Difference between revisions of "2021 AIME I Problems/Problem 9"
Jackshi2006 (talk | contribs) (→Solution 2 (LOC and Trig)) |
MRENTHUSIASM (talk | contribs) (→Solution 4 (Similar Triangles, Cyclic Quadrilaterals, Pythagorean Theorem)) |
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==Diagram== | ==Diagram== | ||
− | + | <asy> | |
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(250); | ||
+ | pair A, B, C, D, E, F, G, H; | ||
+ | A = (-45sqrt(2)/8,18); | ||
+ | B = (45sqrt(2)/8,18); | ||
+ | C = (81sqrt(2)/8,0); | ||
+ | D = (-81sqrt(2)/8,0); | ||
+ | E = foot(A,C,B); | ||
+ | F = foot(A,C,D); | ||
+ | G = foot(A,B,D); | ||
+ | H = intersectionpoint(A--F,B--D); | ||
+ | markscalefactor=0.1; | ||
+ | draw(rightanglemark(A,E,B),red); | ||
+ | draw(rightanglemark(A,F,C),red); | ||
+ | draw(rightanglemark(A,G,D),red); | ||
+ | dot("$A$",A,1.5*NW,linewidth(4)); | ||
+ | dot("$B$",B,1.5*NE,linewidth(4)); | ||
+ | dot("$C$",C,1.5*SE,linewidth(4)); | ||
+ | dot("$D$",D,1.5*SW,linewidth(4)); | ||
+ | dot(E,linewidth(4)); | ||
+ | dot(F,linewidth(4)); | ||
+ | dot(G,linewidth(4)); | ||
+ | draw(A--B--C--D--cycle^^B--D^^B--E); | ||
+ | draw(A--E^^A--F^^A--G,dashed); | ||
+ | label("$10$",midpoint(A--G),1.5*(1,0)); | ||
+ | label("$15$",midpoint(A--E),1.5*N); | ||
+ | Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); | ||
+ | draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
− | + | ==Solution 1 (Similar Triangles and Pythagorean Theorem)== | |
+ | Let <math>\overline{AE}, \overline{AF},</math> and <math>\overline{AG}</math> be the perpendiculars from <math>A</math> to <math>\overleftrightarrow{BC}, \overleftrightarrow{CD},</math> and <math>\overleftrightarrow{BD},</math> respectively. Next, let <math>H</math> be the intersection of <math>\overline{AF}</math> and <math>\overline{BD}.</math> | ||
− | + | We set <math>AB=x</math> and <math>AH=y,</math> as shown below. | |
− | + | <asy> | |
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(250); | ||
+ | pair A, B, C, D, E, F, G, H; | ||
+ | A = (-45sqrt(2)/8,18); | ||
+ | B = (45sqrt(2)/8,18); | ||
+ | C = (81sqrt(2)/8,0); | ||
+ | D = (-81sqrt(2)/8,0); | ||
+ | E = foot(A,C,B); | ||
+ | F = foot(A,C,D); | ||
+ | G = foot(A,B,D); | ||
+ | H = intersectionpoint(A--F,B--D); | ||
+ | markscalefactor=0.1; | ||
+ | draw(rightanglemark(A,E,B),red); | ||
+ | draw(rightanglemark(A,F,C),red); | ||
+ | draw(rightanglemark(A,G,D),red); | ||
+ | dot("$A$",A,1.5*NW,linewidth(4)); | ||
+ | dot("$B$",B,1.5*NE,linewidth(4)); | ||
+ | dot("$C$",C,1.5*SE,linewidth(4)); | ||
+ | dot("$D$",D,1.5*SW,linewidth(4)); | ||
+ | dot("$E$",E,1.5*dir(E),linewidth(4)); | ||
+ | dot("$F$",F,1.5*S,linewidth(4)); | ||
+ | dot("$G$",G,SE,linewidth(4)); | ||
+ | dot("$H$",H,SE,linewidth(4)); | ||
+ | draw(A--B--C--D--cycle^^B--D^^B--E); | ||
+ | draw(A--E^^A--F^^A--G,dashed); | ||
+ | label("$10$",midpoint(A--G),1.5*(1,0)); | ||
+ | label("$15$",midpoint(A--E),1.5*N); | ||
+ | Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); | ||
+ | draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); | ||
+ | label("$x$",midpoint(A--B),N); | ||
+ | label("$y$",midpoint(A--H),W); | ||
+ | </asy> | ||
+ | From here, we obtain <math>HF=18-y</math> by segment subtraction, and <math>BG=\sqrt{x^2-10^2}</math> and <math>HG=\sqrt{y^2-10^2}</math> by the Pythagorean Theorem. | ||
− | -- | + | Since <math>\angle ABG</math> and <math>\angle HAG</math> are both complementary to <math>\angle AHB,</math> we have <math>\angle ABG = \angle HAG,</math> from which <math>\triangle ABG \sim \triangle HAG</math> by AA. It follows that <math>\frac{BG}{AG}=\frac{AG}{HG},</math> so <math>BG\cdot HG=AG^2,</math> or <cmath>\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)</cmath> |
