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| ==Problem 23==
| | #redirect [[2025 AMC 12A Problems/Problem 16]] |
| Triangle <imath>\triangle ABC</imath> has side lengths <imath>AB = 80</imath>, <imath>BC = 45</imath>, and <imath>AC = 75</imath>. The bisector of <imath>\angle B</imath> and the altitude to side <imath>\overline{AB}</imath> intersect at point <imath>P</imath>. What is <imath>BP</imath>?
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| <imath>\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22</imath>
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| ===Diagram===
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| <asy>
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| import olympiad;
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| size(350);
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| pair A = (0,0);
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| pair B = (80,0);
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| pair C = (62.5,41.4578098794425);
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| pair H = (62.5,0);
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| // Angle bisector point D on AC: AD : DC = 48 : 27
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| pair D = (40.0,26.5329983228432);
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| path BD = B--D;
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| pair P = intersectionpoint(BD,H--C);
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| // main triangle
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| draw(A--B--C--cycle, linewidth(1));
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| draw(C--H);
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| draw(B--D);
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| // right-angle mark at H
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| draw((H+(1.8,0))--(H+(1.8,1.8))--(H+(0,1.8)), linewidth(0.8));
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| dot(A); label("$A$",A,SW);
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| dot(B); label("$B$",B,SE);
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| dot(C); label("$C$",C,NE);
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| dot(H); label("$H$",H,S);
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| dot(D); label("$D$",D,NNW);
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| dot(P); label("$P$",P,ENE);
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| // side labels (closer, larger)
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| label("$80$", midpoint(A--B), S+1.5*dir(270), fontsize(11));
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| label("$75$", midpoint(A--C) + (-2,2), fontsize(11));
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| label("$45$", midpoint(B--C) + (2,2), fontsize(11));
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| xlimits(-5,85);
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| ylimits(-5,50);
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| currentpicture.fit();
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| </asy>
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| ~Avs2010
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| ==Solution 1(Takes some time)==
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| Let <imath>CH</imath> be the altitude from vertex <imath>C</imath> to the side <imath>\overline{AB}</imath>, so <imath>H</imath> is a point on <imath>AB</imath> and <imath>\angle CHB = 90^\circ</imath>.
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| Let <imath>BD</imath> be the angle bisector of <imath>\angle B</imath>, where <imath>D</imath> is on <imath>AC</imath>.
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| Point <imath>P</imath> is the intersection of the altitude <imath>CH</imath> and the angle bisector <imath>BD</imath>.
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| We want to find the length of <imath>BP</imath>.
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| Consider the triangle <imath>\triangle BPH</imath>. Since <imath>P</imath> lies on the altitude <imath>CH</imath>, the angle <imath>\angle BHP</imath> is the same as <imath>\angle CHB</imath>, which is <imath>90^\circ</imath>. Therefore, <imath>\triangle BPH</imath> is a right-angled triangle.
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| In the right <imath>\triangle BPH</imath>, the angle <imath>\angle PBH</imath> is the angle formed by the angle bisector <imath>BD</imath> and the side <imath>AB</imath>. By definition of the angle bisector, <imath>\angle PBH = \angle ABC / 2 = B/2</imath>.
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| Using trigonometry in the right <imath>\triangle BPH</imath>, we have:
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| <cmath>\cos(\angle PBH) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BH}{BP}</cmath>
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| <cmath>\cos(B/2) = \frac{BH}{BP}</cmath>
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| Rearranging this gives:
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| <cmath>BP = \frac{BH}{\cos(B/2)}</cmath>
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| To solve the problem, we need to find the lengths of <imath>BH</imath> and the value of <imath>\cos(B/2)</imath>.
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| 1. Find the length of
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| <imath>BH</imath> is the projection of side <imath>BC</imath> onto <imath>AB</imath>. In the right-angled triangle <imath>\triangle CHB</imath>, <imath>BH = BC \cdot \cos(B) = a \cdot \cos(B)</imath>.
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| We can find <imath>\cos(B)</imath> using the Law of Cosines on <imath>\triangle ABC</imath>:
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| <cmath>b^2 = a^2 + c^2 - 2ac \cos(B)</cmath>
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| <cmath>AC^2 = BC^2 + AB^2 - 2(BC)(AB) \cos(B)</cmath>
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| <cmath>75^2 = 45^2 + 80^2 - 2(45)(80) \cos(B)</cmath>
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| <cmath>5625 = 2025 + 6400 - 7200 \cos(B)</cmath>
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| <cmath>5625 = 8425 - 7200 \cos(B)</cmath>
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| <cmath>7200 \cos(B) = 8425 - 5625</cmath>
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| <cmath>7200 \cos(B) = 2800</cmath>
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| <cmath>\cos(B) = \frac{2800}{7200} = \frac{28}{72} = \frac{7}{18}</cmath>
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| Now, we can find <imath>BH</imath>:
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| <cmath>BH = a \cdot \cos(B) = 45 \cdot \left(\frac{7}{18}\right) = \frac{45 \cdot 7}{18} = \frac{(5 \cdot 9) \cdot 7}{2 \cdot 9} = \frac{35}{2}</cmath>
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| 2. Find the value of <imath>\cos(B/2)</imath>}
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| We use the half-angle identity for cosine: <imath>\cos(B) = 2\cos^2(B/2) - 1</imath>.
