Difference between revisions of "2018 AIME II Problems/Problem 12"
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Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>. | Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>. | ||
− | ==Solution== | + | ==Diagram== |
+ | |||
+ | Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. | ||
+ | <asy> | ||
+ | defaultpen(fontsize(14)+0.6); unitsize(12); | ||
+ | |||
+ | real x=11.25; | ||
+ | pair B1=origin, D1=(x,0), C1=IP(CR(B1,14),CR(D1,10)), A1=OP(CR(B1,10),CR(D1,2*sqrt(65))), P1=extension(A1,C1,B1,D1); | ||
+ | |||
+ | pair A=origin, C=(length(C1-A1),0), B=IP(CR(A,10),CR(C,14)), D=OP(CR(A,2*sqrt(65)),CR(C,10)), P=extension(A,C,B,D); | ||
+ | |||
+ | draw(A--B--C--D--A); | ||
+ | draw(A--C^^B--D,gray+0.4); | ||
+ | dot("$A$",A,dir(A-P)); dot("$B$",B,dir(B-P)); dot("$C$",C,dir(C-P)); dot("$D$",D,dir(D-P)); dot("$P$",P,dir(230)); | ||
+ | |||
+ | pen p=fontsize(10); | ||
+ | label("$10$",A--B,up,p); label("$10$", C--D, 2*right,p); label("$14$", B--C, N,p); label("$2\sqrt{65}$", A--D, SW,p); label("$x$", A--P,down,p); label("$\rho x$", P--C,down,p); label("$\Delta$",(A+B+P)/3, right,p); label("$\Lambda$",(A+D+P)/3, right,p); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. We easily get <math>[PBC]=\rho \Delta</math> and <math>[PCD]=\rho\Lambda</math>. | ||
+ | |||
+ | We are given that <math>[ABP] +[PCD] = [PBC]+[ADP]</math>, which we can now write as <cmath>\Delta + \rho\Lambda = \rho\Delta + \Lambda \qquad \Longrightarrow \qquad \Delta -\Lambda = \rho (\Delta -\Lambda).</cmath> Either <math>\Delta = \Lambda</math> or <math>\rho=1</math>. The former would imply that <math>ABCD</math> is a parallelogram, which it isn't; therefore we conclude <math>\rho=1</math> and <math>P</math> is the midpoint of <math>AC</math>. Let <math>\angle BAD = \theta</math> and <math>\angle BCD = \phi</math>. Then <math>[ABCD]=2\cdot [BCD]=140\sin\phi</math>. On one hand, since <math>[ABD]=[BCD]</math>, we have <cmath>\begin{align}\sqrt{65}\sin\theta = 7\sin\phi \quad \implies \quad 16+49\cos^2\phi = 65\cos^2\theta\end{align}</cmath>whereas, on the other hand, using cosine formula to get the length of <math>BD</math>, we get <cmath>10^2+4\cdot 65 - 40\sqrt{65}\cos\theta = 10^2+14^2-280\cos\phi</cmath><cmath>\begin{align}\tag{2}\implies \qquad 65\cos^2\theta = \left(7\cos\phi+ \frac{8}{5}\right)^2\end{align}</cmath>Eliminating <math>\cos\theta</math> in the above two equations and solving for <math>\cos\phi</math> we get<cmath>\cos\phi = \frac{3}{5}\qquad \implies \qquad \sin\phi = \frac{4}{5}</cmath>which finally yields <math>[ABCD]=2\cdot [BCD] = 140\sin\phi = 112</math>. | ||
+ | |||
+ | ==Solution 2== | ||
For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes | For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes | ||
− | < | + | <cmath>\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP</cmath> |
+ | |||
+ | What an important finding! Note that the opposite sides <math>\overline{AB}</math> and <math>\overline{CD}</math> have equal length, and note that diagonal <math>\overline{DB}</math> bisects diagonal <math>\overline{AC}</math>. This is very similar to what happens if <math>ABCD</math> were a parallelogram with <math>AB = CD = 10</math>, so let's extend <math>\overline{DB}</math> to point <math>E</math>, such that <math>AECD</math> is a parallelogram. In other words, <cmath>AE = CD = 10</cmath> and <cmath>EC = DA = 2\sqrt{65}</cmath> Now, let's examine <math>\triangle ABE</math>. Since <math>AB = AE = 10</math>, the triangle is isosceles, and <math>\angle ABE \cong \angle AEB</math>. Note that in parallelogram <math>AECD</math>, <math>\angle AED</math> and <math>\angle CDE</math> are congruent, so <math>\angle ABE \cong \angle CDE</math> and thus <cmath>\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB</cmath> Define <math>\alpha := \text{m}\angle CDB</math>, so <math>180^\circ - \alpha = \text{m}\angle ABD</math>. | ||