+ | Since <math>\angle AHB = \angle FHD</math> by vertical angles, we have <math>\triangle AHB \sim \triangle FHD</math> by AA, with the ratio of similitude <math>\frac{AH}{FH}=\frac{BA}{DF}.</math> It follows that <math>DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.</math> | ||
− | + | Since <math>\angle EBA = \angle ECD = \angle FDA</math> by angle chasing, we have <math>\triangle EBA \sim \triangle FDA</math> by AA, with the ratio of similitude <math>\frac{EA}{FA}=\frac{BA}{DA}.</math> It follows that <math>DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.</math> | |
+ | |||
+ | By the Pythagorean Theorem on right <math>\triangle ADF,</math> we have <math>DF^2+AF^2=AD^2,</math> or <cmath>\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)</cmath> | ||
+ | Solving this system of equations (<math>(1)</math> and <math>(2)</math>), we get <math>x=\frac{45\sqrt2}{4}</math> and <math>y=\frac{90}{7},</math> so <math>AB=x=\frac{45\sqrt2}{4}</math> and <math>CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.</math> Finally, the area of <math>ABCD</math> is <cmath>K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},</cmath> from which <math>\sqrt2 \cdot K=\boxed{567}.</math> | ||
− | + | ~MRENTHUSIASM | |
− | + | ==Solution 2 (Similar Triangles and Pythagorean Theorem)== | |
− | + | First, draw the diagram. Then, notice that since <math>ABCD</math> is isosceles, <math>\Delta ABD \cong \Delta BAC</math>, and the length of the altitude from <math>B</math> to <math>AC</math> is also <math>10</math>. Let the foot of this altitude be <math>F</math>, and let the foot of the altitude from <math>A</math> to <math>BC</math> be denoted as <math>E</math>. Then, <math>\Delta BCF \sim \Delta ACE</math>. So, <math>\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}</math>. Now, notice that <math>[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}</math>, where <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Letting <math>AB = x</math>, this equality becomes <math>AC = \frac{9x}{5}</math>. Also, from <math>\frac{BC}{AC} = \frac{2}{3}</math>, we have <math>BC = \frac{6x}{5}</math>. Now, by the Pythagorean theorem on triangles <math>ABF</math> and <math>CBF</math>, we have <math>AF = \sqrt{x^{2}-100}</math> and <math>CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Notice that <math>AC = AF + CF</math>, so <math>\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}</math>. Squaring both sides of the equation once, moving <math>x^{2}-100</math> and <math> \left( \frac{6x}{5} \right) ^{2}-100</math> to the right, dividing both sides by <math>2</math>, and squaring the equation once more, we are left with <math>\frac{32x^{4}}{25} = 324x^{2}</math>. Dividing both sides by <math>x^{2}</math> (since we know <math>x</math> is positive), we are left with <math>\frac{32x^{2}}{25} = 324</math>. Solving for <math>x</math> gives us <math>x = \frac{45}{2\sqrt{2}}</math>. | |
− | + | Now, let the foot of the perpendicular from <math>A</math> to <math>CD</math> be <math>G</math>. Then let <math>DG = y</math>. Let the foot of the perpendicular from <math>B</math> to <math>CD</math> be <math>H</math>. Then, <math>CH</math> is also equal to <math>y</math>. Notice that <math>ABHG</math> is a rectangle, so <math>GH = x</math>. Now, we have <math>CG = GH + CH = x + y</math>. By the Pythagorean theorem applied to <math>\Delta AGC</math>, we have <math>(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}</math>. We know that <math>\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}</math>, so we can plug this into this equation. Solving for <math>x+y</math>, we get <math>x+y=\frac{63}{2\sqrt{2}}</math>. | |
− | + | Finally, to find <math>[ABCD]</math>, we use the formula for the area of a trapezoid: <math>K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}</math>. The problem asks us for <math>K \cdot \sqrt{2}</math>, which comes out to be <math>\boxed{567}</math>. | |