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| We know <imath>\cos(B) = 7/18</imath>:
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| <cmath>\frac{7}{18} = 2\cos^2(B/2) - 1</cmath>
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| <cmath>\frac{7}{18} + 1 = 2\cos^2(B/2)</cmath>
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| <cmath>\frac{25}{18} = 2\cos^2(B/2)</cmath>
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| <cmath>\cos^2(B/2) = \frac{25}{36}</cmath>
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| <cmath>\cos(B/2) = \sqrt{\frac{25}{36}} = \frac{5}{6}</cmath>
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| (We take the positive root because <imath>B/2</imath> must be an acute angle).
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| Now we substitute our values for <imath>BH</imath> and <imath>\cos(B/2)</imath> into the equation for <imath>BP</imath>:
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| <cmath>BP = \frac{BH}{\cos(B/2)} = \frac{35/2}{5/6}</cmath>
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| <cmath>BP = \frac{35}{2} \cdot \frac{6}{5} = \frac{35 \cdot 6}{2 \cdot 5} = \frac{210}{10} = 21</cmath>
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| The length of <imath>BP</imath> is 21.
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| ~Jonathanmo
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| ==Solution 2==
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| Let <imath>D</imath> be the foot of the altitude from <imath>C</imath> to <imath>AB</imath>. We wish to find <imath>BD</imath> and <imath>DP</imath>.
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| First, notice that <imath>AD^2 + CD^2 = AC^2</imath> and <imath>BD^2+CD^2=BC^2</imath> by the Pythagorean Theorem. Subtracting the second equation from the first, we get <cmath>AD^2-BD^2=(AD-BD)(AD+BD)=AC^2-BC^2</cmath> Plugging in values and simplifying, we see that <imath>AD-BD =45</imath>. Knowing that <imath>AD+BD=80</imath>, we can solve the system of equations to get <imath>AD = \frac{125}{2}</imath>, <imath>BD=\frac{35}{2}</imath>. Plug those values back into their original equations and we find that <imath>CD = \frac{25}{2}\sqrt{11}</imath>.
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| To find <imath>DP</imath>, we use the Angle Bisector Theorem. The ratio between <imath>BD</imath> and <imath>BC</imath> is <imath>\frac{7}{18}</imath>, so <imath>DP = \frac{7}{25} \cdot CD = \frac{7}{2}\sqrt{11}</imath>. Finally, we use the Pythagorean Theorem to get <cmath>BD^2+DP^2=(\frac{35}{2})^2+(\frac{7}{2}\sqrt{11})^2=\frac{1764}{4}=441=BP^2</cmath>
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| so <imath>BP=\boxed{\textbf{(D)}~21}</imath>.
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| ~ChickensEatGrass
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| == Solution 3 (Heron's Formula) ==
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| We are asked to find \(BP\) in the given triangle configuration.
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| Let \(D\) be the intersection of the altitude from \(C\) to \(AB\). To simplify calculations, we divide all side lengths by \(5\), and multiply by \(5\) again at the end.
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| First, we use Heron’s Formula, \(\sqrt{s(s-a)(s-b)(s-c)}\), to find the area. Let \([ABC]\) denote the area of \(\triangle ABC\). By Heron’s Formula,
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| <cmath>[ABC] = \sqrt{20 \cdot 5 \cdot 4 \cdot 11} = 20\sqrt{11}.</cmath>
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| Next, we find the altitude \(CD\) using the formula for the area of a triangle, \(\tfrac{1}{2}bh = \text{area}\):
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| <cmath>\frac{1}{2}(16)(CD) = 20\sqrt{11} \quad \Rightarrow \quad CD = \frac{5\sqrt{11}}{2}.</cmath>
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| We can use the Pythagorean Theorem in \(\triangle CDB\) to find \(DB\):
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| <cmath>DB^2 + \frac{25 \cdot 11}{4} = 81.</cmath>
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| <cmath>4DB^2 + 275 = 324 \quad \Rightarrow \quad DB^2 = \frac{49}{4}.</cmath>
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| Thus, \(DB = \frac{7}{2}.\)
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| Next, we use the Angle Bisector Theorem to find \(PD\). Let \(x = PC\) and \(y = PD\). Since \(x + y = \frac{5\sqrt{11}}{2}\), we have \(x = \frac{5\sqrt{11}}{2} - y.\)
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| From the given ratio,
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| <cmath>\frac{9}{x} = \frac{7}{2y} \quad \Rightarrow \quad 18y = 7x.</cmath>
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| Substituting \(x = \frac{5\sqrt{11}}{2} - y\),
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| <cmath>18y = 7\left(\frac{5\sqrt{11}}{2} - y\right) \quad \Rightarrow \quad 25y = \frac{35\sqrt{11}}{2}.</cmath>
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| Hence, \(y = \frac{7\sqrt{11}}{10}.\)
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| Now, using the Pythagorean Theorem again to find \(BP\):
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| <cmath>\frac{49 \cdot 11}{100} + \frac{49}{4} = BP^2.</cmath>
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| <cmath>100BP^2 = 49(11 + 25) = 49 \cdot 36 \quad \Rightarrow \quad BP = \frac{42}{10} = \frac{21}{5}.</cmath>
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| Finally, multiplying the side lengths by \(5\) again gives \(BP = 21\), or <imath>\boxed{\textbf{(D) 21}}</imath>.