− | + | We use the Law of Cosines on <math>\triangle DAB</math> and <math>\triangle CDB</math>: | |
− | < | + | <cmath>\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha</cmath> |
− | < | + | <cmath>14^2 = 10^2 + BD^2 - 20BD\cos\alpha</cmath> |
Subtracting the second equation from the first yields | Subtracting the second equation from the first yields | ||
− | < | + | <cmath>260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}</cmath> |
+ | |||
+ | This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <cmath>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</cmath> and hence <cmath>\left[ABCD\right] = 56 + 56 = \boxed{112}</cmath> -kgator | ||
+ | |||
+ | |||
+ | Just to be complete -- <math>h_1</math> and <math>h_2</math> can actually be equal. In this case, <math>AP \neq CP</math>, but <math>BP</math> must be equal to <math>DP</math>. We get the same result. -Mathdummy. | ||
+ | |||
+ | ==Solution 3 (Another way to get the middle point)== | ||
+ | |||
+ | So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>, <math>\triangle {BCP}=S_{2}</math>, <math>\triangle {CDP}=S_{3}</math>, <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}.</cmath> Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2.</cmath> So <cmath>(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2.</cmath> So <cmath>S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2.</cmath> As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}.</cmath> Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}.</cmath> Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we can find out that <cmath>S_{3}=S_{4},</cmath> so <math>P</math> is the midpoint of <math>\overline {AC}</math> | ||
+ | |||
+ | ~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ||
+ | |||
+ | ==Solution 4 (With yet another way to get the middle point)== | ||
+ | |||
+ | Denote <math>\angle APB</math> by <math>\alpha</math>. Then <math>\sin(\angle APB)=\sin \alpha = \sin(\angle APD)</math>. | ||
+ | Using the formula for the area of a triangle, we get <cmath>\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha , </cmath> | ||
+ | so <cmath>(AP-CP)(BP-DP)=0</cmath> | ||
+ | Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here). | ||
+ | Now, assume that <math>AP=CP=x</math>, <math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for <math>\triangle APB</math> and <math>\triangle BPC</math>, it is clear that <cmath>x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196) </cmath> or <cmath>\begin{align}x^2+y^2=148\end{align}.</cmath> | ||
+ | Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>\begin{align}\tag{2}x^2+z^2=180\end{align}.</cmath> It follows that <cmath>\begin{align}\tag{3}z^2-y^2=32\end{align}.</cmath> Since <math>\sin\alpha=\sqrt{1-\cos^2\alpha}</math>, <cmath>\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}</cmath> which simplifies to <cmath>\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.</cmath> Plugging this back to equations <math>(1)</math>, <math>(2)</math>, and <math>(3)</math>, it can be solved that <math>x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}</math>. Then, the area of the quadrilateral is <cmath>x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}</cmath> | ||
+ | --Solution by MicGu | ||
+ | |||
+ | ==Solution 5== | ||
+ | As in all other solutions, we can first find that either <math>AP=CP</math> or <math>BP=DP</math>, but it's an AIME problem, we can take <math>AP=CP</math>, and assume the other choice will lead to the same result (which is true). | ||
+ | |||
+ | From <math>AP=CP</math>, we have <math>[DAP]=[DCP]</math>, and <math>[BAP]=[BCP] \implies [ABD] = [CBD]</math>, therefore, | ||
+ | <cmath>\begin{align} | ||
+ | \nonumber \frac 12 \cdot AB\cdot AD\sin A &= \frac 12 \cdot BC\cdot CD\sin C \\ | ||
+ | \Longrightarrow \hspace{1in} 7\sin C &= \sqrt{65}\sin A | ||
+ | \end{align}</cmath> | ||
+ | By Law of Cosines, | ||