− | ~ | + | ~advanture |
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− | ==Solution 3 ( | + | ==Solution 3 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)== |
− | Let the foot of the altitude from A to BC be P, to CD be Q, and to BD be R. | + | Let the foot of the altitude from <math>A</math> to <math>BC</math> be <math>P</math>, to <math>CD</math> be <math>Q</math>, and to <math>BD</math> be <math>R</math>. |
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, <math>A</math> is on the circumcircle of <math>\triangle BCD</math> and we have that <math>PRQ</math> is the Simson Line from <math>A</math>. As <math>\angle QAB = 90^\circ</math>, we have that <math>\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ</math>, with the last equality coming from cyclic quadrilateral <math>APBR</math>. Thus, <math>\triangle QAR \sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can see gives us that <math>QP = 27</math>. Further ratios using the same similar triangles gives that <math>QR = \frac{25}{3}</math> and <math>RP = \frac{56}{3}</math>. | Note that all isosceles trapezoids are cyclic quadrilaterals; thus, <math>A</math> is on the circumcircle of <math>\triangle BCD</math> and we have that <math>PRQ</math> is the Simson Line from <math>A</math>. As <math>\angle QAB = 90^\circ</math>, we have that <math>\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ</math>, with the last equality coming from cyclic quadrilateral <math>APBR</math>. Thus, <math>\triangle QAR \sim \triangle QPA</math> and we have that <math>\frac{AQ}{AR} = \frac{PQ}{PA}</math> or that <math>\frac{18}{10} = \frac{QP}{15}</math>, which we can see gives us that <math>QP = 27</math>. Further ratios using the same similar triangles gives that <math>QR = \frac{25}{3}</math> and <math>RP = \frac{56}{3}</math>. | ||
− | We also see that quadrilaterals <math>APBR</math> and <math>ARDQ</math> are both cyclic, with diameters of the circumcircles being <math>AB</math> and <math>AQ</math> respectively. The intersection of the circumcircles are the points <math>A</math> and <math>R</math>, and we know <math>DRB</math> and <math>QRP</math> are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center A taking <math>\triangle APQ</math> to <math>\triangle APD</math>. Because we know a lot about <math>\triangle APQ</math> but very little about <math>\triangle APD</math> and we would like to know more, we wish to find the ratio of similitude between the two triangles. | + | We also see that quadrilaterals <math>APBR</math> and <math>ARDQ</math> are both cyclic, with diameters of the circumcircles being <math>AB</math> and <math>AQ</math> respectively. The intersection of the circumcircles are the points <math>A</math> and <math>R</math>, and we know <math>DRB</math> and <math>QRP</math> are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center <math>A</math> taking <math>\triangle APQ</math> to <math>\triangle APD</math>. Because we know a lot about <math>\triangle APQ</math> but very little about <math>\triangle APD</math> and we would like to know more, we wish to find the ratio of similitude between the two triangles. |
