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2025 AMC 10A Problems/Problem 23: Difference between revisions

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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>
<imath>\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22</imath>
==Solution 1==
==Solution 1==
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>.
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>.
Line 77: Line 78:
~Jonathanmo
~Jonathanmo


==Solution 1==
==Solution 2==
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>.  
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>.  


Revision as of 14:53, 6 November 2025

Problem 23

Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$. What is $BP$?

$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$

Solution 1

Let $CH$ be the altitude from vertex $C$ to the side $\overline{AB}$, so $H$ is a point on $AB$ and $\angle CHB = 90^\circ$. Let $BD$ be the angle bisector of $\angle B$, where $D$ is on $AC$.

Point $P$ is the intersection of the altitude $CH$ and the angle bisector $BD$.

We want to find the length of $BP$.

Consider the triangle $\triangle BPH$. Since $P$ lies on the altitude $CH$, the angle $\angle BHP$ is the same as $\angle CHB$, which is $90^\circ$. Therefore, $\triangle BPH$ is a right-angled triangle.

In the right $\triangle BPH$, the angle $\angle PBH$ is the angle formed by the angle bisector $BD$ and the side $AB$. By definition of the angle bisector, $\angle PBH = \angle ABC / 2 = B/2$.

Using trigonometry in the right $\triangle BPH$, we have:

\[\cos(\angle PBH) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BH}{BP}\]

\[\cos(B/2) = \frac{BH}{BP}\]

Rearranging this gives:

\[BP = \frac{BH}{\cos(B/2)}\]

To solve the problem, we need to find the lengths of $BH$ and the value of $\cos(B/2)$.

\textbf{1. Find the length of BH} $BH$ is the projection of side $BC$ onto $AB$. In the right-angled triangle $\triangle CHB$, $BH = BC \cdot \cos(B) = a \cdot \cos(B)$. We can find $\cos(B)$ using the Law of Cosines on $\triangle ABC$:

\[b^2 = a^2 + c^2 - 2ac \cos(B)\]

\[AC^2 = BC^2 + AB^2 - 2(BC)(AB) \cos(B)\]

\[75^2 = 45^2 + 80^2 - 2(45)(80) \cos(B)\]

\[5625 = 2025 + 6400 - 7200 \cos(B)\]

\[5625 = 8425 - 7200 \cos(B)\]

\[7200 \cos(B) = 8425 - 5625\]

\[7200 \cos(B) = 2800\]

\[\cos(B) = \frac{2800}{7200} = \frac{28}{72} = \frac{7}{18}\]

Now, we can find $BH$:

\[BH = a \cdot \cos(B) = 45 \cdot \left(\frac{7}{18}\right) = \frac{45 \cdot 7}{18} = \frac{(5 \cdot 9) \cdot 7}{2 \cdot 9} = \frac{35}{2}\]

\textbf{2. Find the value of $\cos(B/2)$} We use the half-angle identity for cosine: $\cos(B) = 2\cos^2(B/2) - 1$. We know $\cos(B) = 7/18$:

\[\frac{7}{18} = 2\cos^2(B/2) - 1\]

\[\frac{7}{18} + 1 = 2\cos^2(B/2)\]

\[\frac{25}{18} = 2\cos^2(B/2)\]

\[\cos^2(B/2) = \frac{25}{36}\]

\[\cos(B/2) = \sqrt{\frac{25}{36}} = \frac{5}{6}\]

(We take the positive root because $B/2$ must be an acute angle).

\textbf{3. Calculate BP} Now we substitute our values for $BH$ and $\cos(B/2)$ into the equation for $BP$:

\[BP = \frac{BH}{\cos(B/2)} = \frac{35/2}{5/6}\]

\[BP = \frac{35}{2} \cdot \frac{6}{5} = \frac{35 \cdot 6}{2 \cdot 5} = \frac{210}{10} = 21\]

The length of $BP$ is 21. This corresponds to answer choice $\textbf{(D)}$. ~Jonathanmo

Solution 2

Let $D$ be the foot of the altitude from $C$ to $AB$. We wish to find $BD$ and $DP$.

First, notice that $AD^2 + CD^2 = AC^2$ and $BD^2+CD^2=BC^2$ by the Pythagorean Theorem. Subtracting the second equation from the first, we get \[AD^2-BD^2=(AD-BD)(AD+BD)=AC^2-BC^2\] Plugging in values, we see that $AD-BD =45$. So, $AD = \frac{125}{2}$, $BD=\frac{35}{2}$, and $CD = \frac{25}{2}\sqrt{11}$. To find $DP$, we use the Angle Bisector Theorem. The ratio between $BD$ and $BC$ is $\frac{7}{18}$, so $DP = \frac{7}{25} \cdot CD = \frac{7}{2}\sqrt{11}$. Finally, we use the Pythagorean Theorem to get \[BD^2+DP^2=(\frac{35}{2})^2+(\frac{7}{2}\sqrt{11})^2=\frac{1764}{4}=441=BP^2\] so $BP=\boxed{21}$.

~ChickensEatGrass

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