisector of $AC$ , so we can let $L = (a, y)$ . Similarly, we’re given $AK = BK$ , so $\triangle ABK$ is also isosceles, and $NK$ is the perpendicular bisector of $AB$ . Therefore, we can let $K = (x, b)$ .

We have $AB = 2b$ and $AC = 2a$ . We’re given that $\angle BAC = 90^\circ$ and $BC = 38$ , so by the Pythagorean theorem, $(2a)^2 + (2b)^2 = 38^2 \implies 4a^2 + 4b^2 = 1444 \implies a^2 + b^2 = 361.$

We now place the diagram onto the complex plane. We use the x-axis of the coordinate plane as the complex plane’s real axis, and we use the y-axis of the coordinate plane as the complex plane’s imaginary axis. So, on the complex plane, $A = 0$ , $L = a + yi$ , and $K = x + bi$ . Also, since we are given $AK = KL = AL$ , $\triangle AKL$ is equilateral. In addition, since $AL = AK$ , $\angle KAL = 60^\circ$ , and because we constructed our diagram with $K$ counterclockwise of $L$ (if it were the other way around, we could go through the same steps as this solution, but with variables switched around), $K$ is a $60^\circ$ counterclockwise rotation of $L$ about $A$ , and $L$ is a $60^\circ$ clockwise or $300^\circ$ counterclockwise rotation of $K$ about $A$ .

Rotation on the complex plane is equivalent to multiplying by a root of unity. Here, $K$ and $L$ are rotated a multiple of $60^\circ$ to each other about $A$ . $60^\circ$ is one-sixth of a full circle, so to go from $L$ to $K$ or $K$ to $L$ , we multiply by a 6th root of unity. Specifically, to go from $L$ to $K$ , we multiply by $\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ , and to go from $K$ to $L$ , we multiply by $\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$ .

We multiply the coordinate of $L$ by $\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ on the complex plane to obtain equations for the coordinates of $K$ : $(a + yi)\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) = \left(\dfrac{a}{2} - \dfrac{\sqrt{3}}{2}y\right) + \left(\dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a\right)i = x + bi.$ Equating real and imaginary parts, we obtain $x = \dfrac{a}{2} - \dfrac{\sqrt{3}}{2}y \text{ and } b = \dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a.$

Similarly, we multiply the coordinate of $K$ by $\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$ to obtain equations for the coordinates of $L$ : $(x + bi)\left(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right) = \left(\dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b\right) + \left(\dfrac{b}{2} - \dfrac{\sqrt{3}}{2}x\right) = a + yi.$ Equating real and imaginary parts, we obtain $a = \dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b \text{ and } y = \dfrac{b}{2} - \dfrac{\sqrt{3}}{2}x.$

We now look back at the problem to see what it asks for: $[BKLC]$ . Looking at the diagram, we see we can express the area of the quadrilateral as the area of the big right triangle $ABC$ minus the two isosceles triangles $ABK$ and $ALC$ minus the equilateral triangle $AKL$ : $[BKLC] = [ABC] - [ABK] - [ACL] - [AKL].$

We are given that $AK = KL = AL = 14$ , so the area of equilateral triangle $AKL$ is $\dfrac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}$ . Also, we can use $AC = 2a$ as the base of $\triangle ABC$ and $AB = 2b$ as the height, so $[ABC] = \dfrac{(2a)(2b)}{2} = 2ab$ . Similarly, we use $AC = 2a$ as the base of $\triangle ACL$ and $ML = y$ ( $M = (a, 0)$ and $L = (a, y)$ , so the distance between the two is equal to $y$ ) as the height, so $[ACL] = \dfrac{(2a)(y)}{2} = ay$ . Finally, we use $AB = 2b$ and $KN = x$ ( $N = (0, b)$ and $K = (x, b)$ , so the distance between the two is equal to $x$ ) as the base and height of $\triangle ABK$ respectively, so $[ABK] = \dfrac{(2b)(x)}{2} = bx$ . Therefore, $[BKLC] = [ABC] - [ABK] - [ACL] - [AKL] = 2ab - bx - ay - 49\sqrt{3}.$

We have already shown that $b = \dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a$ . Substituting this into $a^2 + b^2 = 361$ , we have $a^2 + \left(\dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a\right)^2 = 361.$ Expanding this out, we have $a^2 + \dfrac{y^2}{4} + \dfrac{\sqrt{3}}{2}ay + \dfrac{3}{4}a^2 = 361.$ Multiplying both sides by $4$ and rearranging the left side, we have $7a^2 + y^2 + 2ay\sqrt{3} = 1444.$ We previously showed that $AC \perp ML$ , so $AM \perp ML$ (since $M$ is on $AC$ ). Therefore, $\triangle AML$ has a right angle at $M$ . By the Pythagorean theorem, $AM^2 + ML^2 = a^2 + y^2 = AL^2 = 196$ . Subtracting $a^2 + y^2$ from the left side and $196$ from the right side, we obtain $6a^2 + 2ay\sqrt{3} = 1248.$ Dividing both sides of the equation by $2\sqrt{3}$ and factoring $a$ out of the left side, we have $a(a\sqrt{3} + y) = 208\sqrt{3}.$ However, we have $b = \dfrac{y}{2} + \dfrac{\sqrt{3}}{2}a$ , so the expression inside the parentheses is simply $2b$ ! Therefore, $2ab = 208\sqrt{3}.$