− | To do this, we use the one number we have for <math>\triangle APD</math>: we know that the altitude from <math>A</math> to <math>BD</math> has length 10. As the two triangles are similar, if we can find the height from <math>A</math> to <math>PQ</math>, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that <math>QP = 27</math>. Using this, we can drop the altitude from <math>A</math> to <math>QP</math> and let it intersect <math>QP</math> at <math>H</math>. Then, let <math>QH = x</math> and thus <math>HP=27-x</math>. We then have by the Pythagorean Theorem on <math>\triangle AQH</math> and <math>\triangle APH</math>: <cmath>15^2 - x^2 = 18^2 - (27-x)^2 | + | To do this, we use the one number we have for <math>\triangle APD</math>: we know that the altitude from <math>A</math> to <math>BD</math> has length <math>10</math>. As the two triangles are similar, if we can find the height from <math>A</math> to <math>PQ</math>, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that <math>QP = 27</math>. Using this, we can drop the altitude from <math>A</math> to <math>QP</math> and let it intersect <math>QP</math> at <math>H</math>. Then, let <math>QH = x</math> and thus <math>HP=27-x</math>. We then have by the Pythagorean Theorem on <math>\triangle AQH</math> and <math>\triangle APH</math>: |
+ | <cmath>\begin{align*} | ||
+ | 15^2 - x^2 &= 18^2 - (27-x)^2 \\ | ||
+ | 225 - x^2 &= 324 - (x^2-54x+729) \\ | ||
+ | 54x &= 630 \\ | ||
+ | x &= \frac{35}{3}. | ||
+ | \end{align*}</cmath> | ||
+ | Then, <math>RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}</math>. This gives us then from right triangle <math>\triangle ARH</math> that <math>AH = \frac{20\sqrt{2}}{3}</math> and thus the ratio of <math>\triangle APQ</math> to <math>\triangle ABD</math> is <math>\frac{3\sqrt{2}}{4}</math>. From this, we see then that <cmath>AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}</cmath> and <cmath>AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.</cmath> The Pythagorean Theorem on <math>\triangle AQD</math> then gives that <cmath>QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.</cmath> | ||
+ | Then, we have the height of trapezoid <math>ABCD</math> is <math>AQ = 18</math>, the top base is <math>AB = \frac{45\sqrt{2}}{4}</math>, and the bottom base is <math>CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}</math>. From the equation of a trapezoid, <math>K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}</math>, so the answer is <math>K\sqrt{2} = \boxed{567}</math>. | ||
− | + | ~lvmath | |
− | + | ==Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)== | |
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− | ==Solution 4 ( | ||
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Let <math>E,F,</math> and <math>G</math> be the feet of the altitudes from <math>A</math> to <math>BC,CD,</math> and <math>DB</math>, respectively. | Let <math>E,F,</math> and <math>G</math> be the feet of the altitudes from <math>A</math> to <math>BC,CD,</math> and <math>DB</math>, respectively. | ||
Line 60: | Line 119: | ||
Claim: We have <math>2</math> pairs of similar right triangles: <math>\triangle AEB \sim \triangle AFD</math> and <math>\triangle AGD \sim \triangle AEC</math>. | Claim: We have <math>2</math> pairs of similar right triangles: <math>\triangle AEB \sim \triangle AFD</math> and <math>\triangle AGD \sim \triangle AEC</math>. | ||
− | Proof: Note that <math>ABCD</math> is cyclic. We need one more angle, and we get this from this cyclic | + | Proof: Note that <math>ABCD</math> is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: |
− | + | <cmath>\begin{align*} | |
− | + | \angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\ | |
+ | \angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square | ||
+ | \end{align*}</cmath> | ||
Let <math>AD=a</math>. We obtain from the similarities <math>AB = \frac{5a}{6}</math> and <math>AC=BD=\frac{3a}{2}</math>. | Let <math>AD=a</math>. We obtain from the similarities <math>AB = \frac{5a}{6}</math> and <math>AC=BD=\frac{3a}{2}</math>. | ||
− | By Ptolemy, <math>(\frac{3a}{2})^2 = a^2 + \frac{5a}{6} \cdot CD</math>, so <math>\frac{5a^2}{4} = \frac{5a}{6} \cdot CD</math>. | + | By Ptolemy, <math>\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD</math>, so <math>\frac{5a^2}{4} = \frac{5a}{6} \cdot CD</math>. |
