document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210489>Question 1210489</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In triangle ABC, \angle A = 90^\circ.  Altitude $\overline{AP},$ angle bisector $\overline{AQ},$ and median $\overline{AR}$ are drawn. If $PQ = 3$ and $QC = 4,$ find $AR.$ </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852901 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852901\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a challenging geometry problem that requires using properties of right triangles, angle bisectors, and medians.\r<BR>\n" );
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document.write( "## 1. Determine Side Lengths\r<BR>\n" );
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document.write( "In $\triangle ABC$, $\angle A = 90^\circ$.\r<BR>\n" );
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document.write( "### A. Locate Point C's Position\r<BR>\n" );
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document.write( "The angle bisector $\overline{AQ}$ divides $\overline{BC}$ into segments $\overline{BQ}$ and $\overline{QC}$.<BR>\n" );
document.write( "We are given $QC = 4$. Let $BQ = y$.<BR>\n" );
document.write( "The entire hypotenuse is $BC = BQ + QC = y + 4$.\r<BR>\n" );
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document.write( "### B. Use the Segment $PQ$\r<BR>\n" );
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document.write( "The altitude $\overline{AP}$ and the angle bisector $\overline{AQ}$ are distinct segments on the hypotenuse $\overline{BC}$.<BR>\n" );
document.write( "We are given $PQ = 3$.\r<BR>\n" );
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document.write( "Since $P$ is the foot of the altitude from $A$, $\triangle APB$ and $\triangle APC$ are right triangles.<BR>\n" );
document.write( "Since $Q$ is between $P$ and $C$ (altitude is often closer to the right angle vertex):<BR>\n" );
document.write( "$$PC = PQ + QC = 3 + 4 = 7$$\r<BR>\n" );
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document.write( "Then, the segment $BP$ is $BC - PC$:<BR>\n" );
document.write( "$$BP = (y + 4) - 7 = y - 3$$\r<BR>\n" );
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document.write( "### C. Apply the Geometric Mean Theorem (Leg Rule)\r<BR>\n" );
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document.write( "In right $\triangle ABC$:<BR>\n" );
document.write( "1. $AC^2 = PC \cdot BC$ (Leg $\overline{AC}$ is the geometric mean of $\overline{PC}$ and $\overline{BC}$)<BR>\n" );
document.write( "2. $AB^2 = BP \cdot BC$ (Leg $\overline{AB}$ is the geometric mean of $\overline{BP}$ and $\overline{BC}$)\r<BR>\n" );
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document.write( "Substitute the lengths:<BR>\n" );
document.write( "1. $AC^2 = 7(y+4)$<BR>\n" );
document.write( "2. $AB^2 = (y-3)(y+4)$\r<BR>\n" );
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document.write( "### D. Apply the Angle Bisector Theorem\r<BR>\n" );
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document.write( "The angle bisector $\overline{AQ}$ divides the opposite side $\overline{BC}$ into segments proportional to the other two sides:<BR>\n" );
document.write( "$$\frac{AB}{AC} = \frac{BQ}{QC}$$<BR>\n" );
document.write( "$$\frac{AB}{AC} = \frac{y}{4}$$<BR>\n" );
document.write( "$$4 \cdot AB = y \cdot AC \quad \text{(Equation 1)}$$\r<BR>\n" );
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document.write( "### E. Solve for $y$ (the length of $BQ$)\r<BR>\n" );
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document.write( "Square Equation 1:<BR>\n" );
document.write( "$$16 \cdot AB^2 = y^2 \cdot AC^2$$\r<BR>\n" );
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document.write( "Substitute the Leg Rule expressions for $AB^2$ and $AC^2$:<BR>\n" );
document.write( "$$16 \cdot [(y-3)(y+4)] = y^2 \cdot [7(y+4)]$$\r<BR>\n" );
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document.write( "Since $y+4 = BC$ cannot be zero, we can divide both sides by $(y+4)$:<BR>\n" );
document.write( "$$16(y-3) = 7y^2$$<BR>\n" );
document.write( "$$16y - 48 = 7y^2$$<BR>\n" );
document.write( "$$7y^2 - 16y + 48 = 0$$\r<BR>\n" );
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document.write( "Let's check the position of $P$ and $Q$. The altitude $\overline{AP}$ falls between the angle bisector $\overline{AQ}$ and the median $\overline{AR}$ if the triangle is not isosceles.\r<BR>\n" );
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document.write( "**Revisiting the Geometric Setup:** In a right triangle, the order of the points on the hypotenuse from $B$ to $C$ must be: **$P$ (Altitude), $Q$ (Angle Bisector), $R$ (Median)** *or* **$P, R, Q$** *or* **$R, Q, P$**, etc.\r<BR>\n" );