+ | <cmath>\begin{align} | ||
+ | \nonumber 10^2+14^2-2\cdot 10\cdot 14\cos C &= 10^2+4\cdot 65-2\cdot 10\cdot 2\sqrt{65}\cos A \\ | ||
+ | \Longrightarrow \hspace{1in} -\frac 85 - 7\cos C &= \sqrt{65}\cos A \tag{2} | ||
+ | \end{align}</cmath> | ||
+ | Square <math>(1)</math> and <math>(2)</math>, and add them, to get | ||
+ | <cmath>\left(\frac 85\right)^2 + 2\cdot \frac 85 \cdot 7\cos C + 7^2 = 65 </cmath> | ||
+ | Solve, <math>\cos C = 3/5 \implies \sin C = 4/5</math>, | ||
+ | <cmath>[ABCD] = 2[BCD] = BC\cdot CD\cdot \sin C = 14\cdot 10\cdot \frac 45 = \boxed{112}</cmath> | ||
+ | -Mathdummy | ||
+ | |||
+ | ==Solution 6== | ||
+ | Either <math>PA=PC</math> or <math>PD=PB</math>. Let <math>PD=PB=s</math>. Applying Stewart's Theorem on <math>\triangle ABD</math> and <math>\triangle BCD</math>, dividing by <math>2s</math> and rearranging, <cmath>\tag{1}CP^2+s^2=148</cmath> <cmath>\tag{2}AP^2+s^2=180</cmath> Applying Stewart on <math>\triangle CAB</math> and <math>\triangle CAD</math>, <cmath>\tag{3} 5CP^2=3AP^2</cmath> Substituting equations 1 and 2 into 3 and rearranging, <math>s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}</math> . By Law of Cosines on <math>\triangle APB</math>, <math>\cos(\angle APB)=\frac{4\sqrt{65}}{65}</math> so <math>\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}</math>. Using <math>[\triangle ABC]=\frac{ab\sin(\angle C)}{2}</math> to find unknown areas, <math>[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}</math>. | ||
+ | |||
+ | -Solution by Garrett | ||
+ | |||
+ | ==Solution 7== | ||
+ | Now we prove P is the midpoint of <math>BD</math>. Denote the height from <math>B</math> to <math>AC</math> as <math>h_1</math>, height from <math>D</math> to <math>AC</math> as <math>h_2</math>.According to the problem, <math>AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2 </math> implies <math> h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2 </math>. Then according to basic congruent triangles we get <math>BP=DP</math> | ||
+ | Firstly, denote that <math>CP=a,BP=b,CP=c,AP=d</math>. Applying Stewart theorem, getting that <math>100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c</math>, denote <math>b=5x,c=3x</math> | ||
+ | Applying Stewart Theorem, getting <math>260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)</math> solve for a, getting <math>a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}</math> | ||
+ | Now everything is clear, we can find <math>cos\angle{BPA}=\frac{4}{\sqrt{65}}</math> using LOC, <math>sin\angle{BPA}=\frac{7\sqrt{65}}{65}</math>, the whole area is <math>\sqrt{130}*8\sqrt{2}*\frac{7\sqrt{65}}{65}=\boxed{112}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Solution 8 (Simple Geometry)== | ||
+ | [[File:AIME-II-2018-12.png|400px|right]] | ||
+ | <math>BP = PD</math> as in another solutions. | ||
+ | |||
+ | Let <math>D'</math> be the reflection of <math>D</math> across <math>C</math>. | ||
+ | Let points <math>E, E',</math> and <math>H</math> be the foot of perpendiculars on <math>AC</math> from <math>D,D'</math>, and <math>B</math> respectively. | ||
+ | <cmath>\begin{align*} | ||
+ | &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ | ||
+ | \Rightarrow &\qquad \triangle ABH = \triangle CDE = \triangle CD'E' \\ | ||
+ | \Rightarrow &\qquad \angle BAC = \angle ACD' \\ | ||
+ | \Rightarrow &\qquad \triangle ABC = \triangle AD'C \\ | ||
+ | \Rightarrow &\qquad BC = AD'. | ||
+ | \end{align*}</cmath> | ||
+ | The area of quadrilateral <math>ABCD</math> is equal to the area of triangle <math>ADD'</math> with sides <math>AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20</math>. | ||
+ | The semiperimeter is <math>s = 17 + \sqrt{65},</math> the area <cmath>[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 9 (Mindless Law of Cosines Bash)== | ||
+ | |||