We obtain <math>CD=\frac{3a}{2}</math>, so <math>DF=\frac{CD-AB}{2}=\frac{a}{3}</math>. | We obtain <math>CD=\frac{3a}{2}</math>, so <math>DF=\frac{CD-AB}{2}=\frac{a}{3}</math>. | ||
Line 71: | Line 132: | ||
Applying the Pythagorean theorem on <math>\triangle ADF</math>, we get <math>324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}</math>. | Applying the Pythagorean theorem on <math>\triangle ADF</math>, we get <math>324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}</math>. | ||
− | Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}</math>, yielding <math>\boxed{567}</math>. | + | Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}</math>, yielding <math>\sqrt2\cdot[ABCD]=\boxed{567}</math>. |
− | ==Solution | + | ==Solution 5 (Similar Triangles and Trigonometry)== |
− | Let <math> | + | Let <math>AD=BC=a</math>. Draw diagonal <math>AC</math> and let <math>G</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>, <math>F</math> be the foot of the perpendicular from <math>A</math> to line <math>BC</math>, and <math>H</math> be the foot of the perpendicular from <math>A</math> to <math>DC</math>. |
− | We | + | Note that <math>\triangle CBG\sim\triangle CAF</math>, and we get that <math>\frac{10}{15}=\frac{a}{AC}</math>. Therefore, <math>AC=\frac32 a</math>. It then follows that <math>\triangle ABF\sim\triangle ADH</math>. Using similar triangles, we can then find that <math>AB=\frac{5}{6}a</math>. Using the Law of Cosines on <math>\triangle ABC</math>, We can find that the <math>\cos\angle ABC=-\frac{1}{3}</math>. Since <math>\angle ABF=\angle ADH</math>, and each is supplementary to <math>\angle ABC</math>, we know that the <math>\cos\angle ADH=\frac{1}{3}</math>. It then follows that <math>a=\frac{27\sqrt{2}}{2}</math>. Then it can be found that the area <math>K</math> is <math>\frac{567\sqrt{2}}{2}</math>. Multiplying this by <math>\sqrt{2}</math>, the answer is <math>\boxed{567}</math>. |
− | + | ~happykeeper | |
− | + | ==Solution 6 (Similar Triangles and Trigonometry)== | |
+ | Draw the distances in terms of <math>B</math>, as shown in the diagram. By similar triangles, <math>\triangle{AEC}\sim\triangle{BIC}</math>. As a result, let <math>AB=u</math>, then <math>BC=AD=\frac{6}{5}u</math> and <math>2AC=3BC</math>. The triangle <math>ABC</math> is <math>6-5-9</math> which <math>\cos(\angle{ABC})=-\frac{1}{3}</math>. By angle subtraction, <math>\cos(180-\theta)=-\cos\theta</math>. Therefore, <math>AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}</math> and <math>AD=BC=\frac{27}{\sqrt{2}}</math>. By trapezoid area formula, the area of <math>ABCD</math> is equal to <math>(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}</math> which <math>\sqrt{2}\cdot k=\boxed{567}</math>. | ||
− | + | ~math2718281828459 | |
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==Video Solution== | ==Video Solution== |
Latest revision as of 15:06, 13 December 2021
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Similar Triangles and Pythagorean Theorem)
- 4 Solution 2 (Similar Triangles and Pythagorean Theorem)
- 5 Solution 3 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 6 Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 7 Solution 5 (Similar Triangles and Trigonometry)
- 8 Solution 6 (Similar Triangles and Trigonometry)
- 9 Video Solution
- 10 See Also
Problem
Let be an isosceles trapezoid with and Suppose that the distances from to the lines and are and respectively. Let be the area of Find
Diagram
~MRENTHUSIASM
Solution 1 (Similar Triangles and Pythagorean Theorem)
Let and be the perpendiculars from to and respectively. Next, let be the intersection of and
We set and as shown below. From here, we obtain by segment subtraction, and and by the Pythagorean Theorem.
