document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210478>Question 1210478</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let XYZ be a triangle, and let XP, XQ, XR be the altitude, angle bisector, and median from X, respectively.  If angle YQZ = 90^\circ and angle ZQX = 22^\circ, then what is the measure of angle RZP in degrees?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852846 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=852846\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a geometry problem that requires applying the definitions of altitudes, angle bisectors, and medians, along with standard triangle theorems.\r<BR>\n" );
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document.write( "## 📐 Given Information\r<BR>\n" );
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document.write( "Let the angles of $\triangle XYZ$ be denoted by $X, Y, Z$.\r<BR>\n" );
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document.write( "* **XP** is the **altitude** from $X$, so $\angle XPY = \angle XPZ = 90^\circ$.<BR>\n" );
document.write( "* **XQ** is the **angle bisector** of $\angle YXZ$ (or $\angle X$), so $\angle YXQ = \angle QXZ = \frac{1}{2}X$.<BR>\n" );
document.write( "* **XR** is the **median** from $X$, so $R$ is the midpoint of $YZ$ (i.e., $YR = RZ$).<BR>\n" );
document.write( "* $\angle YQZ = 90^\circ$<BR>\n" );
document.write( "* $\angle ZQX = 22^\circ$\r<BR>\n" );
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document.write( "## 🔍 Step 1: Determine $\angle YXQ$ and $\angle X$\r<BR>\n" );
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document.write( "Since $\angle YQZ = 90^\circ$, the segment $XQ$ is part of a right angle. The angle $\angle YQZ$ is an external angle to $\triangle XQY$ and an internal angle to the larger $\triangle YQZ$ (which is $180^\circ$ on a straight line, but $Q$ is *on* $YZ$).\r<BR>\n" );
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document.write( "**Crucial Insight:** The points $Y, Q, Z$ are collinear because $Q$ is a point on the side $YZ$ since $XQ$ is an angle bisector from $X$. Therefore, $\angle YQZ$ must be $180^\circ$ (a straight line) unless the problem means $\angle YQX + \angle ZQX$ or $\angle YQZ$ is an external angle to $\triangle YXZ$. Given the standard notation, $\angle YQZ = 90^\circ$ implies that **the line segment $XQ$ is perpendicular to $YZ$**.\r<BR>\n" );
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document.write( "If $XQ$ is both the angle bisector and the altitude to $YZ$ (since $\angle XQZ = 90^\circ$ must be true if $Y, Q, Z$ are on a straight line and $\angle YQZ = 90^\circ$ is used in this way), then $\triangle XYZ$ must be an **isosceles triangle** with $XY = XZ$.\r<BR>\n" );
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document.write( "Assuming $XQ \perp YZ$:<BR>\n" );
document.write( "* $\angle XQZ = 90^\circ$.<BR>\n" );
document.write( "* From the given information, $\angle ZQX = 22^\circ$. This contradicts $\angle XQZ = 90^\circ$ unless the angle $\angle ZQX$ is intended to be a different angle, or $Q$ is not on $YZ$.\r<BR>\n" );
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document.write( "Let's assume the question means that $\angle XQY$ and $\angle XQZ$ are the angles formed by the angle bisector and the side $YZ$.\r<BR>\n" );
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document.write( "**If $Q$ is on $YZ$, then $Y, Q, Z$ are collinear, meaning the angle $\angle YQZ$ cannot be $90^\circ$.**\r<BR>\n" );
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document.write( "Therefore, the condition $\angle YQZ = 90^\circ$ **must refer to $\angle XQZ = 90^\circ$** or $\angle XQY = 90^\circ$, which makes $XQ$ an altitude. Given $XQ$ is the angle bisector, this means $XQ$ is also the altitude, implying $XY = XZ$.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Assumption: $\triangle XYZ$ is Isosceles ($XY = XZ$)**\r<BR>\n" );
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document.write( "If $XY=XZ$, then:<BR>\n" );
document.write( "1.  The angle bisector $XQ$ is also the altitude $XP$ and the median $XR$.<BR>\n" );
document.write( "2.  $P=Q=R$.<BR>\n" );
document.write( "3.  The angle we need to find, $\angle RZP$, is $\angle QZQ$ or $\angle Z$.<BR>\n" );
document.write( "4.  If $XQ \perp YZ$, then $\angle XQZ = 90^\circ$. This contradicts $\angle ZQX = 22^\circ$.\r<BR>\n" );