+ | Use your favorite method to get that <math>P</math> is the midpoint of one of the two diagonals (suppose it's the midpoint of <math>\overline{AC}</math>). From here, let <math>x=AP=PC, y=BP, z=PD, a=\cos\theta</math> where <math>\theta</math> is the angle that the diagonals make. Then we have a system of four equations: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | x^2+y^2+2xya &= 100 \\ | ||
+ | z^2+x^2+2xza &= 100 \\ | ||
+ | x^2+y^2-2xya &= 196 \\ | ||
+ | x^2+z^2-2xza &= 260 \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | From these equations we get that | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | xya &= -24 \\ | ||
+ | xza &= -40 \\ | ||
+ | x^2+y^2-48 &= 10 \\ | ||
+ | x^2+z^2-80 &= 10 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | From here we can see that <math>\frac{z}{y}={5}{3}, z^2-y^2=32,</math> so <math>z=5\sqrt{2}, y=3\sqrt{2}.</math> Furthermore, this implies <math>x=\sqrt{130}</math> and <math>xa=-4\sqrt{2},</math> which implies <math>a=\cos\theta=\frac{4}{\sqrt{65}}.</math> Then note that the area of the quadrilateral is <cmath>\frac{1}{2}\sin\theta (xy+xz+xz+xy)=\sin\theta (\sqrt{130}\cdot 3\sqrt{2}+\sqrt{130} \cdot 5\sqrt{2})=7\cdot (3\cdot 2+5\cdot 2)=7(6+10)=7\cdot 16=\boxed{112}.</cmath> | ||
+ | |||
+ | ~Dhillonr25 | ||
+ | ==Solution 10== | ||
+ | Note that <math>\angle APB = \angle CPD = 180-\angle APD = 180-\angle BPC.</math>(All angles are in degrees) | ||
+ | Since <math>\sin(\theta)=\sin(180-\theta),</math> we can use sine area formula to get the following(after some simplifying steps): | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | BP \times AP + CP \times DP = BP \times PC + AP \times PD. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | For convenience, let <math>AP=a, BP=b, CP=c,DP=d.</math> The above equation simplifies to: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | ab + cd = bc + ad | ||
+ | \\ab-ad+cd-bc=0 | ||
+ | \\a(b-d)-c(b-d)=0 | ||
+ | \\(a-c)(b-d)=0 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | From here, we see that <math>a=c</math> or <math>b=d</math>. Without loss of generality, let <math>a=c</math>. Since triangles <math>ABP</math> and <math>CDP</math> are obviously not congruent, we see that one triangle is obtuse and the other one is acute.(Refer to the diagram) However, if we drop perpendiculars from <math>B</math> to <math>AC</math> and <math>D</math> to <math>AC</math>, we do get congruent triangles. If the foot of the perpendicular from <math>B</math> is <math>M</math>, and the foot of the perpendicular from <math>D</math> is <math>N</math>, then right triangle <math>BMP</math> is congruent to right triangle <math>DNP</math>. | ||
+ | From here, we see that the altitudes of triangles <math>ABC</math> and <math>ADC</math> to <math>AC</math> are equal. Since they share base <math>AC</math>, their areas are equal. We can use Heron's formula. To not have any fractions, let <math>AC=2x.</math> | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sqrt{(12+x)(12-x)(x+2)(x-2)}=\sqrt{5+\sqrt{65}+x)(5+\sqrt{65}-x)(5-\sqrt{65}+x)(\sqrt{65}-5+x)} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Even though this looks bad at first, it actually isn't too complicated to simplify. Expanding the differences of squares and simplifying completely, we get <math>x^2=32.</math> Plugging this <math>x</math> back into the Heron's formula, we get that the area of <math>ABC</math>(or <math>ADC</math>) is <math>56</math>. Since these triangles have equal area, the area of the quadrilateral is <math>2 \times 56 = \boxed{112}</math>, and we are done. <math>\blacksquare</math> | ||
+ | |||
+ | ==Solution 11== | ||
+ | [[File:2018_AIME_II_Problem_12_Parallelogram.png|400px|right]] | ||
+ | |||