Since and are both complementary to we have from which by AA. It follows that so or Since by vertical angles, we have by AA, with the ratio of similitude It follows that
Since by angle chasing, we have by AA, with the ratio of similitude It follows that
By the Pythagorean Theorem on right we have or Solving this system of equations ( and ), we get and so and Finally, the area of is from which
~MRENTHUSIASM
Solution 2 (Similar Triangles and Pythagorean Theorem)
First, draw the diagram. Then, notice that since is isosceles, , and the length of the altitude from to is also . Let the foot of this altitude be , and let the foot of the altitude from to be denoted as . Then, . So, . Now, notice that , where denotes the area of triangle . Letting , this equality becomes . Also, from , we have . Now, by the Pythagorean theorem on triangles and , we have and . Notice that , so . Squaring both sides of the equation once, moving and to the right, dividing both sides by , and squaring the equation once more, we are left with . Dividing both sides by (since we know is positive), we are left with . Solving for gives us .
Now, let the foot of the perpendicular from to be . Then let . Let the foot of the perpendicular from to be . Then, is also equal to . Notice that is a rectangle, so . Now, we have . By the Pythagorean theorem applied to , we have . We know that , so we can plug this into this equation. Solving for , we get .
Finally, to find , we use the formula for the area of a trapezoid: . The problem asks us for , which comes out to be .
~advanture
Solution 3 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
Let the foot of the altitude from to be , to be , and to be .
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, is on the circumcircle of and we have that is the Simson Line from . As , we have that , with the last equality coming from cyclic quadrilateral . Thus, and we have that or that , which we can see gives us that . Further ratios using the same similar triangles gives that and .
We also see that quadrilaterals and are both cyclic, with diameters of the circumcircles being and respectively. The intersection of the circumcircles are the points and , and we know and are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center taking to . Because we know a lot about but very little about and we would like to know more, we wish to find the ratio of similitude between the two triangles.
To do this, we use the one number we have for : we know that the altitude from to has length . As the two triangles are similar, if we can find the height from to , we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that . Using this, we can drop the altitude from to and let it intersect at . Then, let and thus . We then have by the Pythagorean Theorem on and : Then, . This gives us then from right triangle that and thus the ratio of to is . From this, we see then that and The Pythagorean Theorem on then gives that Then, we have the height of trapezoid is , the top base is , and the bottom base is . From the equation of a trapezoid, , so the answer is .
~lvmath
Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
Let and be the feet of the altitudes from to and , respectively.
Claim: We have pairs of similar right triangles: and .
Proof: Note that is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: Let . We obtain from the similarities and .
By Ptolemy, , so .
We obtain , so .
Applying the Pythagorean theorem on , we get .
Thus, , and , yielding .
Solution 5 (Similar Triangles and Trigonometry)
Let . Draw diagonal and let be the foot of the perpendicular from to , be the foot of the perpendicular from to line , and be the foot of the perpendicular from to .
Note that , and we get that . Therefore, . It then follows that . Using similar triangles, we can then find that . Using the Law of Cosines on , We can find that the . Since , and each is supplementary to , we know that the . It then follows that . Then it can be found that the area is . Multiplying this by , the answer is .
~happykeeper
Solution 6 (Similar Triangles and Trigonometry)
Draw the distances in terms of , as shown in the diagram. By similar triangles, . As a result, let , then and . The triangle is which . By angle subtraction, . Therefore, and . By trapezoid area formula, the area of is equal to which .
~math2718281828459
Video Solution
https://www.youtube.com/watch?v=6rLnl8z7lnM
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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