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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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| ==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 BH}
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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{(B) 21}}</imath>.
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| ~Voidling
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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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