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document.write( "If $Q$ is between $P$ and $C$, our lengths are:<BR>\n" );
document.write( "* $PC = 7$<BR>\n" );
document.write( "* $BP = y-3$<BR>\n" );
document.write( "* $BC = y+4$\r<BR>\n" );
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document.write( "The discriminant ($\Delta$) of the quadratic equation $7y^2 - 16y + 48 = 0$ is:<BR>\n" );
document.write( "$$\Delta = (-16)^2 - 4(7)(48) = 256 - 1344 = -1088$$\r<BR>\n" );
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document.write( "Since the discriminant is negative, there are **no real solutions for $y$**. This means the geometric configuration $B, P, Q, C$ is impossible with the given lengths.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## 2. Re-evaluate the Geometric Setup (P and Q Order)\r<BR>\n" );
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document.write( "Since $P$ and $Q$ must both be on the segment $\overline{BC}$, and the previous setup failed, let's assume the order of points is $B, Q, P, C$.\r<BR>\n" );
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document.write( "* $BC = y + 4$ (Still the total length)<BR>\n" );
document.write( "* $BQ = y$<BR>\n" );
document.write( "* $QC = 4$<BR>\n" );
document.write( "* $PQ = 3$\r<BR>\n" );
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document.write( "If $Q$ is between $B$ and $P$:<BR>\n" );
document.write( "$$BP = BQ + QP = y + 3$$<BR>\n" );
document.write( "$$PC = BC - BP = (y+4) - (y+3) = 1$$\r<BR>\n" );
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document.write( "Now the geometric relationships are:<BR>\n" );
document.write( "1. $AC^2 = PC \cdot BC = 1(y+4) = y+4$<BR>\n" );
document.write( "2. $AB^2 = BP \cdot BC = (y+3)(y+4)$\r<BR>\n" );
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document.write( "Apply the Angle Bisector Theorem ($\frac{AB}{AC} = \frac{BQ}{QC}$):<BR>\n" );
document.write( "$$\frac{AB^2}{AC^2} = \left(\frac{y}{4}\right)^2$$<BR>\n" );
document.write( "$$16 \cdot AB^2 = y^2 \cdot AC^2$$\r<BR>\n" );
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document.write( "Substitute the Leg Rule expressions:<BR>\n" );
document.write( "$$16 \cdot [(y+3)(y+4)] = y^2 \cdot [y+4]$$\r<BR>\n" );
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document.write( "Divide by $(y+4)$ (since $y+4 = BC \neq 0$):<BR>\n" );
document.write( "$$16(y+3) = y^2$$<BR>\n" );
document.write( "$$16y + 48 = y^2$$<BR>\n" );
document.write( "$$y^2 - 16y - 48 = 0$$\r<BR>\n" );
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document.write( "### Solve the Quadratic Equation for $y$\r<BR>\n" );
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document.write( "Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<BR>\n" );
document.write( "$$y = \frac{16 \pm \sqrt{(-16)^2 - 4(1)(-48)}}{2(1)}$$<BR>\n" );
document.write( "$$y = \frac{16 \pm \sqrt{256 + 192}}{2}$$<BR>\n" );
document.write( "$$y = \frac{16 \pm \sqrt{448}}{2}$$<BR>\n" );
document.write( "$$y = \frac{16 \pm \sqrt{64 \cdot 7}}{2} = \frac{16 \pm 8\sqrt{7}}{2}$$\r<BR>\n" );
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document.write( "Since $y$ must be positive (it is a length $BQ$):<BR>\n" );
document.write( "$$y = 8 + 4\sqrt{7}$$\r<BR>\n" );
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document.write( "## 3. Find the Length of the Median $\overline{AR}$\r<BR>\n" );
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document.write( "The median $\overline{AR}$ goes from the vertex of the right angle to the midpoint ($R$) of the hypotenuse $\overline{BC}$.\r<BR>\n" );
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document.write( "### A. Find the length of the hypotenuse $BC$\r<BR>\n" );
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document.write( "$$BC = y + 4$$<BR>\n" );
document.write( "$$BC = (8 + 4\sqrt{7}) + 4 = 12 + 4\sqrt{7}$$\r<BR>\n" );
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document.write( "### B. Apply the Median to Hypotenuse Theorem\r<BR>\n" );
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document.write( "In a right triangle, the length of the median to the hypotenuse is **half the length of the hypotenuse**.<BR>\n" );
document.write( "$$AR = \frac{1}{2} BC$$\r<BR>\n" );
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document.write( "$$AR = \frac{1}{2} (12 + 4\sqrt{7})$$<BR>\n" );
document.write( "$$AR = \mathbf{6 + 2\sqrt{7}}$$ </SPAN>\n" );
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