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document.write( "**This path leads to a contradiction, so the problem must use $\angle YQZ$ differently.**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Alternative Interpretation (Standard in Contests)**\r<BR>\n" );
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document.write( "In many contest problems, if an angle is given using three points on a line, it usually implies a specific configuration. A common configuration that leads to $\angle YQZ=90^\circ$ being used is when $Q$ is **not** on $YZ$. However, $XQ$ is the angle bisector from $X$, so $Q$ **must** be on $YZ$.\r<BR>\n" );
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document.write( "Let's assume the given angle values are correct, and the label $\angle YQZ = 90^\circ$ is a **typo** for $\angle XQZ = 90^\circ$.\r<BR>\n" );
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document.write( "**If $\angle XQZ = 90^\circ$:**<BR>\n" );
document.write( "* Then $\angle XQY = 180^\circ - 90^\circ = 90^\circ$.<BR>\n" );
document.write( "* Since $XQ$ is both the angle bisector and the altitude, $\triangle XYZ$ is isosceles with $XY=XZ$.<BR>\n" );
document.write( "* This contradicts $\angle ZQX = 22^\circ$.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **The Most Likely Scenario (Angle Bisector and Altitude Formula)**\r<BR>\n" );
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document.write( "The problem setup strongly suggests using the formula for the angle between the altitude and the angle bisector from the same vertex.\r<BR>\n" );
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document.write( "Let's use the given values $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$ to find the angles of $\triangle XYZ$.\r<BR>\n" );
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document.write( "If $Q$ is on $YZ$, $\angle YQZ$ forms a straight line: $\angle YQZ = 180^\circ$.<BR>\n" );
document.write( "The angles formed by the angle bisector $XQ$ and the side $YZ$ are $\angle XQY$ and $\angle XQZ$.<BR>\n" );
document.write( "$$\angle XQY + \angle XQZ = 180^\circ$$\r<BR>\n" );
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document.write( "Let's assume the angles given are $\angle XQY$ and $\angle XQZ$. The angle $\angle YQZ$ is irrelevant if $Q$ is on $YZ$.\r<BR>\n" );
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document.write( "Assume the problem intended to give **$\angle XQZ = 90^\circ$** and **$\angle YXQ = 22^\circ$**.\r<BR>\n" );
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document.write( "* If $\angle YXQ = 22^\circ$, then $\angle X = 2 \times 22^\circ = 44^\circ$.<BR>\n" );
document.write( "* If $\angle XQZ = 90^\circ$, then $\triangle XQZ$ is a right triangle.<BR>\n" );
document.write( "* The sum of angles in $\triangle XQZ$: $90^\circ + Z + \angle QXZ = 180^\circ$.<BR>\n" );
document.write( "* Since $\angle QXZ = \angle YXQ = 22^\circ$ (since $XQ$ bisects $X$):<BR>\n" );
document.write( "    $$90^\circ + Z + 22^\circ = 180^\circ \implies Z = 180^\circ - 112^\circ = 68^\circ$$<BR>\n" );
document.write( "* Angle $Y$: $Y = 180^\circ - X - Z = 180^\circ - 44^\circ - 68^\circ = 68^\circ$.<BR>\n" );
document.write( "* Since $Y=Z=68^\circ$, $\triangle XYZ$ is isosceles with $XY=XZ$.<BR>\n" );
document.write( "* If $XY=XZ$, then the angle bisector $XQ$ is also the altitude $XP$ and the median $XR$.<BR>\n" );
document.write( "* $P=Q=R$. The points $P, Q, R$ are all the same point on $YZ$.<BR>\n" );
document.write( "* The required angle $\angle RZP = \angle QZQ = \angle Z = 68^\circ$.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **The ONLY Interpretation that Uses ALL Given Angles**\r<BR>\n" );
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document.write( "Let $\alpha$ be the angle between the altitude $XP$ and the angle bisector $XQ$.<BR>\n" );
document.write( "The standard formula for this angle is:<BR>\n" );
document.write( "$$\angle PXQ = \alpha = \frac{1}{2}|Y - Z|$$\r<BR>\n" );
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document.write( "The angle $\angle XQZ$ is an external angle to $\triangle XQY$.<BR>\n" );
document.write( "$$\angle XQZ = Y + \angle YXQ = Y + \frac{1}{2}X$$<BR>\n" );
document.write( "The angle $\angle XQY$ is an external angle to $\triangle XQZ$.<BR>\n" );
document.write( "$$\angle XQY = Z + \angle QXZ = Z + \frac{1}{2}X$$\r<BR>\n" );
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document.write( "The given angles are related to $\angle XQZ$ and $\angle XQY$:<BR>\n" );
document.write( "$$\angle XQY + \angle XQZ = 180^\circ$$\r<BR>\n" );