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| ~Voidling
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| == Solution 4==
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| The semiperimeter is <imath>\frac{75+45+80}{2}=100.</imath> Heron's formula for area gives <imath>\sqrt{100\cdot20\cdot25\cdot55}=\left[ABC\right]=500\sqrt{11}\Rightarrow CH=\frac{25\sqrt{11}}{2}.</imath> Pythagorean theorem gives <cmath>BH^2=BC^2-CH^2=5^2\left(\frac{4\cdot 9^2-25\cdot 11}{4}\right)\Rightarrow BH=35/2.</cmath>
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| Angle bisector theorem on <imath>\overline{CH}</imath> gives <imath>\frac{PH}{35/2}=\frac{PC}{45},</imath> giving <imath>PH=\frac7{18}\cdot PC\Rightarrow PH=\frac7{25}\cdot CH.</imath>
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| Angle bisector theorem on <imath>\overline{AC}</imath> gives <imath>AD=48</imath> and <imath>DC=27.</imath> Then, notice that <imath>\frac{27}{45}=\frac{45}{75},</imath> as well as the fact that <imath>\angle DCB=\angle ACB</imath> is shared, so <imath>\triangle ABC\sim\triangle BDC</imath> by SAS. Therefore, <imath>\angle DAB=\angle DBA.</imath>
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| Now we look for another similar pair of triangles, <imath>\triangle CHA\sim\triangle PHB</imath> by <imath>AA.</imath> Hence, <cmath>\frac{75}{CH}=\frac{BP}{PH}=\frac{BP}{\frac{7}{25}\cdot CH}\Rightarrow \frac{25}{7}\cdot BP=75\Rightarrow BP=\boxed{21}.</cmath>
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| -NinJaPro
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| == Solution 5==
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| By angle bisector theorem, we find <imath>AD=48</imath> and <imath>DC=27.</imath>
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| Let <imath>BH=x,</imath> <imath>CH=h,</imath> and <imath>HA=16-x.</imath> We have
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| <cmath>x^2 + h^2 = 45^2 = 2025</cmath>
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| and
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| <cmath>(16-x)^2 +h^2 = 256 - 32x +x^2 + h^2 = 75^2 = 5625</cmath>
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| by Pythagorean Theorem. After some algebra, we find <imath>x = \frac{35}{2},</imath> meaning
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| <cmath>BH = \frac{35}{2}, HA = \frac{125}{2}.</cmath>
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| By Menelaus' Theorem,
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| <cmath>\frac{BH}{HA} \cdot \frac{AC}{CD} \cdot \frac{DP}{PB} = 1.</cmath>
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| We have <imath>\frac{\frac{35}{2}}{\frac{125}{2}} \cdot \frac{75}{27} \cdot \frac{DP}{PB} = 1,</imath> meaning <imath>\frac{DP}{PB} = \frac{9}{7},</imath> or
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| <cmath>BP = BD \cdot \frac{7}{16}.</cmath>
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| Stewart's Theorem on triangle <imath>ABC</imath> with diagonal <imath>BD</imath> gives the equation
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| <cmath>27 \cdot 75 \cdot 48 + d^2 \cdot 75 = 45^2 \cdot 48 + 80^2 \cdot 27.</cmath>
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| This gives <imath>d = 48</imath> after some algebra.
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| Now, <imath>BP = 48 \cdot \frac{7}{16} = \boxed{21}.</imath>
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| -Jayjay2010
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| == Solution 6 (Construction Cheese)==
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| Use the compass and the ruler to construct the diagram accurately. I used <imath>5:1</imath> cm. Use the ruler to measure the distance between <imath>B</imath> and <imath>P</imath> for <imath>BP=\boxed{\textbf{(D)}~21}</imath>.
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| ~dareckolo and friends
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| ~Ben Dong (typo fix, originally said distance between <imath>B</imath> and <imath>C</imath>)
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| == Video Solution (In 3 Mins) ==
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| https://youtu.be/nAimLnvSTwQ?si=cRm14r4GyfgxRq5L ~ Pi Academy
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| ==See Also==
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| {{AMC10 box|year=2025|ab=A|num-b=22|num-a=24}}
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| * [[AMC 10]]
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| * [[AMC 10 Problems and Solutions]]
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| * [[Mathematics competitions]]
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| * [[Mathematics competition resources]]
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| {{MAA Notice}}
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