The algebra’s not over yet. We also showed that $a = \dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b$ , so substituting that into $a^2 + b^2 + 361$ , we obtain $b^2 + \left(\dfrac{x}{2} + \dfrac{\sqrt{3}}{2}b\right)^2 = 361.$ Expanding this out, we have $b^2 + \dfrac{x^2}{4} + \dfrac{\sqrt{3}}{2}bx + \dfrac{3}{4}b^2 = 361.$ Multiplying both sides by $4$ and rearranging the left side, we now have $7b^2 + x^2 + 2xb\sqrt{3} = 1444.$ Does this equation look familiar? We previously showed that $NK \perp AB$ . Therefore, $NK \perp AN$ (since $N$ is on $AB$ ). So, $\triangle ANK$ has a right angle at $N$ . By the Pythagorean theorem, $AN^2 + KN^2 = b^2 + x^2 = AK^2 = 196$ . Subtracting $b^2 + x^2$ from the left side and $196$ from the right side, we have $6b^2 + 2xb\sqrt{3} = 1248.$ We previously also had the equation $6a^2 + 2ay\sqrt{3} = 1248$ , and adding this equation to the above equation and factoring out $2\sqrt{3}$ , we have $6a^2 + 6b^2 + 2\sqrt{3}(bx + ay) = 2496.$ We previously showed $a^2 + b^2 = 361$ , so $6a^2 + 6b^2 = 6 \cdot 361 = 2166$ . Subtracting $6a^2 + 6b^2$ from the left side and $2166$ from the right side, we obtain $2\sqrt{3}(bx + ay) = 330$ . Finally, dividing both sides by $2\sqrt{3}$ , we have $bx + ay = 55\sqrt{3}.$

We previously arrived at this expression for $[BKLC]$ : $[BKLC] = 2ab - bx - ay - 49\sqrt{3}.$ We now know $2ab = 208\sqrt{3}$ and $bx + ay = 55\sqrt{3}$ , so we can simply substitute them in. Therefore, $[BKLC] = 2ab - bx - ay - 49\sqrt{3} = 208\sqrt{3} - 55\sqrt{3} - 49\sqrt{3} = 104\sqrt{3}.$ Finally, we are given $[BKLC] = n\sqrt{3}$ for some integer $n$ . We know $[BKLC] = 104\sqrt{3}$ , so $n = \boxed{104}$ .

Notice that $[BKLC] = 104\sqrt{3} = \dfrac{208\sqrt{3}}{2} = \dfrac{[ABC]}{2}$ . Is this a coincidence?

~V0305

Solution 8

$AK = BK = KL, \angle AKL = 60 ^\circ \implies \angle ABL = 30^\circ.$

Similarly, $\angle ACK = 30^\circ.$ $\angle BAK = \angle ABK, \angle LBK + \angle ABK = 30^\circ.$ $\angle CAL + \angle BAK = 90^\circ - \angle LAK = 30^\circ \implies$ $\angle CAL = \angle LBK \implies \triangle KLB = \triangle LAC \implies$ $AC = BL.$ Similarly, $AB = CK.$

Denote the area of triangle $X$ as $[X].$ $2[ABL] = AB \cdot BL \cdot \sin 30^\circ = \frac{AB \cdot AC}{2} = [ABC].$ Similarly, $[ABL] = [ACK] = \frac {[ABC]}{2}.$ $[BKLC] = [ABC] - [ABL] - [ALC]+ [BLK] = [ABC] - [ABL] = [ABL].$ By applying the Law of Cosines on $\triangle ABL,$ we get $AB^2 + BL^2 - 2 AB \cdot BL \cos 30^\circ = AL^2 \implies AB^2 + AC^2 - AB \cdot AC \sqrt{3} = AL^2 \implies$ $BC^2 - AL^2 = 4 [BKLC] \sqrt{3} \implies$ $[BLKC] = \frac {38^2-14^2}{4 \sqrt{3}} = \frac{19^2-7^2}{3} \sqrt{3} = 4 \cdot 26 \sqrt{3} = 104 \sqrt{3}.$ vladimir.shelomovskii@gmail.com, vvsss

Solution 9

$\triangle KAL$ is clearly equilateral, and $\triangle AKB$ and $\triangle ALC$ are clearly isosceles. Now, we can do a bit of angle chasing. Say $\angle AKB = 2\theta$ . Then, since $\triangle AKB$ is isosceles, $\angle ABK = 90 - \theta$ . $\angle KAL = 60^\circ$ , so $\angle LAC = \theta - 60$ , so $\angle ALC = 300 - 2\theta$ , so $\angle KLC = 2\theta$ . Since the angles are congruent, $\triangle AKB$ is congruent to $\triangle KLC$ . As such, we can rotate it around the center of $\triangle KAL$ onto side AL to produce $\triangle LAD$ . Because all the triangles are congruent, connecting the congruent corners will all produce the same length. From BC, we see that this length is $38$ , so $\triangle BCD$ is equilateral with side length $38$ . This triangle is comprised of 3 quadrilaterals and $\triangle KAL$ in the center, and it's easy to see that each quadrilateral is congruent to $BKLC$ . As such, $[BKLC] = \frac{[BCD]-[KAL]}{3} = \frac{361 \sqrt{3} - 49 \sqrt{3}}{3} = 104 \sqrt{3}$ , giving an answer of $\boxed{104}$ .

~noob1877

Video Solution (Angle chasing and congruent triangles)

https://youtu.be/APaVy8OqJ04?si=MSBv1fS1ajjqcxN6

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).

~Bloggish

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