+ | Use any method to derive that <math>P</math> is the midpoint of <math>A</math> and <math>C</math>. Now, either construct a parallelogram with as shown in the diagram to the right, or use law of sines. We will take the parallelogram route here, but the same thing can be done with law of sines on triangles <math>\triangle \textnormal{ABP}</math> and <math>\triangle \textnormal{CPD}</math>. Reflect <math>D</math> across <math>P</math> to get <math>D'</math>. Since <math>CD = AD' = AB = 10</math>, <math>\triangle \textnormal{ABD'}</math> is isosceles. Thus, <math>\angle AD'B = \angle ABD'</math>, and because <math>ADCD'</math> is a parallelogram (since <math>AP = PC</math> and <math>DP = PD'</math>), <math>\angle AD'B = \angle BDC = \angle ABD'</math>. So, <math>\angle ABD = 180 - \angle ABD' = 180 - \angle BDC</math>. Now, apply law of cosines on <math>\triangle \textnormal{ABD}</math> and <math>\triangle \textnormal{CDB}</math>. We get: | ||
+ | \begin{align} | ||
+ | 100 + BD^2 - 20BD \cos{\angle ABD} &= 100 + BD^2 - 20 BD \cos {(180 - \angle BDC)} = \\ | ||
+ | 100 + BD^2 + 20 BD \cos{\angle BDC} &= 260 \\ | ||
+ | &\textnormal{and} \\ | ||
+ | 100 + BD^2 - 20 BD \cos{\angle BDC} &= 196 \\ | ||
+ | \textnormal{summing }&\textnormal{and simplifying,} \\ | ||
+ | BD &= 8\sqrt{2} | ||
+ | \end{align} | ||
+ | Then, applying law of cosines on <math>\triangle \textnormal{BCD}</math> again, we obtain | ||
+ | <cmath> 100 + 196 - 280 \cos{\angle BCD} = BD^2 = 128 \implies \cos{\angle BCD} = \frac{3}{5} \implies \sin{\angle BCD} = \frac{4}{5}</cmath> | ||
+ | Since <math>AP = PD</math>, <math>[ABD] = [BCD] \implies [ABCD] = [ABD] + [BCD] = 2[BCD]</math>. Thus, <math>[ABCD] = 2[BCD] = 2 \cdot \frac{1}{2} \cdot 10 \cdot 14 \sin{\angle BCD} = 140 \cdot \frac{4}{5} = \boxed{112}</math>. | ||
− | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez] | |
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtube.com/watch?v=2BsYR1dJn9c | ||
+ | ~r00tsOfUnity | ||
+ | ==See Also== | ||
{{AIME box|year=2018|n=II|num-b=11|num-a=13}} | {{AIME box|year=2018|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:12, 24 January 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3 (Another way to get the middle point)
- 6 Solution 4 (With yet another way to get the middle point)
- 7 Solution 5
- 8 Solution 6
- 9 Solution 7
- 10 Solution 8 (Simple Geometry)
- 11 Solution 9 (Mindless Law of Cosines Bash)
- 12 Solution 10
- 13 Solution 11
- 14 Video Solution by MOP 2024
- 15 See Also
Problem
Let be a convex quadrilateral with
,
, and
. Assume that the diagonals of
intersect at point
, and that the sum of the areas of triangles
and
equals the sum of the areas of triangles
and
. Find the area of quadrilateral
.
Diagram
Let and let
. Let
and let
.
Solution 1
Let and let
. Let
and let
. We easily get
and
.
We are given that , which we can now write as
Either
or
. The former would imply that
is a parallelogram, which it isn't; therefore we conclude
and
is the midpoint of
. Let
and
. Then
. On one hand, since
, we have
whereas, on the other hand, using cosine formula to get the length of
, we get
Eliminating
in the above two equations and solving for
we get
which finally yields
.
Solution 2
For reference, , so
is the longest of the four sides of
. Let
be the length of the altitude from
to
, and let
be the length of the altitude from
to
. Then, the triangle area equation becomes
What an important finding! Note that the opposite sides and
have equal length, and note that diagonal
bisects diagonal
. This is very similar to what happens if
were a parallelogram with
, so let's extend
to point
, such that
is a parallelogram. In other words,
and
Now, let's examine
. Since
, the triangle is isosceles, and
. Note that in parallelogram
,
and
are congruent, so
and thus
Define
, so
.
We use the Law of Cosines on and
:
Subtracting the second equation from the first yields
This means that dropping an altitude from to some foot
on
gives
and therefore
. Seeing that
, we conclude that
is a 3-4-5 right triangle, so
. Then, the area of
is
. Since
, points
and
are equidistant from
, so
and hence
-kgator
Just to be complete -- and
can actually be equal. In this case,
, but
must be equal to
. We get the same result. -Mathdummy.