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document.write( "The only way to use $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$ without a contradiction is if $\angle YQZ$ is an external angle to $\triangle XYZ$, which is not possible for an angle bisector $XQ$.\r<BR>\n" );
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document.write( "We must assume a typo and that the two given angles refer to the adjacent angles $\angle XQY$ and $\angle XQZ$. Since $\angle XQY + \angle XQZ = 180^\circ$, it is impossible for $\angle YQZ = 90^\circ$ unless the angle is labeled wrong.\r<BR>\n" );
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document.write( "Let's assume the question meant:<BR>\n" );
document.write( "$$\angle XQY = 90^\circ + 22^\circ = 112^\circ \quad \text{and} \quad \angle XQZ = 180^\circ - 112^\circ = 68^\circ$$\r<BR>\n" );
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document.write( "This is the standard configuration for this problem where $Q$ is on $YZ$.<BR>\n" );
document.write( "$$\angle XQY = 112^\circ \quad \text{and} \quad \angle XQZ = 68^\circ$$\r<BR>\n" );
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document.write( "Using the exterior angle theorem:<BR>\n" );
document.write( "1.  $112^\circ = Z + \frac{1}{2}X$<BR>\n" );
document.write( "2.  $68^\circ = Y + \frac{1}{2}X$\r<BR>\n" );
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document.write( "Subtracting the two equations:<BR>\n" );
document.write( "$112^\circ - 68^\circ = (Z + \frac{1}{2}X) - (Y + \frac{1}{2}X)$<BR>\n" );
document.write( "$$44^\circ = Z - Y$$\r<BR>\n" );
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document.write( "We also know that $X + Y + Z = 180^\circ$.<BR>\n" );
document.write( "From equation (2): $Y = 68^\circ - \frac{1}{2}X$.\r<BR>\n" );
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document.write( "Substitute $Y$ into $44^\circ = Z - Y$:<BR>\n" );
document.write( "$$Z = Y + 44^\circ = (68^\circ - \frac{1}{2}X) + 44^\circ = 112^\circ - \frac{1}{2}X$$\r<BR>\n" );
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document.write( "Substitute $Y$ and $Z$ into $X + Y + Z = 180^\circ$:<BR>\n" );
document.write( "$$X + (68^\circ - \frac{1}{2}X) + (112^\circ - \frac{1}{2}X) = 180^\circ$$<BR>\n" );
document.write( "$$X + 180^\circ - X = 180^\circ$$<BR>\n" );
document.write( "$$180^\circ = 180^\circ$$<BR>\n" );
document.write( "This identity confirms the angles are consistent with *any* value of $X$. **We need more information.**\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "### **Using the ONLY numbers provided: $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$**\r<BR>\n" );
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document.write( "The only non-contradictory interpretation is that $\angle YQZ=90^\circ$ is a **typo** for $\angle **X**QZ = 90^\circ$, and $\angle ZQX = 22^\circ$ is a **typo** for $\angle **P**XQ = 22^\circ$ (the angle between altitude and bisector).\r<BR>\n" );
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document.write( "**Assume $\angle PXQ = 22^\circ$** (The angle between the altitude $XP$ and the angle bisector $XQ$):\r<BR>\n" );
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document.write( "We use the formula: $\angle PXQ = \frac{1}{2}|Y - Z|$.<BR>\n" );
document.write( "$$22^\circ = \frac{1}{2}|Y - Z| \implies |Y - Z| = 44^\circ$$<BR>\n" );
document.write( "Thus, $Z = Y + 44^\circ$ or $Y = Z + 44^\circ$.\r<BR>\n" );
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document.write( "Since $R$ is the median (midpoint of $YZ$), we need to find $\angle RZP = \angle Z$.<BR>\n" );
document.write( "We need $Y, Z$ to solve for $X, Y, Z$.\r<BR>\n" );
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document.write( "The problem is degenerate without assuming a typo. The intended solution is likely one that uses the properties of the median and altitude.\r<BR>\n" );
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document.write( "**Key property of $\triangle XYZ$:** $XR$ is the median, and $XP$ is the altitude. $\triangle XPR$ is a right triangle with the right angle at $P$.<BR>\n" );
document.write( "We need to find $\angle RZP = \angle Z$.\r<BR>\n" );
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document.write( "Given the structure of geometry problems like this, the most common interpretation uses the angle between the altitude and bisector. Let's assume **$22^\circ = \angle PXQ$** (the angle between the altitude and angle bisector).\r<BR>\n" );
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document.write( "If $Y > Z$, then $Y = Z + 44^\circ$.<BR>\n" );
document.write( "$$X + Y + Z = 180^\circ$$<BR>\n" );
document.write( "$$X + (Z + 44^\circ) + Z = 180^\circ$$<BR>\n" );
document.write( "$$X + 2Z = 136^\circ$$<BR>\n" );
document.write( "$$\implies 2Z = 136^\circ - X$$\r<BR>\n" );
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document.write( "The intended solution is likely to show that $\triangle XQR$ is a right triangle.<BR>\n" );
document.write( "If $Q$ is the midpoint of $YZ$, then $Q=R$. Since $R$ is the median, we have $\angle RZP = \angle Z$.\r<BR>\n" );
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document.write( "---\r<BR>\n" );
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document.write( "## 💡 Final Assumption: $\angle PXQ = 22^\circ$ and $X=90^\circ$ (Hypotenuse Median)\r<BR>\n" );
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document.write( "If $X=90^\circ$, then $R$ (the median point on $YZ$) is the circumcenter, and $XR = YR = RZ$. This makes $\triangle XZR$ an isosceles triangle.<BR>\n" );
document.write( "$$\angle RXZ = \angle RZX = Z$$\r<BR>\n" );
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document.write( "Angle bisector $XQ$ bisects $X=90^\circ$, so $\angle QXZ = 45^\circ$.<BR>\n" );
document.write( "The angle between altitude $XP$ and bisector $XQ$ is $22^\circ$.<BR>\n" );
document.write( "$$\angle PXQ = 22^\circ$$\r<BR>\n" );
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document.write( "The position of $P$ relative to $Q$:<BR>\n" );
document.write( "* The angle $\angle PXR$ is the angle between the altitude and the median.<BR>\n" );
document.write( "* The angle $\angle QXR$ is the angle between the angle bisector and the median.<BR>\n" );
document.write( "* $\angle PXR = \angle QXR + \angle PXQ$ (or vice versa).\r<BR>\n" );
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document.write( "If $X=90^\circ$, $Y+Z=90^\circ$.<BR>\n" );
document.write( "$$\angle PXQ = \frac{1}{2}|Y - Z| = 22^\circ \implies |Y - Z| = 44^\circ$$<BR>\n" );
document.write( "Assume $Z > Y$, so $Z = Y + 44^\circ$.<BR>\n" );
document.write( "$$Y + (Y + 44^\circ) = 90^\circ$$<BR>\n" );
document.write( "$$2Y = 46^\circ \implies Y = 23^\circ$$<BR>\n" );
document.write( "$$Z = 23^\circ + 44^\circ = 67^\circ$$\r<BR>\n" );
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document.write( "Now we find $\angle RZP$:<BR>\n" );
document.write( "1.  $P$ is the foot of the altitude, $\triangle XPZ$ is a right triangle.<BR>\n" );
document.write( "2.  $R$ is the midpoint of $YZ$, so $XR = RZ$. $\triangle XZR$ is isosceles.<BR>\n" );
document.write( "3.  $\angle RZX = Z = 67^\circ$.<BR>\n" );
document.write( "4.  Since $\triangle XZR$ is isosceles, $\angle RXZ = \angle RZX = 67^\circ$.<BR>\n" );
document.write( "5.  In $\triangle XZR$, $\angle X R Z = 180^\circ - 2(67^\circ) = 180^\circ - 134^\circ = 46^\circ$.\r<BR>\n" );
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document.write( "The angle $\angle RZP$ is the angle between $RZ$ and $ZP$. Since $R, P, Z$ are all on the line segment $YZ$, $\angle RZP$ is either $\angle Z$ or $0^\circ$, which isn't right. The requested angle must be an angle *involving* the vertices. The only angle that makes geometric sense is $\angle **P R Z**$. Let's assume the question meant $\angle PRZ$.\r<BR>\n" );
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document.write( "If $X=90^\circ$, then $P$ lies between $R$ and $Z$.<BR>\n" );
document.write( "$$\angle PRZ = 180^\circ - \angle XRZ = 180^\circ - 46^\circ = 134^\circ$$\r<BR>\n" );
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document.write( "Given the highly ambiguous nature of the angle labels, the question is likely from a set where $\angle YQZ=90^\circ$ and $\angle ZQX=22^\circ$ *are* the two angles of the two parts of $XQ$.\r<BR>\n" );
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document.write( "If $\angle XQZ = 90^\circ$ (typo for $\angle YQZ$) AND $\angle ZQX = 22^\circ$ (typo for $\angle PXQ$), then $Z = 67^\circ$.\r<BR>\n" );
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document.write( "The measure of the angle $\angle RZP$ in degrees is $\mathbf{67}$.\r<BR>\n" );
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document.write( "* This answer assumes $X=90^\circ$ and that $\angle YQZ = 90^\circ$ and $\angle ZQX = 22^\circ$ are meant to imply $|Y-Z| = 44^\circ$. </SPAN>\n" );
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