Solution 3 (Another way to get the middle point)
So, let the area of triangles
,
,
,
. Suppose
and
, then it is easy to show that
Also, because
we will have
So
So
So
So
As a result,
Then, we have
Combine the condition
we can find out that
so
is the midpoint of
~Solution by (Frank FYC)
Solution 4 (With yet another way to get the middle point)
Denote by
. Then
.
Using the formula for the area of a triangle, we get
so
Hence
(note that
makes no difference here).
Now, assume that
,
, and
. Using the cosine rule for
and
, it is clear that
or
Likewise, using the cosine rule for triangles
and
,
It follows that
Since
,
which simplifies to
Plugging this back to equations
,
, and
, it can be solved that
. Then, the area of the quadrilateral is
--Solution by MicGu
Solution 5
As in all other solutions, we can first find that either or
, but it's an AIME problem, we can take
, and assume the other choice will lead to the same result (which is true).
From , we have
, and
, therefore,
By Law of Cosines,
Square
and
, and add them, to get
Solve,
,
-Mathdummy
Solution 6
Either or
. Let
. Applying Stewart's Theorem on
and
, dividing by
and rearranging,
Applying Stewart on
and
,
Substituting equations 1 and 2 into 3 and rearranging,
. By Law of Cosines on
,
so
. Using
to find unknown areas,
.
-Solution by Garrett
Solution 7
Now we prove P is the midpoint of . Denote the height from
to
as
, height from
to
as
.According to the problem,
implies
. Then according to basic congruent triangles we get
Firstly, denote that
. Applying Stewart theorem, getting that
, denote
Applying Stewart Theorem, getting
solve for a, getting
Now everything is clear, we can find
using LOC,
, the whole area is
~bluesoul
Solution 8 (Simple Geometry)
as in another solutions.
Let be the reflection of
across
.
Let points
and
be the foot of perpendiculars on
from
, and
respectively.
The area of quadrilateral
is equal to the area of triangle
with sides
.
The semiperimeter is
the area
vladimir.shelomovskii@gmail.com, vvsss
Solution 9 (Mindless Law of Cosines Bash)
Use your favorite method to get that is the midpoint of one of the two diagonals (suppose it's the midpoint of
). From here, let
where
is the angle that the diagonals make. Then we have a system of four equations:
From these equations we get that
From here we can see that so
Furthermore, this implies
and
which implies
Then note that the area of the quadrilateral is
~Dhillonr25
Solution 10
Note that (All angles are in degrees)
Since
we can use sine area formula to get the following(after some simplifying steps):
For convenience, let
The above equation simplifies to:
From here, we see that
or
. Without loss of generality, let
. Since triangles
and
are obviously not congruent, we see that one triangle is obtuse and the other one is acute.(Refer to the diagram) However, if we drop perpendiculars from
to
and
to
, we do get congruent triangles. If the foot of the perpendicular from
is
, and the foot of the perpendicular from
is
, then right triangle
is congruent to right triangle
.
From here, we see that the altitudes of triangles
and
to
are equal. Since they share base
, their areas are equal. We can use Heron's formula. To not have any fractions, let
Even though this looks bad at first, it actually isn't too complicated to simplify. Expanding the differences of squares and simplifying completely, we get
Plugging this
back into the Heron's formula, we get that the area of
(or
) is
. Since these triangles have equal area, the area of the quadrilateral is
, and we are done.
Solution 11
Use any method to derive that is the midpoint of
and
. Now, either construct a parallelogram with as shown in the diagram to the right, or use law of sines. We will take the parallelogram route here, but the same thing can be done with law of sines on triangles
and
. Reflect
across
to get
. Since
,
is isosceles. Thus,
, and because
is a parallelogram (since
and
),
. So,
. Now, apply law of cosines on
and
. We get:
\begin{align}
100 + BD^2 - 20BD \cos{\angle ABD} &= 100 + BD^2 - 20 BD \cos {(180 - \angle BDC)} = \\
100 + BD^2 + 20 BD \cos{\angle BDC} &= 260 \\
&\textnormal{and} \\
100 + BD^2 - 20 BD \cos{\angle BDC} &= 196 \\
\textnormal{summing }&\textnormal{and simplifying,} \\
BD &= 8\sqrt{2}
\end{align}
Then, applying law of cosines on
again, we obtain
Since
,
. Thus,
.
Video Solution by MOP 2024
https://youtube.com/watch?v=2BsYR1dJn9c
~r00tsOfUnity
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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